Electromagnetism (B U study material)

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UNIT I

Lesson – 1

ELECTROMAGNETIC THEORY AND OPTICAL PHYSICS

1.0 Aims and Objectives

1.1 (a) Electric potential at a point due to an electric dipole:

(b) Electric field at any point due to an electric dipole

1.2 Dielectric polarization

1.3 Relative permitivity (dielectric constant) and displacement vector

1.4 External field of a dielectric medium

1.5 Relation between electric displacement, electric field and polarisation (D, E and P)

1.6 Molecular field in a dielectric (Clausius-Mossotti relation)

1.7 Polarizaiton of polar-molecules (Debye’s formula)

1.8 Electrostatic energy and energy density in free space and in dielectric

1.9 Let us sum up

1.10 Check your progress

1.11 Lesson-end activities

1.12 Points for discussion

1.13 References


This lesson deals with potential and fields due to electric dipole. The relation

between electric susceptibility, polarization, displacement will be obtained. The molecular

field, derivation of claussion mossotti relation for non-polar molecules, Debge formula for

polar molecules are explained in detail. The derivation of electrostatic energy and energy

density who has has discussed.

1.1(a) ELECTRIC POTENTIAL AT A POINT DUE TO AN ELETRIC DIPOLE:

Two charges –q at A and +q at B separated by a small distance 2d constitutes an electric

dipole and its dipole moment is p (Fig 1.1a).

Let r1 and r2 be the distances of the point P from +q and –q charges respectively. Let P

be the point at a distance r from the midpoint of the dipole O and q be the angle between PO

and the axis of the dipole OB.

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Fig 1.1(a)

Potential at P due to charge (+q) = 104

1

r

q

pe

Potential at P due to charge (-q) = ÷÷ø

öççè

æ-

204

1

r

q

pe

Total Potential at P due to dipole is, V = 2010

4

1

4

1

r

q

r

q

pepe-

÷÷ø

öççè

æ-=

210

11

4 rr

qV

pe (1)

Applying cosine law,

÷÷ø

öççè

æ+-=

-+=

2

2

22

1

222

1

cos21

cos2

r

d

rdrr

rddrr

q

q

Since d is very smaller than r, d2/r2 can be neglected.

2/1

1

cos21 ÷

ø

öçè

æ-=\

rdrr

q



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or

2/1

1

cos2

111

-

÷ø

öçè

æ-= q

r

d

rr

Using the binomial theorem and neglecting higher powers

÷ø

öçè

æ+= qcos1

11

1r

d

rr (2)

Similarly,

( )q

q

cos2

180cos2222

2

222

2

rddrr

rddrr

++=

--+=

2/1

2cos

21 ÷

ø

öçè

æ+= q

r

drr (Since d2 / r2 is negligible)

or

2/1

2cos

21

11-

÷ø

öçè

æ+= q

r

d

rr

Using the Binomial theorem and neglecting higher powers,

÷ø

öçè

æ-= qcos1

11

2r

d

rr (3)

Substituting equation (2) and (3) in equation (1) and simplifying

÷ø

öçè

æ+-+= qq

pecos1cos1

1

4 0 r

d

r

d

r

qV

2

0

2

0

cos.

4

1

.4

cos2

r

p

r

dqV

q

pepe

q== (4)

Special cases:

1. When the point P lies on the axial line of the dipole on the side of +q, then q = 0.

\ V = p / (4pe0r2)

2. When the point P lies on the axial line of the dipole on the side of –q, then q = 180.

V = -p / (4pe0r2)



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3. When the point P lies on the equatorial line of the dipole, then, q = 90,

V = 0.

1.1(b) Electric field at any point due to an Electric dipole

AB is an electric dipole of dipole moment p. O is its midpoint. R is a point with polar

coordinate (r, q) (Fig 1.1(b))

Fig. 1.1(b) Electric field at any point due to an Electric dipole

\ OR = r, LROX = q

The dipole moment p is resolved into two components along the horizontal and

vertical directions. The point R is on the axial line of a dipole of moment p cosq and on the

equatorial line of the dipole of moment p sinq.

Electric field at R along the axial line at a distance r from O

3

0

3

0

1

cos2

4

12

4

1

r

p

r

pE

q

pepe== (along RA) ---(1)

Electric field at R along the equatorial line at a distance r from O is

3

0

3

0

2

sin

4

1

4

1

r

p

r

pE

q

pepe== (along RB) ---(2)



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\ Magnitude of the resultant electric field is,

E = Ö (E12 + E2

2) = p / (4pe0r3) . Ö (4 cos2q + sin2q)

E = p / (4pe0r3) . Ö (1 + 3 cos2q) ---(3)

If f is the angle between E1 and resultant, then

qpeq

peqf tan

2

1

4cos2

4sintan

3

0

3

0 ==rp

rp ---(4)

This gives the direction of the resultant electric field.

The unit of potential difference is volt. The potential difference between two points is

1 volt if 1 joule of work is done in moving 1 Coulomb of charge from one point to another

against the electric force.

The electric potential in an electric field at a point is defined as the amount of work

done in moving a unit positive charge from infinity to that point against the electric forces.

Relation between electric field and potential

Let the small distance between A and B be dx. Work done in moving a unit positive

charge from A to B is dV = E.dx.

The work has to be done against the force of repulsion towards the charge +q. Hence,

dV = -E.dx

E = - (dV/dx)

The change of potential with distance is known as potential gradient, hence the

electric field is equal to the negative gradient of potential.

The negative sign indicates that the potential decreases in the direction of electric

field. The unit of electric intensity can also be expressed as Vm-1.

1.2 DIELECTRIC POLARIZATION In substances, called insulators or dielectrics, which do not have free electrons, or the

number of such electrons is too low, the electrons are tightly bound to the atom. When

potential difference is applied to insulators no electric current flows, even then their



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behaviour in fields is very important because the presence of the field may change the

behaviour of an insulator. The insulators whose behaviour gets modified in the electric field

are called as dielectrics. When the change in the behaviour of the dielectric is independent of

the direction of applied field, the dielectric is called as isotropic. On the other hand, if the

change in behaviour of the dielectric depends on the direction of applied field then the

dielectric is called as anisotropic.

If we consider a dielectric in an electric field it exerts a force on each charged particle.

The positive particles are pushed in one direction (direction of field) while the negative

particles in the opposite direction. As a result the positive and negative parts of each molecule

are displaced from their equilibrium positions in opposite directions. The overall effect is a

displacement of the entire positive charge in the dielectric relative to the negative charge. The

relative displacement of the charges is called polarization, and the dielectric is said to be

polarized.

The molecules of dielectrics are classified into two classes:

(1) polar molecules, and

(2) non-polar molecules.

The positive charge of the nucleus may be supposed to be concentrated at a point, say

the centre of gravity of positive charge. Similarly, the negative charge due to electrons in

orbit may be supposed to act at a point, known as centre of gravity of negative charge. When

two centres of gravity coincide in a molecule, the molecule as a whole possesses no resultant

charge and it is said to be non-polar, e.g., carbon tera-chloride. If two centres of gravity are

displaced from each other, the molecule as a whole possesses polarity and has permanent

electric moment and the molecule is said to be a polar molecule, e.g., CHCL3 etc. A polar

molecule in an electric field exhibits both permanent and induced dipole moments while non-

polar exhibits induced dipole moment only.

When either of these types of molecules is placed in an external field, the small

displacements of the orbital electrons will cause the distance between the centres of gravity to

change. As a result non-polar molecules become induced dipoles whereas polar molecules

will have both permanent and induced dipoles. The orientation of the induced dipoles or of



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permanent dipoles in an external electric field will be such as to align the dipole along the

field lines. Thus resultant dipole moment even in case of polar molecules gets modified.

Electric polarization effects in simple models of non-polar and polar dielectric materials are

shown in Fig 1.2

Fig 1.2 Electric polarization effect in simple modes of non-polar and polar

dielectric materials (a) Non-polar (b) Polar.

Definition of Dielectric Polarisation: In a homogeneous isotropic dielectric, the

polarisation (i.e., reorientation of electronic orbits) depends directly on the total electric fields

E in the medium. All the molecules are polarized to the same extent. The polarisation charges

within the body of the medium neutralize each other except at the surface of end faces



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A

B

C

D

E

G

F

H

E

-σP +σP

(ABCD and EFGH Fig 1.2(a). The electric field due to induced charges is opposite to the

applied (external) electric field E0. Thus resultant field E is less than the applied field E0.

Now the average dipole moment p induced in each molecule of the dielectric is proportional

to E provided E is not too great i.e.,

p µ E or p = aE ---(1)

where a is a constant called molecular polarizability and E is the intensity of field. If there

are n molecules per unit volume in a dielectric and p of each molecule be independent, we

have electric moment per unit volume = naE, and is represented by a vector P, given by

P = n a E ---(2)

Electric moment per unit volume is known as polarization.

Consider a rectangular block of dielectric material of length l [Fig 1.2a]. Let a be the

area of cross-section of the block perpendicular to the direction of the field (face ABCD or

EFGH). Let the surface density of charge appearing on these faces be sp due to polarization.

If l be the length of the block, then the charge induced on each face = spa

Therefore total induced electric dipole moment

=spa.l………..(3)

Fig.1.2(a): Electric moment Calculation

Again the total induced dipole moment = P. a l

Because P is the polarization i.e., electric dipole moment per unit volume and volume of the

block is a l.

From equations (3) and (4) we have



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sp.al = P. a l

Or P =sp

Thus the polarization may also be defined as the surface density of charge appearing at faces

perpendicular to the direction of applied field.

Electrical Susceptibility:

It has been found experimentally that in a large number of dielectrics, polarization P

is proportional to the applied electric field E, atleast for field strengths that are not too large.

Thus,

POOE

= ke0E,

Where k is dimensionless constant called electrical susceptibility of the dielectric and e0 the

permittivity of free space, has been included to make k dimensionless. Equation is true when

the polarization properties of medium are independent of direction of polarization which is

the case for liquids, gases, amorphous solids and cubic crystals.

1.3 RELATIVE PERMITTIVITY (DIELECTRIC CONSTANT) AND

DISPLACEMENT VECTOR

The capacity of a parallel plate capacitor increases if the gap between the plates is

filled with a dielectric. The increase in capacity of capacitor with a dielectric compared to

capacity of the same capacitor with vacuum between plates, measures a constant known as

dielectric constant, relative permittivity or specific inductive capacity (S.I.C.), i.e.,

,0e

ee ==

vacuumwithcapacity

presentdielectricwithcapacityr

where e and e0 are absolute permittivities of the dielectric and vacuum.

Or e = er . eo

is absolute permittivity of dielectric.



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The dielectric constant is defined in terms of electric intensity as follows: When a

dielectric is polarized, the charges appearing on the surfaces perpendicular to the field are

opposite to the real charges present on the plates. Hence the field decreases in presence of

dielectric. Therefore relative permittivity is also defined as the ratio of two fields; that is

E

E

dielectricinfield

vacuuminfieldr

0==e

e

e

pe

pe 0

2

2

0

4

4==

rq

rq

because 3

0

0.

4

1

r

qrE

pe= and 3

.4

1

r

qrE

pe=

Thus the dielectric constant may defined as ratio of the absolute permitivity of the dielectric

to that of the free space.

Since the electric intensity depends on the medium, we define another electrical

vector which depends only on the magnitude of the charge and its distribution but

independent of the nature of medium. This vector is termed as displacement, D, and is given

by

3.

4

1

r

qrD

p=

Er

qrD e

pee =÷

ø

öçè

æ=

3.

4

1

3.

4

1

r

qrE

pe=

vector D is analogous to E. Its unit is coul. /m2.

While E is defined as force experienced by a positive charge in any medium, the product

D.dS gives the flux of normal displacement through surface dS and remains unaltered in any



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medium. The direction of the two vectors being the same in isotropic homogeneous

dielectric.

1.4 EXTERNAL FIELD OF A DIELECTRIC MEDIUM

In the presence of an electric field, a dielectric becomes polarized and aquires a dipole

moment P, per unit volume. The calculation of electric field intently produced by elementary

dipoles though out the space at an external point will be presented below.

Consider a finite piece of polarized dielectric. Let it be characterized at each point r

by a polarization P(r). The polarization gives rise to an electric field. The aim is to calculate

the potential and field at point r[denoted by Q(r’) in Fig.1.4] outside the dielectric. Let the

volume of the dielectric is dividd into small elementary volumes and

Fig.1.4 Dielectric body with external point of observations.

consider one such volume element dV of the dielectric.

The potential due to dielectric body occupying volume dV as observed

outside dV for single dipole potential can be expressed as,



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( ) ( ) ( )( )ò

-

-=

VdV

rr

rrrPr

3

0 '

'.

4

1'

pef

( )( ) ( )

( )( )ò

-

-

-

=V rr

dVrr

rrrP

r2

0 '

'

'.

4

1'

pef

( ) ( ) ( )( )ò

-

-=

VdV

rr

rrrPr

2

0

0 '

'.

4

1'

pef

where (r-r’)0 represents unit vector directed from Q towards dV. It is obviously equal

to

( ) ( ) ( )

( ) ( ) ( ){ }222'''

'''

zzyyxx

kzzjyyixx

-+-+-

-+-+-

where x’, y’,z’ are the Cartesian coordinates of point r’ and (x,y,z)those at r. Further,

( ) ( ) ( ) ( )úû

ùêë

é-

¶

¶+-

¶

¶+-

¶

¶-=- ''''

0rr

xkrr

xjrr

xirr

and ( ) ( ) ( ) ( )zyxkPzyxjPzyxiPrP zyx ...... ++=

substituing these values in equation (1), we get

( )( )

( )( )

( )( )

( )þýü

îíì

-¶

¶

-+-

¶

¶

-+-

¶

¶

-= ò òò V V

zy

V

x dVrrzrr

PdVrr

yrr

PdVrr

xrr

Pr '.

''.

''.

'4

1'

222

0pef (2)

321fff ++=

Now we shall consider the integrals separately. It is obvious that

( )( )

( ) 22

'

1'.

'

1

rrxx

rr

rr -¶

¶=

¶

-¶

--

and dxdydzdV =



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Fig.1.4 (a) Integration with respect to ‘x’

Therefore

( )òòò-¶

¶= dxdydz

rrxPx 2

0

1

'

1

4

1

pef

Let us first integrate with respect to x. If we consider integration along x-axis only, it would

be simply a contribution of a straight rod of material with area of cross-section dydz. If such

an elements cuts the surface of the body at x1 and x2, coordinates of respective points, then x

integral would be

( )( )

( )( )

( ) ( )dx

x

P

rrrr

zyxp

rr

zyxpdx

rrxP x

x

xxx

X

xx

x

x

X

x¶

¶

--

úúû

ù

êêë

é

--

úúû

ù

êêë

é

-=

-¶

¶òò

==

2

112

2

2

1

2

'

1

'

,,

'

,,

'

1

( ) ( ) ( ) ïþ

ïý

ü

ïî

ïí

ì

¶

¶

--

úú

û

ù

êê

ë

é

--

úú

û

ù

êê

ë

é

-= òòòòòòò

==

dxdydzx

P

rrdydz

rr

Pdydz

rr

P x

xx

x

xx

x

222

0

1

'

1

''4

1

12

pef

if n is unit outward normal at dA then

dy dz = n.dA2 = nxdA2

and dy dz = n.dA1 = nxdA1

since at x1 which is on the part S1, the x-component of n, i.e., nx has negative value.

dydz

x2 x1



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Now the integral of Φ1 can be written as

( ) ( ) ( ) ïþ

ïýü

ïî

ïíì

¶

¶

--

-+

-= ò òò

12 '

1

''4

112

0

1S

x

V

xx

S

xx dVx

P

rrdA

rr

nPdA

rr

nP

pef

Since the two integrals over S1 and S2 cover whole boundary S, we have

( ) ( )dV

x

P

rrdA

rr

nP x

VS

xx

¶

¶

--

-= òò '

1

4

1

'4

1

00

1pepe

f

similar integrals for Φ2 and Φ3 can be written and hence value of Φ will be

( )( ) ( ) ïþ

ïýü

ïî

ïíì

÷÷ø

öççè

æ

¶

¶+

¶

¶+

¶

¶

--

-

++= òò dV

z

P

y

P

x

P

rrdA

rr

nPnPnPr zyx

VS

zzyyxx

'

1

'4

1'

0pef

or ( )( )

( )( )òò -

--

-=

VS rr

dVdivPdA

rr

nr

'4

1

'

.

4

1'

00pepe

fP

(3)

Equation (3) obviously shows that the potential at point r’ due to dielectric body is the same as if it

were replaced by a system of bound charges in empty space. A part of these bound charges appears

on the dielectric surfaces as a surface density σP’ given by

σP’=P.n

while the remaining bound charge appears throughout the volume V as a volume density rp’ given

as

÷÷ø

öççè

æ

¶

¶+

¶

¶+

¶

¶-=

z

P

y

P

x

Pzyx

P

'r

in the case of homogeneous and isotropic dielectrics, where the polarisation of each molecule is to

the same extent, the volume distribution of polarized charge is zero because the positive and

negative charges of dipoles neutralize each other. Hence the second term does not contribute

anything to the potential, i.e., -div P = 0. The potential is due to the surface distribution of charges

only. The intensity of electric field due to the polarized volume may be expressed as



P a g e | 15

( )( ) ( )

úú

û

ù

êê

ë

é

-

-+

-

-= ò ò

S V

PP dVrr

rrdA

rr

rrrE

3

'

3

'

0 '

'

'

'

4

1'

rs

pe

this gives the outer field due to polarisation.

The total bound charge of a dielectric body of volume V and complete boundary surfaces S

is zero as the total charge is

ò ò ¢+¢=¢S V

total dVdAq rs

òò =-=VS

divPdVndAP 0.

since by divergence theorem of vector algebra,

òò =VS

dVPdivndAP.

Equation (3) is in M.K.S system units. If it is to be expressed in e.s.u then

( )( ) ( )( )

ò ò-

--

-= dVPdiv

rrdA

rr

nPr

rr '00 '

11

'

.1'

eef

1.5 RELATION BETWEEN ELECTRIC DISPLACEMENT, ELECTRIC FIELD AND

POLARISATION (D, E and P)

(1) Electric Intensity: The electric intensity E at a point in the electric field is

numerically equal to the force experienced by a unit positive charge placed at that point. The

direction of E being the same as that of the field.

(2) Dielectric Polarisation: When a dielectric is placed between the two charged

metallic plates, the dielectric is polarized. The distorted atom is called as an electric dipole or

an electric doublet. The electric dipole moment per unit volume is called the dielectric

polarisation P. It is a vector parallel to the electric field.



P a g e | 16

(3) Electric Displacement: When no electric field is present, Gauss’ law gives

0

0.e

qAEdSE ==ò

A

qE

0

0e

=

where E0 is the electric field and q is the free charge.

When dielectric is present, induced surface charges say-q’ appears, so that Gaussian

surface encloses (q-q’) charge . Gaussian’s law then becomes

0

.e

qqAEdSE

¢-==ò

A

q

A

qE

00 ee

¢-=

Now the field when dielectric is present is given by

A

qEE

rr 0

0

eee==

A

q

A

q

A

q

r 000 eeee

¢-=

A

q

A

q

A

q

r 000 eeee

¢+=

A

q

A

q

A

q

r

¢-÷

÷ø

öççè

æ=

0

0ee

e

PEA

q+= 0e

As the quantity on left hand side is termed as electric displacement D, we write



P a g e | 17

PED +=0

e

It can be noted that D is connected with free charge, q only. P is connected with

polarisation charges, q¢ only and E is connected with all charges that are actually present

whether free or polarised.

The relationship between D, E and P can also be obtained as follows:

The differential form of Gauss law is

0

.e

r tE =Ñr

(1)

where rt is the total charge density. For dielectrics, rt must include charge densities of real

and polarized charges, that is,

rt = r + r¢,

where rt is volume density of real charge and r¢ that of polarized charges.

Therefore eq. (1) becomes

0

.e

rr ¢-=Ñ E

r

00

.e

r

e

r=

¢Ñ Er

(2)

We have shown that P.Ñ-=¢r

r which when substituted in eq.(2) gives

00

.e

r

e=

Ñ¢+Ñ¢

PE

( ) PPE =+Ñ 0. er

r=Ñ D.r

PED += 0e (3)



P a g e | 18

and defines electric displacement vector D. Eq. (3) shows that D is responsible for the partial

field due to real charges, and that D and E are additive. This result is called differential form

of Gauss’ theorem in a dielectric. The curl of the displacement vector, on the other hand, can

be written as

( )PED +´Ñ=´Ñ0

err

(4)

PE ´Ñ+´Ñ=rr

0e (5)

in case of electric field of solenoidal type

E´Ñr

=0

in case of polarisation of isotropic homogeneous dielectric

0=´Ñ Pr

so that 0=´Ñ Dr

in such cases. But curl of displacement vector D does not vanish in the case of non-solenoidal

type of fields and anisotropic dielectric.

Further relation(4) can be put in the form,

PED += 0e

÷÷ø

öççè

æ+=

0

0e

eP

E

÷÷ø

öççè

æ+=

0

0

0e

ee

PkE

( )kE += 10e (6)

Eree 0=

where ( )rk e=+1



P a g e | 19

is the relative permitivity of medium defining dielectric constant by 0e

ee =

r , so that

D = ε E (7)

Now P is related to E by the relation

P = D- ε0E = ε0 εr E - ε0E

= ε0 (εr-1)E (8)

A relation can also be developed between r and r’ as follows:

P.Ñ-=¢r

r

( )Er 1. 0 -Ñ-= eer

÷÷ø

öççè

æ-Ñ-=

r

rE

eee

11.

0

r

( ) ÷÷ø

öççè

æ-Ñ=

r

De

11.

r by eq.(5)

( )Dr

.1

1 Ñ÷÷ø

öççè

æ--=

r

e

÷÷ø

öççè

æ--=

rer

11

As er > 1, r¢ < -r that is bound charge density is always opposite in sign and less in

magnitude than free charge density.

1.6 MOLECULAR FIELD IN A DIELECTRIC (CLAUSIUS-MOSSOTTI RELATION)

The aim of this section is to examine the molecular nature of the dielectric and to

consider how the electric field & responsible for polarizing the molecule is related to the

macroscopic electric filed. The electric field which is responsible for polarizing a molecule of



P a g e | 20

the dielectric is called the molecular field. This is the electric field at a molecular field or

local field or local field is produced by all external sources and by all polarized molecules in

the dielectric, except one molecule under consideration.

Fig.1.6(a) Fig.1.6(b)

The molecular field, Ein, may be calculated in the following way. Let us cut a

spherical cavity of radius r (such that its dimensions are very great as compared to the

molecular dimensions and very small as compared to the volume of the dielectric)

surrounding the point at which the molecular field is to be computed. The dielectric which is

now left, will be treated as a continuum. The cavity is put in its original position in the

dielectric (molecule by molecule) except the molecule where the molecular field is to be

computed. The molecules which are just replaced in the cavity are treated as individual

dipoles and not as a continuum.

Let the dielectric be placed in the uniform electric field between two parallel plates

(of a condenser) as shown in fig (1.6b). The dotted lines show the boundary of the dielectric.

Let the surface density of real charges on the capacitor plates be s. Again let the surface of

cavity has polarized charges of surface density sp.



P a g e | 21

The field experienced by the molecule of the dielectric at the centre of the cavity C,

Ein, given by

Ein = E1 + E2 + E3 + E4

Where

(i) E1 is the field between two plates with no dielectric, so that E1 = s / e0,

(ii) E2 is the field at C due to polarized charges on the plane surfaces of the

dielectric facing the capacitor plates and is given by E2 = sp / e0.

(iii) E3 is the field at C due to polarized charges on the surface of cavity, to be

calculated.

(iv) E4 is the field at C due to permanent dipoles. But in present case for non-polar

isotropic dielectrics E4 = 0.

Thus, 3

00

EEp

in +-=e

s

e

s (1)

Evaluation of E3 : Consider a small elemental area ds on the surface of cavity of an

angular width dq and an angle q with the direction field E. The vector P shows the direction

of displacement at the centre of ds, Fig (1.6a). The normal component of displacement is

PN = P cos q.

By definition of polarization, it is the surface charge per unit area. Such a charge on ds to

provide flux normal to ds is,

PN = P cos q ds

and the electric intensity at C due to this charge is given by,

2

04

cos

r

dsP

pe

q=

where r is the radius of cavity. The field is directed along the radius CA. Resolving the

intensity along and perpendicular to the applied field, we have the components



P a g e | 22

qpe

qcos

4

cos2

0r

dsP= along the field

qpe

qsin

4

cos2

0r

dsP= perpendicular to field

If the area ds be taken round through 2p radians about LM, it will describe a ring, the

surface area of which is given by,

qqp drr ..sin2=

qqp dr sin2 2=

Therefore intensity at C due to ring in the field direction is

qqppe

qdr

r

dsPsin2

4

cos 2

2

0

2

=

qqqe

dr

Psincos

22

2

0

=

while the normal components of intensity due to the ring cancel each other.

Integrating the intensity at C due to charges on the surface of cavity, we have

3

2

2 0

´=e

P

therefore the total intensity or Ein at C is given by eq.(1) as

000 3ee

s

e

s PE p

in +-= (2)

The resultant field, E between the plates is

ò=p

qqqe 0

2

0

3 sincos2

dP

E

03e

P=



P a g e | 23

00e

s

e

s p-=

Therefore 0

3e

PEEin +=

Further we know that PEED +== 0ee

So that 0ee -

=P

E

Putting this value in eq.(2) for Ein, we get

00 3eee

PPE

in+

-=

÷÷ø

öççè

æ

-

+=

0

0

0

2

3 ee

ee

e

P (3)

if the number of the molecules per unit volume , a molecular polarizability then polarization,

P, is defined as electric moment per unit volume . That is

inEnP a=

Putting this value in eq,(3), we get

÷÷ø

öççè

æ

-

+=

0

0

0

2

3 ee

ee

ea

P

n

P

÷÷ø

öççè

æ

+

-=

0

0

0 23 ee

ee

e

an

2

1

0

0

+

-=

ee

ee

2

1

3 0 +

-=

r

rn

e

e

e

a



P a g e | 24

where er is the relative permittivity or dielectric constant. Equation (6) is well known

Claussius-Mossotti relation.

We know that er = mg2 where mg is the refractive index of gas. Hence equation (6) can

be expressed in terms of refractive index. The relation in that case is

2

1

3 2

2

0 -

-=

g

gn

m

m

e

a (4)

called Lorentz formula and is valid only as long as er is frequency independent.

Validity of Claussius-Mossotti Relation:

The number of molecules per unit volume is proportional to the density of gas. If

molecular polarizability a is taken to be a constant, the number n is proportional to 2

1

+

-

r

r

e

e.

Also the constant a is proportional to the cube of the radius of molecule. Hence finding e r

experimentally and calculating n from the density at definite temperature and pressure, the

value of the radius of the molecule of dielectric may be reckoned. The values, so obtained,

fairly agree with the values obtained by other methods and proves the validity of the relation.

It is true only in case of monatomic gases and weak solutions. The experimental and

theoretical values disagree in case of strong solutions and solids. It is due to the fact that in

these cases the interaction forces among molecules are sufficiently great, the account of

which has not been taken.

Molar Polarizability:

Multiplying Eq. (4) by M*/d on both sides, where M* is the molecular weight and d

the density of the dielectric, we have

d

nM

d

M

r

r

*

0

*

2

.3

.1

e

a

e

e=

-

+

But (M*n / d) = N and is called the Avogadro’s number. Therefore



P a g e | 25

d

MN

r

r

*

20

.1

3 +

-=

e

e

e

a

where Na is known as molar polarizability.

1.7 POLARIZAITON OF POLAR-MOLECULES,(DEBYE’S FORMULA):

Dependence of polarizability on temperature.

While discussing the polarization of non-polar molecules, it has been observed that

the molecular polarizability, a, does not depend on the temperature. But such is not the case

with polar molecules. If polar molecules are subjected to electric field then in addition to

induced polarization, the permanent polarization of molecules is also present. In absence of

electric field, the permanent dipoles point in random directions and the net polarization is

zero. On application of field, the permanent dipoles are forced to orient in the direction of the

field. But the orientations along the field direction are continuously disturbed by thermal

agitation. Therefore an equilibrium is attained and the dipoles make

Fig.1.7

zero to p angles with field direction. The polarization of such permanent molecules may

therefore be resolved in the direction of the field and the average value of Pp cos q may be

found (fig 1.7). Thus we have Pp the permanent polarization, and Pi the induced polarization.

Therefore total polarization is given by

PT = Pp + Pi



P a g e | 26

If Ein is the inner field acting on the dielectric molecules, the potential energy of

permanent dipoles per unit volume of moment Pp is given by

Ein . Pp = Ein . Pp cos q.

Calculation of the average value of Pp cos q:

Following the statistical method of Boltzmann used in kinetic theory of gases, the

probability that a molecule is found in an energy state W, at temperature T is proportional to

e-W/KT, where K = R/N is the Boltzmann constant, R and N are gas constant and the Avogadro

number respectively. In present case, the potential energy of dipoles per unit volume is given

by - Ein . Pp. Hence probability of finding dipoles at an angle q with the field direction is

given by

Pq µ eEin . Pp /KT

Or Pq µ eEin . Pp cos q /KT

The number of dipoles at an orientation will be proportional to its probability. Hence

if Nq is the number of electric dipoles per unit volume at an angle q and Np/2 at q/2, we have

Nq µ eEin . Pp cos q /KT

Np/2 µ 1

\ (Nq / Np/2) = eEin . Pp cos q /KT

or Nq = Np/2 eEin . Pp cos q /KT

Differentiating, we obtain the number of dipoles in an angular width dq as

qqq

pqqq dkT

PEeNdN

pinkTPE

d

pin sincos

2-=+®

The average moment per unit volume in the direction of field is given by

ò

ò=

p

q

p

qq

q

0

0

cos

cos

dN

dNP

Pp

p



P a g e | 27

Substituting the value of dNθ, we get

( )

( ) qq

qqqq

pq

p

pq

p

deNKT

EP

deNKT

EP

PkTPEinp

kTPEinp

p

pin

pin

ò

ò=

0

cos

2

0

cos

2

2

sin

sincoscos

( )( ) qq

qqqq

pq

pq

de

dePP

kTPE

kTPE

p

ppin

pin

ò

ò=

0

cos

0

cos

sin

sincoscos

KT

EPu inp=

qq

qqqq

pq

pq

du

duPP

p

p

ò

ò=

0

cos

0

cos

sin

sincoscos

qcosuKT

PEt in ==

qq dudt sin=

we get

( )

ò

ò

-

-=u

u

t

u

u

t

p

p

dte

dttuePP qcos

[ ][ ]u

u

t

u

u

tt

p

eu

eteP

-

--=

( )úû

ùêë

é

-

--

+

+=

-

-

-

-

uu

uu

uu

uup

ee

ee

ee

eeu

u

P



P a g e | 28

÷ø

öçè

æ-=

uuP

p

1coth

÷÷

ø

ö

çç

è

æ-=

inp

inp

pEP

KT

KT

EPP coth

which is called Langevin equation

when the inner field, Ein , is small and temperature T is large

i.e., 1ááKT

EP inp

then we can expand the exponentials in Langevin equation as

÷÷ø

öççè

æ-=

uu

uPP pp

1

sinh

coshcosq

úúúú

û

ù

êêêê

ë

é

-

÷÷ø

öççè

æ++

ççè

æ++

=uu

u

uu

Pp

1

.........!3

..........!4!2

1

3

42

úúû

ù

êêë

é-÷÷

ø

öççè

æ++÷÷

ø

öççè

æ++=

-

u

uu

uPp

1....

!3....

!21

132

3.

2u

u

Pp=

3

uPp

=

putting KT

EPu inp=

we get average polarisation of permanent molecules in the direction of the field as



P a g e | 29

KT

EPP inp

p3

cos2

=q

qcosPPPiT+=

KT

EPPP inp

iT3

2

+=

inii EP a=

KT

EPEP inp

iniT3

2

+= a

in

p

i EKT

P÷÷ø

öççè

æ+=

3

2

a

inT Ea=

where Ta is the total polarizability per unit volume is

KT

Pp

iT3

2

+= aa

where ia is induced polarizability per unit volume and

KT

Pp

3

2

is permanent polarizability per

unit volume.

Study of Molecular Structure and Debye’s Relation:

The Claussius-Mossotti relation, modified for polar molecules, shows that the

constant a is temperature dependent. The modification was utilized by Debye to study

molecular structure. The modified Claussius-Mossotti relation or Debye’s relation is

02 3

1

e

a

e

e T

r

r º-

+



P a g e | 30

úû

ùêë

é+º

-

+

i

p

r

r

KT

Pa

ee

e

33

112

02

00

2

3

1.

9 e

a

eip

TK

P+=

2

1

+

-=

r

rye

e

K

Pm

p

0

2

9e=

Tx

1=

03e

aic =

For non-polar molecules Pp = 0 and therefore m = 0, ie., the straight line is parallel to

x-axis. Therefore, if 2

1

+

-

r

r

e

e is plotted as ordinate against 1/T as abscissae, a straight line

parallel to x-axis is obtained.

The plot of 2

1

+

-

r

r

e

e against 1/T for CHCl3, the chloroform, and CCl4, the carbon

tetrachloride, is shown in fig 1.6a. We interpret that:

(i) The line for CCl4 is parallel to x-axis which shows that its molecules are

non-polar. It implies that the chlorine atoms in CCl4 are symmetrically

placed with respect to carbon atom, so that the center of gravity of negative

charges coincides with that of the positive charges and no permanent

dipoles are present.

(ii) The straight line for CHCl3 is inclined and shows the presence of polar

molecules. The polarity appears due to the charge asymmetry of the

molecule. The permanent dipole moment per unit volume is calculated

from the slope of the line. The dipole moment is then calculated on

dividing it by the number of dipoles per unit volume n. Constant ai is

determined from the intercept on y- axis.



P a g e | 31

Fig.1.6(a)

The study has been extended to a great number of molecules of monatomic gases like

Argon and molecules like H2 which are made up of two similar atoms. They do not

exhibit charge asymmetry and are therefore non-polar. But for molecules made up of

different atoms the charge asymmetry comes in and the molecules act like permanent

dipoles. The examples are NH3, H2O etc.

1.8 ELECTROSTATIC ENERGY AND ENERGY DENSITY IN FREE SPACE AND

IN DIELECTRIC

We shall first calculate the potential energy of a group of n point charges. It will be

equal to the work done in assembling the charges, bringing in one at a time.

If in the field of a stationary charge q1, another charge q2 is brought from infinity to a

distance 2112 rrr -= then work done is

12

21

0

24

1

r

qqW

pe=D (1)

=If a third charge q3 is brought from infinity to a point distant r13 and r23 from charges q1 and

q2 respectively then work done is

23

21

013

21

0 4

1

4

1

r

qq

r

qqW

pepe+=D (2)

From eqs. (1) and (2), the total work done in assembling the three charges is



P a g e | 32

32 WWW D+D=D

úû

ùêë

é++=

23

21

13

21

12

21

04

1

r

qq

r

qq

r

qq

pe

which can be extended for an assembly of n charges as

úúû

ù

êêë

é+++++= ...............

4

1

23

32

13

31

12

21

0 ij

ji

r

qq

r

qq

r

qq

r

qq

pe (3)

ij

jin

j

n

i r

qqåå ==

=1

01 4

1

2

1

pe

where the prime on the second summation implies that the terms i = j are to be excluded from

the sum. The factor 2

1 has been introduced because of the fact that each pair of charges

occurs twice in the summation.

Equation (3) then represents the total electrostatic energy U of the assembly of n charges and

if we put

i

n

jj

i

q

qj

pe=å =1

04

1

In eqn. (3), we get

i

n

i iqU jå ==

12

1

If the point charges have been assembled in a linear dielectric medium, instead of vacuum or

free space, then free space permittivity e0 is to be replaced by absolute permittivity e = e0 er ,

where er is the relative permittivity or dielectric constant of the medium.

Arbitrary Charge Distribution:

The expression for the field energy, U can be expressed with much advantage in terms of

volume and surface integrals involving the useful quantities electric field E and displacement

D when there is arbitrary charge distribution.

Let us consider a system of arbitrary charge distribution with density of initial charge

distribution asp. Let us remove the charge to infinity so that at any time the charge density is

ar where a is a parameter with values is ranging from unity to zero. At a point, where

density is ar, the charge dq’ in a volume element ‘dv’ will be

dq’ = (ar) dv (1)

and the potential f’ = af (2)

because f µ r and f is the initial potential at that point.



P a g e | 33

As the charge is removed to infinity, charge density at every point in the distribution will fall,

say, from ar to (a - da) r so that decrease in the charge contained in the volume element dv

will be

dq” = (da) r dv (3)

And the energy withdrawn from the system in removing this charge from volume element dv

is

dU = f’ dq” (4)

= (af) (da) r dv (5)

On using Eqs. (2) and (3).

If the whole charge from the system is removed to infinity i.e., the charge density is

reduced to zero everywhere then the total energy withdrawn, equal energy of the system, will

be

òò=v

dvdU fraa1

0

ò

ò

=

÷÷ø

öççè

æ=

v

v

dv

dvU

fr

fra

2

1

2

1

0

2

We know that r = div D

So that field energy is ò= dvDdivU f2

1

Further

div (fD) = f div D + D grad f

so that

dvDD

dvgradDDdivU

v

v

ò

ò

Ñ-Ñ=

-=

ff

ff

rr.)(.

2

1

.)(2

1

But E = - Ñr

f,

so that òò +Ñ=vv

dvEDdvDU .2

1)(.

2

1f

r



P a g e | 34

Changing the first volume integral into surface integral by Gauss divergence theorem, we get

òò +=vs

dvDEdSDU .2

1

2

1f (6)

as v can be any volume which includes all the charges in the system, we can choose the

bounding surface S at large distance from the charge distribution. At large distances

r

1af

2

1

rD a

2rdS a

so that r

dSDS

1. =òf

¥®® rwhen0

Eq. (6) is then left with

ò=v

dvDEV .2

1 (7)

predicting that the energy is distributed with a density ).(2

1DE per unit volume. Hence the

energy density i.e., the ener gy per unit volume in an electrostatic field is

).(2

1DEu = (8)

Eq. (7) holds for linear dielectrics.

Form linear dielectric,

D = e E

So that energy density is 2

2

1Eu e= (9)

In free space e = e0 (since er = 1), we have energy density of electrostatic field in free

space as given by 20

2

1Eu e= (10)

Eqs. (8) and (9) hold good for isotropic homogeneous dielectric in which D and E are

proportional to each other. But if the medium is anisotropic, the permittivity e has different

values in different directions and consequently e which relates D and E is a tensor e0b . The

relation can be written as



P a g e | 35

baba e ED =

babee Er )(0= (11)

where (er)ab is the relative permittivity or the dielectric constant of the medium. Putting eq.

(11) in eq. (7), we get

dvEEU r ababee )(2

10ò= (12)

The dielectric constant (er)ab is a symmetric tensor which can be judged from the integrand of

eq. (12),

abbebaababab eee EEEEEE rrr )()()( == (13)

where the indices of the last step step have been interchanged. Therefore,

[ ] 0)()( =-ò dvEEv

rr babaab ee (14)

which shows that for arbitrary values of Ea and Eb , the tensor

(er)ab = (er)ba (15)

except when field energy vanishes identically. Physically it means that it does not matter

whether the displacement Da is produced by field Ea, the value of (er)ab or (er)ban is the same

for the specified components.

When Charge Distribution is Discrete:

From eqs. (10), (8) and (7), we get for free space

dvEUv

20

2

1ò= e (16)

If the distribution is discrete then

E2 = E.E = (E1 + E2 + E3 + …...) . (E1 + E2 + E3 ……)

ji

n

j

n

i

n

j j EEE .11

1

2 ååå==

=+= (17)

The term for which i = j are included in first term. Prime stands for which ji ¹ . Putting eq.

(17) into eq. (16), we get

dvEEdvEUv

ji

n

j

n

iv

n

jj ò ååò å ÷

÷ø

öççè

æ+÷

÷ø

öççè

æ=

===

.22 11

0

1

20 ee (18)

Obviously for the first term the position of other charges is immaterial for any charge i.e., this

term represents the work done in the creation of the charges and is known as the self energy

of the system, U0. Thus



P a g e | 36

dvEEUUv

ji

n

j

n

iò åå ÷

÷ø

öççè

æ+=

==

.2 11

00

e (19)

The second term represents the work done by one charge in the field of others and so on i.e.,

it is the work done in bringing the charges from infinity to the space to constitute the given

distribution of charge.

1.9 Let us sum up

In this unit we described the terms potential and fields due to an electric

dipole.

Given the relation between electric susceptibility, Polarization, Displacement.

Explained the molecular field and the derivation of Mosotti relation for non-polar

molecules and Debye formula for polar molecules.

Finally the derivation for electrostatic energy and energy density has been done.

1.10 Lesson end activities:-

(i) What is electric dipole and dielectric polarization?

(ii) Define the terms Electric Susceptibility,Polarization(P) and

Displacement(D).

1.11 Points for Discussion

(iii) Prove D = e0 E + P

(iv) Explain Gauss law in Dielectrics

(v) What are dielectrics under what conditions do they differ from conductors?

Is dielectric an insulator? Explain.


(vi) Derive Classius-Mosotti relation for Non-Polar molecules.

(vii) Derive Langevin Debye formulae for Polar molecules.

1.13 Sources/References

1. Electromagnetic Theory by K.K Chopra and G.C. Agrawal.

2. Electricity and Magnetism by R.Murugesan.



P a g e | 37

UNIT – I Lesson – 2

2.0 aims and objectives

This lesson dealer with the discussion about Bio and savant law Lorenz force law, ampere’s

circuital law and its application in magne to statics. The divergence and civil of the magnetic

Induction B and the derivation for the scalar and vector potentiah is abo discussed.

2.1 Magnetic Induction

let us consider a +ve test chare q0, with a velocity v through a point

P. If this charge experiences a side-way deflecting force F, then a magnetic

field is said to exist at that point. This field is defined by means of a vector

quantity B and is called magnetic Induction, shown in Y axis.

If a charge moving through a point P in a magnetic field experiences

no side way deflecting force then the direction of motion of the charge is

defined as the direction of B. Conditions are

(1) when V is parallel to B (in the same direction) F is minimum. (2) when V

is perpendicular to B, F is maximum.

Definition: If a +ve test charge q0 moving with velocity v through point P in

a magnetic field experiences a deflecting force F, then magnetic induction B

at P is defined by

F = q0 V x B ---------(1)

The above relation defines both the direction and magnitude of B is F = q0 V B sinθ

B = F / q0 V sinθ ; θ is angle between V x B.

From the above V is parallel to B, θ = 0 ; F = 0 (min) v is perpendicular

θ = 90 ; F = Bq0 V (max)

Units; Newton / amp –turn

Z F

P



P a g e | 38

B +q0 θ Y

V

X

2.2 Ampere’s force law (Force on current element)

The concept of magnetic field is introduced by considering a test

charge q moving in a region of space with velocity V. Suppose the charge

experiences a force F, then the region is said to be having magnetic field B,

we write

F = qV x B -----------(1)

The above equation in terms of current i.e. the current crossing a surface is

defined as the rate at which charge flows across the surface,

ie I = dq / dt ---------(2)

ie, the force experienced (dF) by the charge dq moving with velocity V then

Eq. (1) becomes

dF = dq VxB

= I dt VxB.

Suppose in the time dt, charge dq travels along the length ‘dl’ of the

conductor then

V = dl / dt ie., dF = I dt (dl /dt) x B

dF = I dl x B --------(3)

this is Ampere’s force law

The total force experienced by the total volume containing the charge can be

calculated by integrating the above equation

F = v0l ( J xB ) dv [ie, I = J.ds and ds.dl = V]

2.3 Ampere’s circuital law



P a g e | 39

The law states that the line integral of the magnetic induction vector B

around a closed path is equal to µ0 times the total current crossing any

surface bounded by the line integral path, ie

c B . dl = µ0 s J.ds = µ0 I ---------(1)

we know curl of the magnetic induction

x B = µ0 J ( the differential form of Amperes’

Circuital law) -----(2)

Integrating over an open surface S, bounded by a closed curve C we get

s ( x B).ds = µ0 s J.ds ------(3)

But according to Stoke's theorem surface integral can be changed to line

integral as,

s ( x B) . ds = c B dl ---------(4)

Substituting Eq.[4] in Eq.[3] we get

c B dl = µ0 s J.ds =MOI

µ0 I - which is Amperes Circuital law. ( s J.ds = I ) this is integral form of

Ampere’s circuital law.

2.4 Biot & Savart law – Magnetic Induction.

Ampere performed series of experiments to find the forces between two

current elements. He observed that force between two current elements dl 1

and dl 2 carrying steady currents I1 and I2 depends on

(1) It varies directly as the magnitude of each current (I1, I2 )

(2) It varies inversely as the square of the distance between the two circuit

elements (r² 21)

(3) It depends upon the lengths and orientation of the two current elements

(dl 1, dl 2 )



P a g e | 40

(4) It depends upon the nature of the medium

- +

- + dl

r r1

dl 2

Fig. 2.4 Two currents I1 and I2

The force exerted on current element dl 2 by current element dl 1 is given by

dF21 = (µ0 / 4 ) (I1, I2) ( 1 / r²21) [dl 2 x (dl 1 x r21 /r21) ]

(µ0 / 4 ) ---------- arises due to (4) -------nature of the media

I1, I2 --------------- due to (1)

(1 / r²21) ------------due to (2)

[dl 2 x ((dl 1 x r21 /r21) ] arise due to (3) r2 1 / r21

represents unit vector along r21

The above equation dF21 can be written for whole lengths of the conductors

as

F21 = (µ0 / 4 ) (I1, I2) 1 2 [dl 1 x (dl 1 x r2) ] / r3 21 -------(2)

The above equations not does have much practical value because of r321 The

force cannot expressed as the interaction of current I1 with field current I2 .

However the above equation. can be written as,

F21 = I1 1 dl 1 x [ µ0 / 4 I2 2 (dl 2 x r2 1 )/ r³2 1 ------(3)

F21 I2 dl 2 x B1

Where B1 =( µ0 / 4 ) 2 dl 2 x r2 1 / r³2 1 --------(4)

B1 is called the magnetic induction, magnetic flux density or magnetic field

current. Unit is web / m2 or Tesla. In general the magnetic induction B at a

position r due to a current carrying circuit of element I dl will be

B =(µ0 / 4 ) I dl x r / r3



P a g e | 41

This is Biot savart law

Point (1) If the current I is distributed in space with a current density J

then I dl = Jd

Here B = µ0 / 4 J x r / r3 d

Point (2) If a single charge q moving with velocity v then

B = µ0 / 4 q ( v x r / r3)

or

B =(µ0 0 v x q r) / 4 0r3

But µ0 0 = 1 / c2 and E = qr/4 0r3

B = v x E / c2

The above gives the relation between electric (E) and magnetic fields (B) of a

uniformally moving charge as v<<c.

2.5 THE DIVERAGENCE OF THE MAGNETIC INDUCTION B

We had demonstrated that the magnetic field of moving charges were

such that . B = 0. It is also possible to arrive at this same result for

steady currents starting from the Biot – Savart law. We know that,

B = (µ0 / 4 ) r' (J1 x r1) / r3 d ' ---------(1)

. B =(µ0 / 4 ) r' (Jf r1) / r2 * dr' = (µ0 / 4 ) r' . ( Jf x r1/r2) d '

. ( Jf x r1/r2) r 1 / r2 .( x Jf ) - Jf. ( x r1/r2)

Where the first term on the right is zero because Jf is a function of the

source point x' , y' , z' while the del operator involves derivatives with respect

to the field point x , y , z. The second term on the right is also zero because

i j k

( x r1)/r2 =( x r )/ r3 = / x / y / z 0

(x - x')/r3 (y - y')/r3 (z - z')/r3

Then



P a g e | 42

.B = 0 -------------(2)

This equation follows form the definition of B given in Eq.(1) we also know

that it is a consequence of Coulomb’s law and of the Lorentz transformation.

The fact thatÑ. . B is zero means that these cannot be sources of B.

The net flux of magnetic induction through any closed surfaces is

equal to zero since

B.da = r . B dr = 0 › B.da = r . Bdr=0 ----------(3)

2.6 The Vector Potential A (Magnetic Vector Potential)

The calculation of electric fields was much simplified by the

introduction of the electrostatic potential. For an electrostatic field, the

relation between electrostatic field E and electrostatic potential V is given by

E = - V

Here V is a scalar quantity

In the case of a magnetic field,

div B = 0

Since the divergence of any curl is zero, it is reasonable to assume that the

magnetic induction may be written as,

B = Curl A = x A

'A' refers to magnetic potential and is called the magnetic vector potential.

Therefore the magnetic vector potential A can be defined as the vector,

whose curl at any point gives the vale of the magnetic fields B at that point.

The only other requirement placed on A is that

x B = x [ x A] = µ0 J

The unit of A is Wb / m

Derivation of the magnetic vector potential of a current loop:-

According to the Biot-Savart law, the magnetic induction at a distance

r from the element of length l carrying a current I (Fig. 2.6) is given by

B = µ0I / 4 . l x r / r3 -------(1)



P a g e | 43

dI dB

›

r p(x,y,z)

Fig. 2.6

We have (1 / r) = - r /r3 ----(2)

B =(µ0I / 4 ) l x { - (1/r)}

= (µ0I / 4 ) { x1/r) x l } ------(3)

We have the vector identity.

x ( A) = ( ) x A + ( x A)

Where is a scalar and A is a vector

Or ( ) x A = x ( A ) - ( x A)

So (1/r) x l = x ( l / r) – 1/r ( x l ) ------(4)

Using Eq. (4) in Eq.(3) we get

B = (µ0I / 4 ) [ x ( l / r) – 1/r ( x l ) -----(5)

In this equation, x l = 0 because the operator is a function of (x,y,z)

and the current element is not a function of (x,y,z) as shown in the Fig. 2.6

Then Eq. (5) reduces to

B = (µ0I / 4 ) x ( l / r) ----- (6)

Therefore the total magnetic induction at the given point by a closed loop

carrying current is given by

B = (µ0I / 4 ) x (dI/r) -------(7)

The operation x is independent of the integration of dI /r around the

closed loop.

The Eq.(7) can be rewritten as

B = x [µ0I / 4 dI/r ] = (curl µ0I / 4 ) dI/r ------(8)



P a g e | 44

Thus, we conclude that a vector exists such that by taking its curl, the

magnetic induction produced at any point by a closed loop carrying current

may be obtained. This vector is known as magnetic vector potential

A. Thus

B = curl A ------(9)

Where magnetic vector potential A = (µ0I / 4 ) dI/r ---------(10)

Eq. (9) is frequently used to derive magnetic induction B at any point from

the magnetic vector potential A at that point.

If the current is flowing through the length element is distributed over

a cross-sectional area a, we write l = Ja. Eq (10) is thus written as

A = µ0 / 4 v (J /r) x dV ----------(11)

The vector potential defined by Eq. (11) is not uniquely defined. We find

that we can add any term, whose curl is zero to the vector potential and it

still gives the same magnetic field. Unlike V, A does not have a physical

significance. It serves as a convenient intermediate step for the computation

of B.

2.7 THE CURL OF THE MAGNETIC INDUCTION B

We have shown that the magnetic induction is always equal to the

curl of the vector potential : B = x A. We shall now show that

x B = µ0 Jf ---------(1)

assuming a steady state and the absence of magnetic materials. In terms of

A,

x B = x x A = ( .A) - 2 A ----(2)

We have already shown that . A is proportional to the time derivative of

the electric potential V, then, with the above assumptions.

. A = 0 -----------(3)

For the second term we have from the definition of A that



P a g e | 45

2 A = (µ0 / 4 ) ' 2 (Jf / r) x d ' ---------(4)

Where we have interchanged the order of differentiation and integration.

Let us imagine At the field point P (x,y,z) where we wish to compute 2 A,

we form the vector J d '/ r, where Jf of and d ' are respectively the current

density and the volume element at the source point P', and where r is the

distance form P' to P. We compute the Laplacian of this vector at P by

taking the appropriate derivatives with respect to the coordinates x, y, z of P.

We then sum the contributions from all such sources in the volume ' which

includes all points at which Jf exists. The volume ' may include the field

point P, where r = 0.

Figure 2.7 Source point P' and field point P for the calculation of 2 A

Since Jf is not a function of the coordinates of P, we can write the

integral as

2 A = (µ0 / 4 ) ' Jf 2 (1/r) d ' -------(5)

Now, by differentiation of

(1/r) = 1/[ (x-x')2 + (y - y')2 + (z - z')2 ]½ -----(6)

we find that 2 (1/r) = 0 if r 0. There can thus be no contribution to the

integral from any element d ' except possibly if P and P' coincide and r is

zero

To investigate the integral at r = 0 we consider a small volume

enclosing the point P, where we wish to calculate 2 A, situated inside the

current distribution.

We take the volume so small that Jf does not change appreciably

within it; Jf may then be removed from the integral:



P a g e | 46

2 A = (µ0 Jf / 4 ) ' › 0 2 (1/r) d ' -------(7)

The meaning of this integral is as follows. For each element of volume d '

centered at the point P' within ' we calculate

2 (1/r) = ( 2/ x2 + 2/ y 2 + 2/ z2) * 1 /[ (x-x')2 + (y - y')2 + (z - z')2 ]½ -----(8)

Multiply by d ' and sum the results. Since 2 (1/r) = ' 2 (1/r)

2 A = (µ0 Jf / 4 ) ' › 0 2 (1/r) d ' ---------(9)

= (µ0 Jf / 4 ) r' › 0 ' . ' (1/r) d ' ---------(10)

(µ0 Jf / 4 ) s' › 0 ' (1/r) . da --------(11)

from the divergence theorem,

2 A = -(µ0 Jf / 4 ) s' › 0 (r1 . da) / r2 –----(12)

where r1 is the unit vector from the source point to the field point. In this

case r1 points inward toward the point P, Thus

2 A = - (µ0 Jf / 4 ) s' › 0 d -----(13)

where d is the element of solid angle subtended at the point P by the

element of area da. Since the surface S' completely surrounds P,

2 A = - µ0 Jf -------(14)

and x B = µ0 Jf ------(15)

This result is again valid only for static fields and in the absence of

magnetic materials.

2.8 THE FORCE ON A POINT CHARGE MOVING IN A MAGNETIC FIELD

The force on a current element I dl is I dl x B. Now, if the cross

sectional area of the wire is da

I = n (da v )Q -------(1)



P a g e | 47

Where n is the number of carriers per unit volume, v is their average drift

velocity, and Q is the charge on one carrier. The reason for this relation is

that the total charge flowing per second is the charge on the carriers that

are contained in the length v of the wire

Then the force on the element dl is

η da d Qv X B

And the force on a single charge Q moving at a velocity v in a field B is

Qv x B

This force is perpendicular both to the velocity v and to the local magnetic

induction B.

More generally if there is also an electric field E, the force is

Q[E + (vxB)

This is the Lorentz force.

2.9 Magnetic Scalar Potential

Consider a closed current loop carrying current I (Fig 2.9). Consider a

point P (r) having position vector r relative to current element I dI. From

Biot - Savart law, the magnetic induction B at P due to whole loop is

B =(µ0/ 4 ) I dI x r / r3

Let the point of observation P(r) be moved through an infinitesimal distance

dr say from P (r) to Q (r+dr). then,

B.dr = (µ0/ 4 ) (I dI) x r / r3 .dr

(µ0I/ 4 ) dr. (dI x r) / r3

µ0I/ 4 (dr x dI) . r / r3

When the point P is shifted to Q, the solid angle subtended by the loop at P

changes by d . But we can also get the same change in solid angle d

keeping P fixed and giving every point of the loop the same but opposite

displacement (-dx). Then, the above equation becomes

B.dr = - µ0I/ 4 (- dr x dI ) . r / r3



P a g e | 48

But, -dx * dI = dS = area traced out by current element dI during the

displacement (-dx)

Therefore B. dr = -µ0I/4 dS . r / r3

But, (dS . r) / r 3 = d = Change in solid angle subtended by current loop

when point P is displaced to Q

Therefore B.dr = - (µ0I/4 ) d ---------(1)

Since is a scalar function of (x,y,z) ›

d = .dr

Hence Eq. (1) becomes

B.dr = - (µ0I/4 ) . d

Or

B = - (µ0I/4 ) = - [ µ0I / 4 ] ----(2)

The direction of B is that of - , so that B points away from the loop along

its positive normal.

Comparing Eq.(2) with B = - Vm we get

Magnetic Scalar Potential Vm = (µ0I )/ 4

= µ0 /4 x current x solid angle ------(3)

Negative gradient of Vm gives the magnetic induction B.

Q (r + dr)

dr

P(r) d › dl ds

dr

Fig. 2.9

dr

I



P a g e | 49

2.10 Equivalance of a small current loop and a magneticdipole

(Application of magnetic vector potential)

Consider a small current loop of area S carrying current I (Fig. 2.10).

Let P be a point having position vector r relative to centre of loop. Then

magnetic scalar potential at P = Vm = (µ0I / 4 ) * .

Here = solid angle subtended by current loop at P

= (S.r / r3 ) = S cos θ / r2

therefore Vm = (µ0I / 4 ) S cos θ / r2 = (µ0 /4 ) * (IS) cos θ / r2 -----(1)

The expressions similar to that for potential due to electric dipole

V = (1/4 0) þ cos θ / r2 -------(2)

Thus magnetic field due to a small current loop is similar to electric field of

an dipole

The product (IS) is called magnetic dipole moment,

P

m

› r S θ

s

Fig. 2.10

m = IS ----(3)

Thus a current loop of area S carrying current I is equivalent to a magnetic

dipole of dipole moment m = IS.

s



P a g e | 50

2.11 ELECTRIC FIELD VECTOR IN TERMS OF SCALAR AND VECTOR

POTENTIAL

In the case of magnetic field,

div. B = 0 ------------(1)

Since the divergence of any curl is zero

B = curl A = x A ---------(2)

Here A is called the magnetic vector potential. It has physical importance

where magnetic field varies with time

The time variation on magnetic field, from Eq. (2) is

B / t = / t ( x A) ---------(3)

Interchanging the space and time operator, we get

B/ t = x A/ t -----------(4)

The differential form of Faraday’s law

x E = - B / t --------(5)

therefore Eq. (4) becomes x ( E+ A / t) = 0 -----(6)

From vector identity, we find that the curl free field must be gradient of a

scalar potential φ. We write

E + A / t = - grad φ ------(7)

or E = - A/ t – grad φ -------(8)

Eq. (8) suggests that for time dependent magnetic field, we may think of –

grad φ as the contribution to E due coulomb field and - A / t as the

contribution due to electromagnetic induction

2.12 Let us sum up

From this unit you have learnt the Biot-savart law, Lorentz force

law,Ampere’s circuital law and its applications in Magnetostatics. Also it dealt

with divergence and curl of B and the derivation for scalar and vector potentials.



P a g e | 51

2.13 Lesson end Activities

(i) Mention the importance of Scalar and Vector potentials in Magnetostatics

(ii) State Biot-Savart law

(iii) From Biot-Savart law derive Ampere’s circuital law

(iv) Write the equation for Lorentz force.


(i) From Biot-Savart law derive B=m0 H

(ii) Find the magnetic induction at the centre of a square current loop of side

I metre carrying a current of 1 ampere.( Ans. 8Ö2 * 10-7 wb/m2)

(iii) In the Bohr model of the hydrogen atom, the electron revolves round

the nucleus in a path of radius 5.29 * 10-11 m at a frequency of 6.58 *

1015 Hz.

What is the dipole moment and Magnetic induction?

[Hint: m=Ai ; B =m0 i / Za ; i= e n ]

[ Ans: 9.266 * 10 -24 Am2 ; 12.52 T]


(i) Derive curl and divergence of B

(ii) Derive the equations for scalar and vector potentials

2.16 Source/Reference

(i) Electricity and Magnetism by R.Murugesan

(ii) Electrodynamics by Gupta, Kumar and Singh



P a g e | 52

UNIT – II - Field Equations

Lesson - 3

3.0. Aim and Objectives: In this unit you study the meaning of

continuity equation, displacement current and also the derivation of

Maxwell’s equations in differential and integral form also you study about

the scalar (Φ) and vector potential (A) and its relation to Lornetz and

Coulomb gauge.

3.1 Equation of Continuity

According to the principle of conservation of charge the net amount of

charge in an isolated system remains constant. The principle can be stated

as follows.

In the net charge crossing a surface bounding a closed volume is not

zero, then the charge density within the volume must change with time in a

manner that the time rate of decrease of charge within the volume equals

the net rate of flow of charge out of the volume. This statement can be

expressed by the equation of continuity.

Derivation: Let us consider that charge density ρ, is a function of time. The

transport of charge constitutes the current i.e.,

I = dq /dt = d/dt v ρ. dV ------------ (1)

Here we have considered that the current is extended in space of volume V

closed by the surface 'S'. The net amount of charge which crosses a unit

area (normal to the direction of charge flow) of a surface in unit time is

defined as the current density J. We know, if a net amount of current is

flowing outward closed surface the charge contained within that volume

should decrease at the rate

-dq / dt = I ----------- (2)

Where I is the total current flowing through surface S. if J is the current

density, then by definition, total current I will le

I = s J. ds------- (3)

From equations (2) and (3) we get



P a g e | 53

s J. ds = - dq/dt

= -d/dt v ρ. dV ---------- (4)

{Using equ. (1) }

Because it is ρ which is changing with time, we can write

d/dt v ρ.dV = v ρ./ t.dV

so that equ. (4) becomes

s J.ds = - v ( ρ/ t). dV ----------(5)

From divergence theorem, we have

s J.ds = - v (div J) dV

so that equ (5) becomes

v (div J) dv = - v ( ρ/ t). dV

(or)

v (div J + ρ/ t)dV = 0 ------(6)

Since Eq. (6) holds for any arbitrary volume, we can put integral equal to zero. i.e.,

div J + ρ/ t = 0 --------(7)

It is referred to as the equation of continuity. It is the mathematical

expression for the conservation of charge. It states that the “ total current

flowing out of some volume must be equal to the rate of decrease of charge

within the volume, assuming that charge cannot the created or destroyed.

i.e., no sources and sinks are present in that volume”. In case of stationary

currents, charge density at any point within the region remain constant

i.e.., ρ/ t = 0

› So that div J=0 or . J = 0 which express the fact – that there is no net

outward flux of current density J.

3.2 Displacement current (D)

Maxwell changed the definition of total current density to adapt the

equation of continuity to time dependent fields

Ampere’s circuital law is

s B. dl = µ0I

s H. dρ = I = s J.ds

Changing line integral into surface integral, by stoke`s theorem,



P a g e | 54

s curl H. ds = s J.ds (or)

curl H = J ----------- (1)

Let us substitute it in equation of continuity, then

div J = - ρ / t

we get div ( curl H) = - ρ/ t

0 = - ρ / t

Here equ (1) leads to steady state condition in which charge density is not

changing. Therefore for time dependent (changing) fields, Eq (1) should be

modified. Maxwell suggested that the definition of total current density is

incomplete and advised to add something to it. Let it be J' then Eq (1)

becomes

Curl H = (J+ J') ---------- (2)

In order to identify J', we take divergence of Eq (2) That is

div (curl H) = div(J+ J')

0=div J+div J'

(or)

div J' = -div J = ρ/ t -----------(3)

We know that ›

ρ. = . D So that Eq. (3) becomes

› div J' = / t = ( .D )

› = . D / t

= div ( D / t)

(or)

div [J' - ( D / t)]= 0 --------- (4)

Eq. (4) is true for any arbitrary volume, we can have

J' = ( D / t) -------(5)

Therefore the modified form of the ampere's law is

Curl H= J+( D / t) --------- (6)

Note: 1. Since J' arises due to the variation of electric displacement D

with time, it is termed as displacement current density . According to



P a g e | 55

Maxwell it is just as effective as J, the conduction current density in

producing magnetic field. (2) The important inference that we get from Eq.

(6) is that, since displacement current J' is related to the electrtic field vector

D (as D = εE ) it is not possible in case of time varying fields to deal

separately with electric and magnetic fields but, instead the two fields are

interlinked giving rise to electromagnetic fields. Thus J' results into

unification of electric and magnetic phenomenon.

3.3 The Maxwell’s equations ( Differential form)

The four equations of Maxwell’s are.,

› (i) .D = ρ ---- obtained by the application of Gauss theorem in

electrostatics. D is the electric displacement in coulomb / meter2 and ρ is

the free charge density in coulomb/meter3.

› (ii) . . B = 0; obtained by the application of Gauss theorem to magnetic

field and B is the magnetic induction in Weber/mt2

› (iii) .E = - B/ t; obtained by Faraday’s and Lenz`s law in

electromagnetic induction and E is the electric intensity in volt /

meter

› (iv) x H = J + D/ t : obtained by Maxwell’s modification of Ampere’s

law in a circuital form for magnetic field accompanying an electric

current and H is magnetic field intensity in ampere/meter and. J is

the current density in ampere/meter2 .

(A) Derivation of Maxwell’s equations

(i) div D= ρ

Consider a surface S bounding a volume V in a dielectric medium. From

Gauss theorem the integral E.ds of the normal component of E over any

closed surface is equal to the total charge enclosed within the surface. Also

we know that the total charge must include both the free and the



P a g e | 56

polarisation charges or the total charge density ρp = -div p and ρ is the free

charge density at a point in a small volume element dV. Thus total charge

density at that point will be, ρ - (divp) then Gauss law can be expressed as

s E.ds = v div E.dV= 1/ 0 (ρ-div p) dV

(or)

div ( 0E+P)dV = ρdV

the quantity ( 0E+P) is D called electric displacement, so that

div D dV = ρdV

(or)

(div D - ρ)dV = 0

Since this equation is true for all volume, the integrand in this equation

must vanish i.e., Div D= ρ

when the medium is isotropic the three vectors D,E, P are in the same

direction and for small field, D is proportional to E, that is

D = εE

Where ε is called dielectric constant of the medium.

(ii)

Since the magnetic lines of force are either closed or go off to infinity, the

number of magnetic line of force entering any arbitrary closed surface is

exactly the same as leaving it. It means that the flux of magnetic induction

B across any closed surface is always zero i.e..,

B.ds = 0

Transforming the surface integral into volume integral, we have

div B dV = 0

The integrand should vanish for the surface boundary as the volume is

arbitrary, i.e.,

div B=0

(iii) curl E - B/ t

By Faraday law we know that emf induced in a closed loop is given by

e = - / t = B/ t . ds

div B= 0



P a g e | 57

Since the flux = s B.ds where S is any surface having the loop as

boundary. E.m.f `e` can also be found by calculating the work done in

carrying a unit charge completely around the loop. Thus e = E. dl where E

is the intensity of the electric field associated with induced em f

Therefore, equating above two equations, we get

E.dl = - s B/ t . ds

Applying stokes` theorem, the line integral can be transformed into surface

integral i.e.,

› s ( x E) . ds = - s ( B/ t) . ds

This equation must be true for any surface whether small or large in the

field. Therefore the two vectors in the integrands must be equal at every

point, i.e.,

› x E = - B/ t

Crul E = - B/ t

(IV) --- Curl H = J+ D/ t

Ampere’s law in the circuital form gives this equation. According to this law,

the work done in carrying a unit magnetic pole once round closed arbitrary

path linked with the current is expressed as

H. dl = I (or) = J. ds

where the integral on the right is taken over the surface through which the

charge flow corresponding to the current I take place. Now changing the

line integral into surface integral by stoke`s theorem

s curl H.ds = J.ds

Curl H= J

The above relation, derived on the basis of Amperes` law, stands only for

steady closed current. But for the changing electric fields, the current

density should be modified. The divergence of the above equation is



P a g e | 58

div (curl H ) = 0 (or)

div J=0 which conflicts with the equation of continuity div J = (- ρ/ t.)

Adding J we get curl H= (J+J'). Taking divergence of the above equation, we

get

div (curl H) = (divJ+divJ') (or)

0= div J+divJ' (or)

div J' = -divJ = + ρ / t.

› we know that ρ = . D

Substituting this value in the expression for Div J', we get

› Div J' = / t ( . D)

› › . J`= ( . D/ t)

therefore the Maxwell’s fourth relation can be written as

Curl H = J+ ( D/ t)

3.4 Maxwell’s equation in free space

In the free space, where the current density J and volume charge

density ρ are zero, Maxwell’s equations reduce to

› .D = 0

› .B = 0

› x E = - B/ t

› x H = D/ t

Maxwell’s equations in linear Isotropic Media

In linear isotropic media

D = E

and H= B/µ

Where is the dielectric constant, µ permeability of the medium.

The Maxwell’s equations become

› . E = ρ /



P a g e | 59

› . H = 0

› x E + µ( H/ t) = 0 › x H - ( E/ t) = J.

3.5 Energy in electromagnet fields: Poynting vector – ( Poynting

theorem)

Energy may be transported through space by means of e.m. waves.

Let the material inside S be isotropic homogenous and characterised by

permeability µ, permittivity and conductivity . For derivation, consider a

volume V bounded by a closed surface S.

Maxwell’s third and fourth relations are

Curl E = - B/ t

Curl H = J + D/ t

Taking scalar product of both sides of the above equations with H and E

respectively and subtracting, we get

E. curl H . H curl E = J. E + [(E.( D/ t) + H ( B/ t)] ---(1)

But we know that

H. curl E-E. curl H = div (ExH)

So that equ. (1) becomes

-div (E x H) = J. E[E.( / t) ( . E) + H( / t) (µH)]

(or) J.E + (1/2. / t ( E2) + 1/2. / t(µH2) + div (ExH) = 0

(or) J.E +(1/2. / t (E. E) +1/2. / t (H. µH) + div (ExH) = 0

(or) J.E +(1/2. / t (E.D) + 1/2. / t(H.B) + div (ExH) = 0 --(2)

Integrating over the volume V bounded by the surface S, we get

v (J.E) dv + v 1/2. / t(E.D+H. B) + vdiv (ExH) dv = 0 ----(3)

But as v div (ExH) dv = (ExH).ds

We write equation (3) as

v (J.B) dv + v 1/2 / t (E.D + H.B) = - s (ExH).ds ---- (4)

Integrating the second term of Equ. (4) we get

1/2 / t(E.D+H.B)dv = 1/2 . / t(E. E + H .µH )dv



P a g e | 60

= / t (1/2 . E2 + 1/2 µH2) dv -----(5)

The first and second term on right hand side represent the time rate of

increase of energy stored in the electric and magnetic fields respectively in

the volume V. Considering the Eq.(4), LHS of this represents the sum of the

power expended by the fields due to the motion of charge and the time rate

of increase of stored energy in the fields. On the other hand RHS of Eq. (4)

must represent the power flow into the volume V across the surface S, or the

power flow out of the volume V across the surface S

= s(ExH) . ds

= s P.ds

where P= (ExH) ----(6)

It then follows that the vector P has the meaning of power density associated

with the electromagnetic filed at that point. The statement represented by

Eq.(6) is known as poynting theorem and the vector P is known as the

poynting vector.

3.6 Electromagnetic potentials – Maxwell’s equations is terms of

electromagnetic Potentials

Consider the Maxwell’s equations

µCurl H = µJ+µ D/ t -------(1)

Curl B = µJ + µ E/ t -----(2)

Where and µ are permittivity and permeability, Substituting for B (i.e. B =

Curl A) and E(-grad φ- A/ t) (where A and φ are electromagnetic potentials),

we get

Curl (CurlA) = µJ +µ / t (-grad φ- A/ t)

i.e.., grad div A- 2A=µJ-µ / t (grad φ)- µ 2A/ t2)

i.e., 2A -µ 2A/ t2 –grad (divA+µ φ/ A) = -µJ ----(3)

Considering other Maxwell’s equations, namely

Div D = ρ

div E = ρ

i.e., div (- grad φ- A/ t) = ρ/

i.e., 2 φ = / t(div A) = - ρ/

Adding and subtracting µ φ 2/ t2 it becomes



P a g e | 61

2φ - µ 2φ/ t2 - / t(div A+µ φ/ t)= - ρ/ --(4)

Equations (3) and (4) are field equations in terms of electromagnetic

potentials. Here Maxwell’s equations are reduced from four to two by

electromagnetic potentials,

Note: Electromagnetic potential define the field vectors uniquely though

they themselves are non-unique. We get the same field vectors when we use

the set (A, φ) or (A', φ’). These transformations are called gauge

transformations.

3.7 Lorentz Gauge

Maxwell’s field equations in terms of electromagnetic potentials are

2A- µ 2A/ t2-grad (divA+µ φ/ t) = -µJ ---(1)

2 φ - µ 2φ/ t2 + / t(div A+µ φ/ t)= -ρ/ –(2)

The above equations may be simplified as

div A +µ φ/ t = 0 ----(3)

This requirement is called the Lorentz condition and when the vector and

scalar potential satisfy it, the gauge is known as Lorentz gauge.

So with Lorentz condition field equations reduce to

2A-µ ( 2A/ t2) = -µJ ----- (4)

and 2φ -µ ( 2φ/ t2) = -ρ/ -----(5)

But we know µ = 1/v2 (i.e., V= 1 µ )

Hence Equations (4) and (5) can be written as

2A = -µJ ------(6)

2 φ = - ρ/ -----(7)

and 2 = 2 – 1/v2 * ( 2 Λ / t2)

Equations (6) and (7) are inhomogenous wave equations and are known as

D' Alembertian equations and can be solved. The potentials obtained by

solving these equations are called retarded potentials.

To determine the requirement that Lorentz condition Λ, we substitute A' and

φ' from equations already given earlier.

Div (A' –grad Λ) + µ / t (φ' + Λ / t) = 0

i.e., div A' = µ ( φ'/ t) = 2 Λ - µ ( 2 Λ / t2)

Hence A' and φ' will satisfy the equations (3) i.e.., Lorentz condition provides



P a g e | 62

2A - µ ( 2 Λ / t2) = 0 ----- (8)

i.e., 2A = 0

Lorentz condition is invariant under those gauge transformations for which

the gauge functions are solutions of the homogeneous wave equations.

Advantages:

(1) It makes equations for A and φ independent of each other.

(2) It leads to the wave equation.

(3) It is a concept which is independent of Co – ordinate system.

3.8 Coulomb Gauge

Consider the field equations in terms of electromagnetic potentials we get,

2 φ + / t (div A) = - ρ/ -----(1)

If we assume div A = 0

The above equation (1) reduces to Poisson’s equation

2 φ(r,t) =ρ(r' t)/ ---(2)

Whose solution is

φ (r,t) = 1/4 (ρ (r' t)/R) d '----(3)

i.e., the scalar potential is just the instantaneour Colombian potential due to

charge ρ (x', y', z',t). This is the origin of the name coulomb gauge.

From equations (2) (3) we get

2 { 1/4 (ρ (r' t)/R) d ' } = -ρ ( r' , t)/ ---(4)

As Poisson’s equations holds good for both scalar and vectors replacing

ρ (r' t) by J' we get

2 [1/4 (J' /R) d ' ] = -J' / --------(5)

As J' is confined to the volume ' the surface contributions will vanish, so

(J' /R) d ' = ( 'J' /R) d ' --------(6)

(Since J' = -(1/4 ) . J' /R) d ' +1/4 x x (J' /R*) d ’]

We know x (J' /R) = 1/R ' x J' x ' (1/R)

= ( ' x J') /R) d 2 + (J' /R) x ds

(as x V d 2 = - sV x ds )

As J' confined to volume ' surface contribution will vanish so

x (J' /R) d ' = ( ' x J' /R) *d ' --------(7)



P a g e | 63

Now as ' x J' = x [-1/4 ( '. J'/R) d ' ----------- (8)

i.e., ' x J' = 0 (as curl grad φ = 0)

and . J' = [ x ( ' x J' /R) d '

i.e. . J = 0 (as div curl V = 0)

In terms of vector potential

2 A – (1/v2) 2A/ t2 = -µJ

2A = -µJ

i.e.., the equation for A can be expressed entirely interms of the transverse

current.

The Coulomb gauge has a certain advantage. In it the scalar potential is

exactly the electrostatic potential and electric field is given by

E = - grad φ - A/ t

It is separable into an electrostatic field V = φ and a wave field given by -

A/ t. This gauge is used when no sources are present. If φ = 0 and A

satisfies the homogeneous wave equation, the field is given by

E = - A/ t and B = x A

3.9 Radiation produced by a low velocity accelerated charged particle

(Larmor`s formulae)

If a charge is accelerated but is observed in a reference frame where

its velocity is small compared to that of light, then in that coordinate frame

the acceleration field is given by

. Ea = e/c [n x (nx β) / R ]ret -----(1)

Where β = dβ / dt is the ordinary acceleration divided by C. n is the unit

vector, `ret` mean that the quantity in the brackets is to be evaluated at

retarded time. The instantaneous energy flux is given by the poynting vector

S = C/4 (ExB) = C/4 -Ea- 2 n---- (2)

It means that the power radiated per unit solid angle d

dP /d = C/ 4 -REa - 2 = [e2/4 c [nxnxβ2) ] ----(3)

If θ is the angle between the acceleration V and n, then the power radiated can be written as . dP/d = (e2 /4 c3) -V - 2 sin2θ ---- (4)

This exhibits the characteristic sin2θ is angular dependence. We note from



P a g e | 64

. equ.(1) that the radiation is polarised in the plane containing V and n. The

total instantaneous power radiated is obtained by integrating equ. (4) Over

all solid angle. Thus

. P= 2/3 (e2/c3) -V - 2 -----(5)

This is Larmor result for a non-relativistic accelerated charge

Larmor`s formulae can be generalised by arguments about covariance

under Lorentz transformations to yield a result which is valid for arbitrary

velocities of the charge. (Radiated electromagnetic energy behaves under

Lorntz transformation like the fourth component of a 4 –vector)

3.10 Radiating systems.

The charges are the ultimate source of electromagnetic fields. Bohr`s

theory of hydrogen spectrum and the allied theories of radiation emission by

atoms and molecules exhibit the basic radiating system. Here the radiant

component of electric and magnetic fields associated with accelerated charge

e are given by the relation

› › › Er = (e/4 0c) [rx(r-rβ) x β / (r-r. β) 3 ----- (1)

and Br = (nxEr)/C -----(2)

The flow of the electromagnetic energy is given by poynting vector

Pr = Er x Hr ----(3)

The radiating system is normally considered as an oscillating dipole. Now

we discuss of half-wave for antenna array systems employed for propagation

of electromagnetic energy into space.

3.10 (a) Radiation due to an Oscillating Electric dipole.

A pair of charges equal in magnitude and opposite in nature separated by a

small distance constitute an electric dipole. Such a dipole is shown below

dl

P

+ +q

-

-q



P a g e | 65

The dipole moment p is defined by p = q.dl --- (1)

Consider the dipole is acted upon by a field of frequency ω varying sinusoid

ally. The separation dl varies according to applied field and varies the

dipole moment. The instaneous value of dipole moment may be written as

þ = þ0e-iωt ----(2)

Where þ0 represent the amplitude of dipole moment. Considering volume

distribution of charge density ρ, the dipole contribution is

þ = (ρ dV') r --------(3)

Where dv' is the source volume separated by a distance r from the origin.

Considering time rate variation of dipole moment,

. dp/dt = p = d/dt (q.dl) = dq/dt x dl = Idl ----(4)

where dq/dt = I, the current. This implies that the time rate variation of

dipole moment is equivalent to a current element I.dl. An oscillatory motion

of a system is an accelerated type of motion and acceleration associated with

such a motion is responsible for radiation.

Poynting vector and Radiated power

The poynting vector P is given by

P= Er x Hr = Er x (Br /µ0) .. ..

þ › (þ)2sin2ø / 16 2 0c3r2 (nθ x n) ---(1)

Where nθ and n are the transverse components of unit normal vector

(þ)2sin2 θ /16 2 0c3r2 . nr ------(2)

Where nr is the radial components of unit normal vector



P a g e | 66

The figure shows P varies as sin2θ. Let us take [þ] = þ0 cosω (t-r/c)

..

[þ]= ω2 þ0 cosθ (t-r/c) ----(3)

Substituting this value in equation (2) we get

P = ω4 þ02sin2θ/16 2 0r2c3 x cos2ω (t-r/c)nr ----(4)

The average of cos2ω (t-r/c) over a period of oscillation is 1/2. Therefore

average power radiated per unit area over a time period is

-P- = ω4 þ02sin2θ/32 2 0c3r2 x nr -----(5)

The power radiated through area ds is

dP =-P-ds

but if ds subtends solid angle d at the source distant r from ds, then

d = ds/r2 = r2sinθ dθ dφ / r2 -----(6)

and therefore power radiated per unit solid angle is

dp/d =-P-ds / (ds/r2) =-P1-r2

= ω4 þ02sin2θ/32 2 0c3 -----(7)

the total radiated power is given by

PT = (dp/d ) x d 2

0 0 ω4 þ02sin2θ/32 2 0c3 (sinθdθ)dφ



P a g e | 67

= ω4þ02/16 0c3 0 (1-cos2θ) sinθdθ

= ω4þ02/16 0c3 (- cosθ + cos3θ/3]0

PT = ω4þ02/12 0c3 ----(7)

This relation shows that the total power depends upon fourth power of the

exciting wavelength and þ02 . The dipole moment amplitude þ0 i s a

characteristic of the substance under test. The theory of dipole radiation is

used in explaining the radiation by atoms, molecules and by the antenna.

3.11 Radiation due to a small current element.

An oscillating dipole may be treated as a small current element, hence the

radiation field equations developed for a dipole may be applied to the case of

linear small current element. The case of the linear small current element

may be further extended to the case of lengthy current elements like linear

antenna and antenna arrays. The radiation components of electric and

magnetic fields due to an oscillating dipole are given by

.. E = [(þ) sinθ/4 0c2r] x nθ

.. B =[µ0(þ) sinφ/4 cr] x nφ -------(1)

The dipole moment of an oscillating dipole, which is treated as small current

element, is given by

þ = q.dl .

dp/dt = þ = (dq/dt) x dl = Idl ..

Also d2þ/dt2 = þ = (dI/dt) x dl ------(2)

Taking Eq. (2) becomes ..

þ = -i ωI0e-i ωtdl

Whose retarded value for a field and source point distant r is written as ..

[þ] = - iωI0dleiω(t-r/c) -----(3)

with real components of current, the substitution of Eq.(3) in Eq. (1) gives



P a g e | 68

E = [iωI0dlsinθ/4 0c2r] x e-i ω (t-r/c)nθ ----(4)

B =[iµ0ωI0dlsinθ/4 cr] x e-i ω (t-r/c)nφ

= - iωI0dlsinθ/4 0c3r x e-i ω (t-r/c)nφ since µ0 0 = 1/c2

[ -E- /C ] x nφ -----(5)

Where -E-is the numerical value of electric field given by equ. (4). Knowing

the components of electric and magnetic radiation fields, we can compute

the poynting vector. The average value of poynting vector for complex fields

is given by

P = (ExH)/2 = (ExB)/2µ0

Where the real part of E and B have to be taken into account

-P-= [ω2I02dl 2sin2θ/2(16 2 0c3r2) x nθ x nφ

= [ω2I02dl 2sin2θ/32 2 0c3r2] x nr -----(6)

Total radiated power is given by

PT = -P-ds = φ=2 θ=

= φ=0 =0 [ω2I02dl 2sin2θ /32 2 0c3r2 ] x (r2sinθdθdφ)

= ω2I02dl 2/32 2 0c3[8 /3]

φ=2 θ = ( Since φ=0 =0 sin3θdθdφ = [8 /3] )

Using ω = 2 c/λ,I0 = Irms 2 and µ0 0=1/c2 we get

PT = [(2 c/λ)2 ( 2 Irms)2.dl 2 /12 c] x µ0

=(2 µ 0c/3) (dl/ λ) 2 I2rms

Substituting µ0 = 4 x10 -7met/sec and C = 3x108 met/sec we get

PT = 80 2(dl/λ)2 I2rms -----(7)

The radiation resistance may be calculated by comparing equ. (7) with the

equation for the loss of power i.e.,



P a g e | 69

P = RI2rms ------(8)

It gives radiation resistance as

Rr = 80 2(dl/λ)2 ohms ---(9)

This shows that the radiation resistance is equal to the resistance in which

the power consumed and converted into heat is equal to the power radiated

in the form of E.M waves be a current element, provided the same current is

maintained.

Note: A short linear antenna is an example of linear current radiator. A

centre fed linear antenna is shown below.

Z F(x )

Coaxial θ r

X

φ

Y

Fig. Short linear antenna.

In case of short linear antenna dl<<λ. In such cases the radiation

resistance is usually quite low as compared to the ohmic resistance. For

efficient radiation of electromagnetic energy it is essential that the length of

the antenna is comparable with wavelength. A half-wave antenna i.e.

dl=λ/2 is commonly used to radiate electromagnetic energy into space.

3.12 Linear Half –wave antenna.

A linear half wave antenna is a straight conductor of length equal to

half the free space wavelength. If a linear conductor is set vertically on the

ground then only quarter free space wavelengths (λ/4) of the conductor

serves the purpose of half wave antenna as additional λ/4 length is

furnished by the ground reflection. Such antenna is used for radio

broadcast purpose and is shown below



P a g e | 70

z

dl F(r,θ)

λ/4 r' λ/4

l r

λ/4

λ/4

Fig. 3.12 Linear half wave antenna

Let the current fed to the antenna be I = I0cosωt and at centre I= I0 at t=0

consider an element dl at a distance l from the centre, then the amplitude

of current in dl is

I0cosωt ( /c) = I0cos (2 l / λ)

Since t = (l/c) and ω = 2 f = 2 c/ λ

The current in dl at any moment t is

I = [I0cos (2 l /λ)]cosωt

= I0/2 [cosωt - 2 l / λ) + cos(ωt + 2 / λ)

Writing this equation in exponential form we get

I = I0/2 [e-i(ωt – (2 l/λ) + e-i(ωt + (2 l/λ)] -----(1)

Now with this value of current in dl at the time t, we can calculate electric

intensity and it can be related to B (magnetic induction ). It can be written

as

E = -(µ0c/2 r) I0eiωI(t-r/c) . cos [ /2 cosθ] /sinθ * nθ ------(2)

The above equation shows that E is independent of frequency. Now B and

E are related as shown below.

B = -E-/C * nφ substituting -E-by equation (2) we get

B = (iµ0I0/2 r) * e-iω(t-r/c) . cos [ /2 cosθ] /sinθ * nφ web/m2 ----(3)

Poynting vector and Radiated power

Knowing E and B poynting vector and radiated power can be evaluated. The

average of Poynting vector or complex field is given by



P a g e | 71

P = ExH/2 = ExB/2µ0= ExE/2µ0c * (n θ x nφ ) = (E2/2µ0c) * nr

Substituting the value of E from (2) we get

P =(µ0c I20/8 2r2) * e-2 iω(t-r/c)] cos² [( /2 cosθ] / sin²θ * nr

Using I0 = 2 Irms we get

P = µ0c/4 2 *I2rms/r2 * cos² [( /2 cosθ] / sin²θ * nr -----(4)

Total radiated power is given by

φ=2 θ= PT = µ0c/4 2 *I2rms θ=0 θ=0 cos2 [( /2) cosθ]/r2 sin2θ * r2 sinθdθdφ

=µ0c/2 *I2rms θ=0 cos2 [( /2) cosθ/sinθ) .dθ -----(5)

Now let ( /2) cosθ = (β/2 - /2) so that cos2 [ /2 cosθ] = cos2θ [β/2 - /2]

=sin2 β/2 = 1-cos β/2

and ( /2) sinθdθ = 1/2dβ (or) dθ = d β/( s inθ) and µ0c/2 =60

Substituting these values in (5) we get

2 PT = 60 I2rms 0 (1-cos β)/β(4 - 2 β)] xd β

But 1/ β(4 - 2 β) = 1/4 [1/β + 1/2 - β]

2 2 PT = 15I2rms[ 0 (1-cos β/β) *dβ + 0 (1-cos β/2 - β) * dβ The integrals within the square brackets are evaluated graphically within

limits 0 to 2 . The plots of (1-cos β) against β and 2 - β are the same as

shown in figure below. The plot (1-cos β) is symmetrical about β = .

Therefore the areas expressed by two integrals within limits 0 to 2 . are

equal and

2 PT = 30I2rms 0 (1-cos β/β) *dβ

The value of the integral as obtained graphically is

2 0 (1-cos β/β) *dβ = 2.44

Therefore PT = 73.2 I2rms ------(6)



P a g e | 72

2 -β

1-cos β

0 2

Fig- 3.13 Current Let sum up

This shows that the radiation resistance of half wave antenna is much

higher than the dipole antenna. This implies that in case of such antenna

most of the power is radiated and ohmic losses are negligibly small. Hence

the efficiency of such antenna is large. It is for this reason that such

antenna are commonly used for radio broadcasting.

-------

3.13 Let us sum up

In this unit we described the terms and derivation for equations of

continuity, displacement current, Lorentz gauge and Coulomb gauge.

Also explained the Maxwell’s equations along with the

derivations.

Finally you have learnt about the low velocity acceleration charged

particle and the radiation due to a small current element and its application

to linear half wave antenna.

3.14 Lesson end activities

(i) What is the displacement current?

(ii) Write the fundamental Maxwell’s filed equations in differential

form?

(iii) Define Poynting vector.


(i) Find the magnitude of displacement current.

(ii) Assuming the total energy Ze of an atomic nucleus is uniformly

distributed within a sphere of radius a, show that the potential

energy of such a nucleus is



P a g e | 73

n = a

Ze2

0 5

3

4

1

pe

[Hint : U = ½ òe0 E2 dv ; Ein = 2

0

2

04

1

R

e

pe; Eout=

2

2

04

1

r

e

pe

Substituting q = Ze and Ro = a and solving the integral we

get the result.]


§ Derive Maxwell’s field equations.

§ Derive equations for

a) Lorentz gauge

b) Coulomb gauge

§ What is oscillating dipole? Derive an equation for radiation due to a

small current element.

§ Write a short note on linear half-antenna.

3.17 References

o Electrodynamics by Gupta ,Kumar Singh

o Electromagnetic Fields and waves by Pauli Lorrain and Dale Corson.

o Electromagnetic theory by K.K. Chopra and G.C.Agrawal.



P a g e | 74

UNIT – II Lesson - 4

INTERACTION OF ELECTROMAGNETIC WAVES WITH MATTER

4.0 Aim and Objectives:

In this chapter, we shall consider the interaction of electromagnetic waves with

matter, i.e., the phenomena of reflection, refraction, scattering and dispersion. We sha ll

see that at the boundary between two dielectrics, the electromagnetic waves obey the

familiar laws of reflection and refraction. The derivations in these phenomena will be

b a s e d o n general electromagnetic equations. In order to discuss the behaviour of

electromagnetic waves at the boundary, we shall first discuss the boundary conditions

which the electric and magnetic fields must satisfy at the surface of discontinuity

between the two media.

4.1.Boundary Conditions For The Electromagnetic Field Vectors: B, E, D, And H, (At

The Interface Between Two Media)

We shall now investigate the boundary conditions which the time dependent

electromagnetic field vectors B, E, D and H sa tisfy at the interface between two

different media. Boundary conditions are:

(1) The normal component of magnetic induction B is continuous across

boundary, i.e.,

B1n=B 2n

(2) The tangential component of E is continuous across the i nterface, ie.,

E1t=E2t.

(3) The normal component of electric displacement D is discontinuous across

the interface, i.e.,

D1n - D2n = s.

(4) The tangential component of magnetic intensity H is continuous across the surface

separating two dielectrics, i.e.,

H1t =H2t ·

(1) Boundary condition for B : The magnetic induction B satisfies the Maxwell's equation

div B = O. .. (1)

At the interface between two media, we construct a pill box like surface as.

shown in figure 4.1.



P a g e | 75

Fig.4.1. Pill box shaped surface at the interface of two media.

The surface is composed of four surfaces i.e., S1, S2 , S1 ’ and S2 ’. We apply the

divergence theorem to the divergence of B over the volume enclosed by this surface and

obtain

0 div =ò dvV

B …(2)

where dV is the infinitesimal volume element. Transforming volume integral into

surface integral by Gauss theorem, we get

ò =S

ndSB 0. …(3)

where n is an outwardly drawn unit vector normal to the infini tesimal area element dS of

the surface. Applying equation (3) to the whole surface of pill box, we have

0....'2

'121

22112211 =+++ òòòò dsnBdsnBdsnBdsnBSSSS

…(4)

The term( the third and fourth) in equation (4) give the contribution to the surface integral

from the walls of the pill box. If B is finite every where, then making the height of pill box to

approach zero. i.e, when S1 and S2 approach each other towards the interface, i.e, in the limit

0®hd , we have,

0)..( 2211 =+ò dAnBnBA

as S1 and S2 approach the area A.

Since the area A is quite arbitrary, the above equation becomes



P a g e | 76

2211 .. nBnB -= …(5)

If n12 is the unit normal vector pointing from the first medium into the second medium, then

n1=-n12 and n2=n12. Equation(5) gives

122121 .. nBnB -=-

or 122121 .. nBnB =

or nn BB 21 = …(6)

i.e., the normal component of magnetic induction is continuous across the boundary.

(2) Boundary condition for E:

The boundary condition, which the tangential component of electric field must

satisfy, may be obtained from Maxwell’s equation

t

BcurlE

¶

¶-= …(7)

Fig.4.1(a) Rectangular loop at the interface between two media.

At the interface, we construct a rectangular loop ABCD bounding a surface S as shown in

Figure 4.1(a). Integrating equation (7) over the surface bound by rectangular loop ABCD, we

have,

ò ò ¶

¶-=

S SdSn

t

BdS E.n curl …(8)

Transforming the surface integral of the left hand side into a line integral over the

path ABCD with the help of Stoke’s theorem, equation(8) yields,

ò ò ¶

¶-=

ABCD SdSn

t

BdlE 1

Or

ò òò ¶

¶-=++

AB SCDdSn

t

BdlEdlE DA and BC sides from onsContributi21 ..(9)



P a g e | 77

If the loop is now shrunk ( letting 0®hd ), the contribution from sides BC and DA will

vanish. Provided tB ¶¶ is every where finite, the right hand side of equation (9) also vanishes

because the area of integration also vanishes i.e, no flux can be enclosed. Thus when 0®hd .

ò ò =+AB CD

dlEdlE 021

or E1AB+E2CD=0

or E1AB-E2AB=0

or E1t=E2t …(10)

where E1t and E2t are the tangential components of the electric field in the two media.

Equation(10) shows that the tangential components of E are continuous across the interface.

(3) Boundary Condition of electric Displacement:D

The boundary condition for the electric displacement D can be obtained from

Maxwell’s equation

divD=r …(11)

integrating equation(11) over the pill box shaped volume V,we obtain

ò ò=V V

dVdV rdivD …(12)

Transforming the volume integral of left hand side of equation(12) in to surface integral with

the help of Gauss divergence theorem, we get,

ò ò=S V

dVdS rD.n

or òòò =++VSS

dVdSndSn r wallsfrom onsContributi.D.D21

2211 …(13)

Now let us suppose that the surfaces S1 and S2 approach each other towards the

interface so that S1 = S2=A, i.e, 0®hd the contribution from walls tends to zero. Now

instead of volume charged density r, the concept of surface charge density s must be used,

i.e,

òò ==®VV

AdAdVhLim ssrd 0 …(14)

\ AdSndSnSS

s=+ òò .D.D21

2211 …(15)



P a g e | 78

or D1.n1S1+D2.n2S2=sA

or D1.n1+D2.n2=s as S1=S2=A

or D1.n1+D2.n1=s

or D1n-D2n=s, …(16)

where D1n and D2n are components of the electric displacement in the two media normal to

the interface in the direction of n1. Evidently, the normal component of electric displacement

is not continuous across the interface and charges by an amount equal to the free surface

density of charge at the interface.

(4) Boundary Condition for H:

Finally, consider the behaviour of the tangential component of magnetic field

intensity. According to Maxwell’s equation, we have,

t

D

¶

¶+= JH curl …(17)

Now consider a small rectangular contour (as shown in figure 4.1(a)) enclosing an

area S and integrate equation(17) over this area we get,

òò ÷ø

öçè

æ

¶

¶+=

SSndS

t

DJdS .H.n curl …(18)

Transforming the surface integral into line integral with the help of Stoke’s theorem,

we have

ò ò ÷ø

öçè

æ

¶

¶+=

ABCD SndS

t

DJdlH ..

or òòò ÷ø

öçè

æ

¶

¶+=++

SCDABndS

t

DJdldl .DA and BC sides from onsContributiHH 21 (19)

If the loop is shrunk in the limit 0®h , we note that the contribution to line integral

along sides BC and DA vanishes. That is

0.DA and BC0

®ò®dlHLim

h

ò ®¶

¶®

0.0

ndSt

DLimh

i.e, t

D

¶

¶ is bound every where

and ò ^®

®S

Sh

lJdSLim J.0



P a g e | 79

where ^SJ represents the component of surface current density perpendicular to the

direction of H-component which is being matched. The idea of surface current density

is closely analogous to that of a surface charge density - it represents a finite current in

an infinitesimal layer. Then in the limit 0®h eq. (19) takes the form

lJdSdl SCDAB

^=+ òò J..H

or lJlHlH Stt ^=- 21

or ^=- Stt JHH 21 …(20)

The surface current density is zero unless the conductivity is infinite, for finite conductivity,

we write

tt HH 21 = …(21)

i.e, the tangential component is continuous across the surface separating the two dielectrics.

4.2. General Treatment of Reflection and Refraction:

Now we shall consider the reflection and refraction. Consider the case of a plane

interface between two different isotropic, homogeneous, stationary, charge free, linear and

non-conducting media of finite extensions. Let the medium of one side (medium-1) has the

permittivity ε1 and permeability m1 and the medium of other side ( medium-2) has the

permittivity ε2 and permeability m2. Consider that an electromagnetic wave is propagating in

medium-1 and is incident at a point P(figure 4.2) on the interface. This wave gives rise to

both a reflected and a transmitted wave. In the figure n i,nr and nt are unit vectors normal

to the respective wave- fronts and point in the direction of propagation. The angles qI,qr

and qt are the angles of incidence, reflection and refraction respectively. The electric field

and magnetic field intensities of the incident, reflected and transmitted waves at the point of

incidence P are described by the following equations:

( ){ }[ ]( ){ }[ ]ïî

ïíì

-=

-=

trnjHH

trnjEE

iiiioii

iiiioii

wmew

wmew

. exp

. exp

11

11

( ){ }[ ]( ){ }[ ]ïî

ïíì

+-=

+-=

'11

11

. exp

. exp

AtrnjHH

AtrnjEE

rirrorr

rirrorr

wmew

wmew

( ){ }[ ]( ){ }[ ]ïî

ïíì

+-=

+-=

'11

11

. exp

. exp

BtrnjHH

BtrnjEE

tittott

tittott

wmew

wmew

1(a)

1(b)

2(a)

2(b)

3(a)

3(b)



P a g e | 80

where r i is the position vector of P and A, A', B and B' allow for possible phase

differences with the incident wave at interface.

Fig.4.2. Showing the phenomena of reflection and refraction.

We can investigate the characteristics of both the r eflected and transmitted waves

from the fact that the tangential components of E and the tangential components of H

must both be continuous across the interface. Thus (E i + Er ) must be equal to E t . The

reflected and the transmitted waves could also be obta ined from the continuity of the

normal components of D and B across the interface,

In order to have the continuity of tangential components of E and H at the interface,

it is essential that there must be some valid relation between E i E r , and Et , for all time t

and for all vectors r i which terminate on the interface. There are two possibilities to

have such a relation, (i) if all three vectors E i E r, and Et are identical function of the

time t and of position ri and (ii) there exist certain relations betwe en E 0 i E 0r, and E0t ,

Applying the condition (i), we have

( ){ }[ ] ( ){ }[ ]

( ){ }[ ]Btrn

Atrntrn

titt

rirriiii

+-=

+-=-

wmew

wmewwmew

.

. .

22

1111 (4)

which must be true for all time t and for all vectors ri. It follows that

tri www == …(5)

This shows that all three waves have the same frequency. Again from equations(4), at any

point ri on the interface, we must also have

( ) ( ) ( ) BrnArnrn itirii +=+= ... 221111 mewmewmew …(6)

Evidently, from equation (6) we get



P a g e | 81

( )( )11

.mew

Arnn iri =- …(7)

where the term ( )11mew

A is a constant for all ri. The projection on (ni-nr) of any vector ri

terminating on the surface must be a constant. The vector (ni-nr) is also a constant for any

given incident plane wave. Then (ni-nr) must be normal to the interface, i.e, parallel to the

unit vector n normal to the interface at the point P, hence ni,nr and n are coplanar. The plane

of the three vectors is called the plane of incidence.

Since (ni-nr) is normal to the interface, the tangential components of these two vectors

must be equal and opposite sign then

ri qq = …(8)

i.e., the angle of reflection is equal to the angle of incidence. These are the laws of reflection.

Again considering equation(6) we obtain,

( ) ( ) Brnn iti =- ].[ 2211 mewmew …(9)

Thus the vector ( ) ( ) ][ 2211 ti nn mewmew - must be normal to the interface, so that

ni,nr, nt and n are coplanar. In this way all the four vectors ni,nr, nt and n are in the plane of

incidence. Moreover, the tangential components of ( ) in11mew and ( ) tn22mew must be

equal but opposite in sign i.e.,

( ) ( ) ti qmewqmew sinsin 2211 =

( )( ) 1

2

0011

0022

11

22

)(

)(

sin

sin

n

n

t

i ===meme

meme

me

me

q

q

or ti nn qq sinsin 21 = …(10)

where n1,n2 are the indices of refraction of the two media. This is Snell’s law.

The constants A and B are related to the choice of the origin. At a given point of interface,

and at a given time, the reflected and transmitted waves have definite phases and if the origin

is displaced, A and B must be adjusted accordingly. If the interface is chosen as the origin

then from Equations (7) and (9) A=B=0, since the ri will be normal to

(ni-nr) and ( ) ( ) ][ 2211 ti nn mewmew -

Taking the axes as shown, we have,



P a g e | 82

ttt

rrr

iii

kjn

kjn

kjn

qq

qq

qq

cossin

cossin

cossin

+-=

+-=

+-=

…(11)

The electric field of the incident wave at a point r =jy+kz in the plane of incidence is given by

{ }[ ]tzyjEE iiii wqqmew -+-= )cossin()(.exp 110 …(12a)

{ }[ ]tzyjEE rrrr wqqmew -+-= )cossin()(.exp 110 …(12b)

{ }[ ]tzyjEE tttt wqqmew -+-= )cossin()(.exp 110 …(12c)

4.3. Fresnel’s Equations(Dynamic Properties):

Let us find the relation between the equations iE0 rE0 and tE0 . The formulae relating

the amplitude of the reflected and transmitted waves with that of incident wave are known as

Fresnel formulae. We know that E and H vectors in a plane electromagnetic wave are always

perpendicular to the direction of propagation and to each other. The vector E does not fix the

direction uniquely and it is said that the wave is polarized in a particular direction of its

electric vector. The vector E of incident wave can be oriented in any direction perpendicular

to vector ni.

It is convenient to consider two cases: (i) in which the incident wave is polarized such

that its vector E is normal to the plane of incidence (ii) in which the vector E is parallel to the

plane of incidence.

Case I. Incident wave polarized with its vector E normal to the plane of incidence:

In this case the electric and magnetic field vectors E and H of the incident wave are

perpendicular to the direction of propagation ki, as shown in figure 4.3(a).

The pictorial diagram of this case is shown in figure 4.3(b).Since the media are

isotropic, the electric field vectors of the reflected and transmitted waves will also be

perpendicular to the plane of incidence.



P a g e | 83

Fig.4.3(a). Reflection and Refraction with Fig 4.3(b). Showing incident, reflected Polarization perpendicular to the plane of and transmitted waves when the incident incidence. wave is polarized with the E vector normal to the plane of incidence.

The continuity of the tangential component of electric field intensity at the interface requires

that

tri EEE 000 =+ …(1)

Similarly, the continuity of magnetic field intensity at the interface requires that

ttrrii HHH qqq coscoscos 000 =+ …(2)

at any given time at any given point on the interface. We know that,

11wmii

i

EkH

´= úû

ùêë

é ´==

wm

EkBandHBQ

or ( )[ ]

1

11

11 m

me

wm

iiiiii

EnEnKH

´=

´= ( )÷÷

ø

öççè

æ= 11me

wi

iKQ

Hence, ( )

ii EH 0

1

11

0m

me=

( )rr EH 0

1

11

0m

me= and

( )tt EH 0

2

22

0m

me=

Substituting these values in Equation (2), we have

( ) ( ) ( )ttirii EEE q

m

meq

m

meq

m

mecoscoscos 0

2

22

0

1

11

0

1

11 =-

or ttriii EEE 0

2

20

1

10

1

1 coscoscos qm

eq

m

eq

m

e÷÷ø

öççè

æ=÷÷

ø

öççè

æ-÷÷

ø

öççè

æ …(3)

Eliminating E0t from equations (1) and (3), we get

)(coscoscos 00

2

20

1

10

1

1ritriii EEEE +÷÷

ø

öççè

æ=÷÷

ø

öççè

æ-÷÷

ø

öççè

æq

m

eq

m

eq

m

e



P a g e | 84

or úú

û

ù

êê

ë

é

÷÷ø

öççè

æ-÷÷

ø

öççè

æ=

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ+÷÷

ø

öççè

ætiitir EE q

m

eq

m

eq

m

eq

m

ecoscoscoscos

2

2

1

10

2

2

1

10

ti

ti

Ni

r

E

E

qm

eq

m

e

qm

eq

m

e

coscos

coscos

2

2

1

1

2

2

1

1

0

0

÷÷ø

öççè

æ+÷÷

ø

öççè

æ

÷÷ø

öççè

æ-÷÷

ø

öççè

æ

=÷÷ø

öççè

æ\ …(4)

The equation gives the amplitude of electric field of reflected wave w.r.t that of incident

wave. Similarly, eliminating E0r from equations(1) and (3), we have

ttitiii EEEE 0

2

200

1

10

1

1 cos][coscos qm

eq

m

eq

m

e÷÷ø

öççè

æ=-÷÷

ø

öççè

æ-÷÷

ø

öççè

æ

ttiii EE 0

2

2

1

1

1

10 coscoscos2

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ+÷÷

ø

öççè

æ=÷÷

ø

öççè

æq

m

eq

m

eq

m

e

or

ti

i

i

t

E

E

qm

eq

m

e

qm

e

coscos

cos2

2

2

1

1

1

1

0

0

÷÷ø

öççè

æ+÷÷

ø

öççè

æ

÷÷ø

öççè

æ

=÷÷ø

öççè

æ

Therefore,

ti

i

Ni

t

E

E

qm

eq

m

e

qm

e

coscos

cos2

2

2

1

1

1

1

0

0

÷÷ø

öççè

æ+÷÷

ø

öççè

æ

÷÷ø

öççè

æ

=÷÷ø

öççè

æ …(5)

Where the index N indicates that iE0 is normal to the plane of incidence. Equations (4) and

(5) are the two Fresnel’s equations. Equation (5) gives the amplitude of electric field of

refractive wave w.r.t that of incident wave.

4.4.Brewster Angle and Degree of Polarization:

According to the Fresnel’s equation, we have

i

t

ttii

ttii

ti

ti

Pi

r

n

n

nn

nn

E

E

q

q

qqqq

qqqq

qq

qq

sin

sin as

cossinsincos

cossinsincos

cos)(cos

cos)(cos

2

1

21

21

0

0

=+

+-=

+

+-=÷÷

ø

öççè

æ



P a g e | 85

)tan(

)tan(

it

it

qq

qq

+

-= …(1)

Equation (1) shows that when ¥=+ )tan( it qq i .e., 2)( pqq =+ it , 00

0 =÷÷ø

öççè

æ

Pi

t

E

E. Thus,

when reflected and refracted rays are perpendicular to each other, there is no energy carried

by the reflected ray. The condition of continuity at the interface are then satisfied by two

waves only-the incident and transmitted waves- instead of the usual three. The angle of

incidence is then called the Brewster angle Bq . At the Brewster angle

( )

B

B

B

t

i

t

n

n

q

qp

q

q

q

q

sin

2sin

sin

sin

sin

sin

2

1 -===

Bqcot=

÷÷ø

öççè

æ=\ -

2

11cotn

nBq …(2)

If an unpolarized wave is incident on the interface at the Brewster angle, then the only

portion of the wave with electric vector perpendicular to the plane of incidence will be

reflected. Thus the reflected wave will be linearly polarized with vibrations perpendicular to

the plane of incidence and hence the Brewster angle is sometimes called as polarization

angle. For light incident on glass with refractive index 1.6, Brewster angle is about 58° and

at tq = 32°. Thus if l ight incident on glass at an angle 58°, the reflected light will be plane

polarized with vibrations perpendicular to the plane of incidence and the transmitted light is

also plane polarized with vibrations parallel to the plane of incidence as shown in fig.4.4.

Fig.4.4. Showing the reflection and transmission at Brewster angle.



P a g e | 86

At radio frequencies the refractive index of water is 9 and the Brewster angle is 84°

and tq = 6°, so water will not reflect a vertically polarized radio wave when the angle of

incidence is 84°.

It is observed that even if the light is unpolarised and is incident at angles other than

Brewster’s angle, there is a tendency of reflected wave to be predominantly polarized with

vibrations perpendicular to the plane of incidence. In such a case we define the degree of

polarization as

Degree of Polarization =PN

PN

RR

RR

-

+ …(3)

Where NR and PR are reflection coefficients for the normal and parallel components

of reflected light. These coefficients are given by,

( )( ) )(sin

)(sin

coscos

coscos2

22

21

21

ti

ti

ti

tiN

nn

nnR

qq

qq

qq

qq

+

-=ú

û

ùêë

é

+

-= ….(4)

and ( )( ) )(tan

)(tan

coscos

coscos2

22

12

21

ti

ti

i

tiP

nn

nnR

qq

qq

qq

qq

+

-=ú

û

ùêë

é

+-

+-= …(5)

4.5. Total Internal Reflection:

Suppose a plane electromagnetic wave is incident on the interface from a denser

medium to a rarer medium. The angle of incidence iq is in denser medium and angle of

refraction tq is rarer medium. Here the word internal implies that the incidence and reflected

waves are in a medium of large refractive index than the refracted wave, i.e, 21 nn > .

Now according to Snell’s law,

ti

t

i

n

n

n

nqq

q

qsinsinor

sin

sin

1

2

1

2 == …(1)

ti qq < because 12 / nn is less than one. The value of tq becomes 90° for a particular value of

angle of incidence iq . This value of angle of incidence for which iq becomes 90° is known

as critical angle and is denoted by cq . Hence, from equation(1),

1

2

1

2 90sinsin n

n

n

nc =°=q …(2)



P a g e | 87

Thus for waves incident at ci qq = , the refractive wave is propagated parallel to the

surface. Now the problem is that what will happen if the angle of incidence iq is greater than

tq . To answer it let us consider equation(1), i.e.,

)(1sin

sinsinsin

12

ci

c

iit as

nnqq

q

qqq >>== …(3)

As we know that the sine of any real angle can never be greater than one, equation(3)

suggests that tq is imaginary, i.e., tq is complex angle with a purely imaginary cosine.

( ) jbjc

itt =

ïþ

ïýü

ïî

ïíì

-÷÷ø

öççè

æ=-=

21

2

2 1sin

sinsin1cos

q

qqq …(4)

In order to find the meaning of imaginary tq , let us consider the transmitted electromagnetic

wave

( ){ }[ ]trnjEE ittt wmew -= .exp 220 …(5)

Consider that the transmitted wave is in x-z plane. Then,

tt

ttii

zx

kzjyixkirn

qq

qq

cossin

)).(cossin(.

+=

+++=

Now eq.(5) can be written as,

( ){ }[ ])cossin(exp 220 tttt zxtjEE qqmeww +--= …(6)

Using equations (3) and (4) equation (6) can be written as

( ) ( )úúû

ù

êêë

é

þýü

îíì

-÷÷ø

öççè

æ--= zjbxtjEE

c

itt 22220

sin

sinexp mew

q

qmeww

or ( )[ ] ( )úúû

ù

êêë

é

þýü

îíì

÷÷ø

öççè

æ--´=

c

itt xtjbzEE

q

qmewwmew

sin

sinexpexp 22220 …(7)

Equation (7) shows that for ci qq > , the transmitted wave is propagated only parallel to the

surface and is attenuated exponentially beyond the interface.

Energy flow in second medium:

The time-averaged normal component of Poynting vector just inside the surface gives

the energy flux carried by transmitted wave per unit area per second. Here we will show that

the lime averaged normal component of the Poynting vector just inside the surface is zero,

Hence the net energy flow through the surface into second medium is zero. Now, we have



P a g e | 88

{ }[ ]

part) real no hasit 0(as

).()(expRe. 2

1

.cos .Re. 2

1

.cos)( ofpart Real 2

1.

2

220

0

1

*

0

1

*

=

-÷÷ø

öççè

æ=

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ=

´=Ù

njbbzE

nEE

nHEnP

t

ttt

tttt

mewm

e

qm

e

q

Intensity ratio of reflected wave to incident wave:

Now let us consider about the fields with the help of Fresnel's equation using the

values of sin tq and cos tq given by equations (3) and (4) respectively. In the first case when

E is normal to the plane of incidence, we have

2222

22

2

12

i

21

21

21

0

0

2

)(

)(

)(

)(

1sin

b

cos)( where

cos)(

coscos)(

ba

abj

ba

ba

jba

jba

jba

jba

nn

nnajba

jba

osnn

nn

E

E

i

ti

ti

Ni

r

+-

+

-=

-

-´

+

-=

úú

û

ù

êê

ë

é-÷÷

ø

öççè

æ=

=+

-=

+

-=÷÷

ø

öççè

æ

q

q

qq

qq

Let us consider

),sin(cos]exp[ NNN rjrjrjba

jbafff -==

+

- …(11)

Comparing equations (10) and (11), we have

22

22

22cos and

2sin

ba

bar

ba

abr NN

+

-=

+= ff

Which shows that r=1 and 2

tan22 ba

abN

+=f

Thus from equation(11), we have

a

be

jba

jba Nj N =÷ø

öçè

æ=

+

- -

2with tan

ff …(12)

Similarly when E is parallel to the plane of incidence, we have

…(10)



P a g e | 89

cos)( where

coscos)(

coscos)(

cos)(cos

cos)(cos

12

12

12

21

21

0

0

i

ti

ti

ti

ti

Pi

r

nnajba

jba

nn

nn

nn

nn

E

E

q

qq

qq

qq

qq

=¢+¢

-¢=

+

-=

+

-=÷÷

ø

öççè

æ

or

a

b

2with tan P

0

0

¢=÷

ø

öçè

æ=÷÷

ø

öççè

æ ffPj

Pi

r eE

E

….(13)

It is clear from equation(12) and (13) that the amplitude of reflected wave is the same as that

of incident wave, i.e, the intensity of reflected wave is same as that of the incident wave,

showing thus that the wave is totally reflected. The phases of normal and parallel reflected

waves are denoted by Nf and Pf respectively, and both are the function of angle of incidence.

4.6. Reflection and Refraction at the boundary of a non-conducting and a conducting

media (i.e., metallic reflection) :

Let us consider the case of reflection and refraction at the boundary of non-

conducting and a conducting media. Here we shall modify the results of reflection and

refraction at the boundary of two non-conducting media by changing the second non-

conducting medium by a conducting medium. Let the non-conducting medium

(medium 1) has permittivity 1e and permeability 1m and the conducting medium bas

permittivity 2e permeability 2m and electric conductivity 2s . When a monochromatic

electromagnetic wave is propagating in medium 1, it is observed that there exists two more

waves (i) reflected wave propagating in medium 1 and (ii) a transmitted wave propagating in

medium 2. The electric and magnetic field intensities are described by the following

equations:

For non-conducting medium

( ){ }[ ]( ){ }[ ]ïî

ïíì

-=

-=

trnjHH

trnjEE

iiiioii

iiiioii

wmew

wmew

. exp

. exp

11

11

( ){ }[ ]( ){ }[ ]ïî

ïíì

+-=

+-=

AtrnjHH

AtrnjEE

rirrorr

rirrorr

wmew

wmew

. exp

. exp

11

11

For conducting medium

1(a)

1(b)

2(a)

2(b)



P a g e | 90

( ){ }[ ]( ){ }[ ]ïî

ïíì

+-=

+-=

BtrnjHH

BtrnjEE

tittott

tittott

wmew

wmew

. exp

. exp

11

11

where ri is the position vector of the point of incidence, ni,nr and nt are the unit vectors in the

directions of incidence, reflection and transmission. A and B are the phase difference of

reflected and transmitted waves with incident wave.

We can investigate the properties of both the reflected and transmitted waves using

the fact that the tangential components of both electric field E and magnetic field intensity H

must be continuous across the interface. Following points are observed :

tri www == …(4)

i e., all the three waves must be of the same angular frequency.

ni,nr and nt are coplanar

ri qq = …(5)

ti qqmew sink sin( 2)11 =

or 1

2

)11

2

)11

2

((sin

sin

n

n

c

ckk

t

i ===me

w

mewq

q …(6)

where c is the phase velocity of the wave in free space, n2 is the refractive index of medium

2, and n1 refractive index of medium -1. Equation (6) is known as Snell's Jaw of refraction. It

should be noted that n2 is a complex quantity and is given by

)( 222 baw

jc

n += …(7)

Where

21

21

222

2222

2 112 ú

ú

û

ù

êê

ë

é+

þýü

îíì

+÷ø

öçè

æ=

we

smewa ….(8)

and

21

21

222

2222

2 112 ú

ú

û

ù

êê

ë

é-

þýü

îíì

+÷ø

öçè

æ=

we

smewb …(9)

The values of 22 and ba are taken from the articles of propagation of E.M.W. in conducting

media. Substituting these values in equation(7) we have

…(10)

In case of good conducting medium equation (10) reduces to

3(a)

3(b)

úúú

û

ù

êêê

ë

é

úú

û

ù

êê

ë

é-

þýü

îíì

+÷ø

öçè

æ+

ïþ

ïý

ü

ïî

ïí

ì

úú

û

ù

êê

ë

é+

þýü

îíì

+÷ø

öçè

æ=

21

21

222

2222

21

21

222

2222

2 112

112 we

smew

we

smew

wj

cn



P a g e | 91

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ÷ø

öçè

æ+÷÷

ø

öççè

æ÷ø

öçè

æ=

we

smew

we

smew

w 2

222

2

2222

22j

cn

or )1(2

222 jcn +÷÷

ø

öççè

æ=

w

ms …(11)

From Snell’s law, we have

itn

nqq sinsin

2

1=

and ïþ

ïýü

ïî

ïíì

÷÷ø

öççè

æ-= it

n

nqq sin1cos

2

2

1 …(12)

since, the refractive index of a conducting medium is very large as compared to the refractive

index of a non conducting medium(n2>>n1) hence inn q2221 sin)( is negligible as compared

to unity and we obtain

1cos =tq …(13)

i.e, tqcos is real and is approximately equal to unity. Expression (13) shows that the angle of

refraction in a good conducting medium is very close to zero irrespective of the angle of

incidence. Thus in a good conducting medium the transmitted wave is propagated n the

normal direction.

The transmitted wave can be represented by

[ ] [ ][ ] [ ]î

íì

+--=

+--=

).(exp).(exp

).(exp).(exp

2

22

BtrnjrnHH

BtrnjrnEE

ititott

ititott

wab

wab

This shows that the transmitted wave is a highly damped wave.

The depth of penetration is given by

÷÷ø

öççè

æ=

ïþ

ïý

ü

ïî

ïí

ì-÷÷

ø

öççè

æ+÷÷

ø

öççè

æ=

wms

me

s

mewb

22

21

21

222

22

222

2

11211

and is independent of angle of incidence.

4.7.BASIC CONCEPTS OF WAVE GUIDES

In this section we seek the solution of Maxwell's equations in a cylindrical region of

space of arbitrary cross-section. Let a region of non-conducting space be bound by an

14(a)

14(b)

…(15)



P a g e | 92

infinitely long perfectly conducting tube of an arbitrary cross- sectional shape, the cross

sectional area and shape of the tube being uniform throughout, The bounded region is

assumed to be lossless region. Such a structure which guides the transmission of

electromagnetic energy in the direction of the axis is called a waveguide. Thus by the term

waveguide we mean a cylindrical pipe bound by a conductor of high conductivity filled with

a dielectric (of low loss) having arbitrary cross section. The two most common waveguides

are those with rectangular and circular cross sections, but in this article we shall consider a

general case.

When an electromagnetic wave is introduced in the bound region, it undergoes

multiple reflections from the wall of the tube and propagates along its axis, say, along z-axis

as shown in fig. 4.7

Fig.4.7.Cylindrical Waveguide

The electric field E and magnetic field intensity H at a point in the bound region

satisfy the following Maxwell's equations

t

HcurlE

¶

¶-= m …(1)

t

EcurlH

¶

¶= e …(2)

Div E=0 …(3)

DivH=0 …(4)

Here E and H are the functions of position coordinates a 1, a2, z and time t.

In rectangular coordinates a 1, a2, are x and y and, in cylindrical coordinate system, they

are r and q respectively, Evidently a1, and a2, are the coordinates lying in a plane normal

to z direction, i.e., in the cross section of the tube. The time dependence of E and H is

often described by tje w- and hence Maxwell's equations reduce to



P a g e | 93

t

tzaaHtzaaCurlE

¶

¶-=

),,,( ),,,( 21

21 m

or tjtj et

zaaHezaaCurlE ww m --

¶

¶-=

),,,( ),,,( 21

21

or tjtj ezaaHjezaaCurlE ww wm -- --= ),,( )(),,( 2121

or ),,( ),,( 2121 zaaHjzaaCurlE wm= …(5)

Similarly,

),,( ),,( 2121 zaaEjzaaHCurl we-= …(6)

div 0),,( Div 21 =zaaE .(7)

div 0),,(H Div 21 =zaa …(8)

Taking the curl of equation (5) we have,

)],,( curl ),,(curl[ 2121 zaaHjzaaCurlE wm=

),,( )( 21 zaaEjj wewm -= (from Eq(6))

),,( 212 zaaEmew=

),,(),,(curl curl 212

21 zaaEzaaE mew=\

),,(),,(),,( div grad 212

212

21 zaaEzaaEzaaE mew=Ñ-

or 0),,( div as ),,(),,( 21212

212 ==Ñ- zaaEzaaEzaaE mew

0),,(),,( 212

212 =+Ñ\ zaaEzaaE mew ….(9)

Similarly, 0),,(),,( 212

212 =+Ñ zaaHzaaH mew ….(10)

It is again useful to single out the spatial variation of ),,( 21 zaaE and ),,( 21 zaaH in the z-

direction. We assume that

jkz

jkz

eaaHzaaH

eaaEzaaE

),(),,(

),(),,(

2121

2121

=

=

where k is guide propagating constant. Applying this equation to equation(9) we get,

0),(),( 212

212 =+Ñ jkzjkz eaaEeaaE mew

or 0),(),( 212

212

22 =+÷÷

ø

öççè

æ

¶

¶+Ñ jkzjkz

t eaaEeaaEz

mew

or 0),()(),( 2122

212 =-+Ñ jkzjkzt eaaEkeaaE mew

or 0),(),( 212

212 =¢+Ñ aaEkaaEt …(11)

where 2tÑ is transverse component of 2Ñ and )( 222 kk -=¢ mew .



P a g e | 94

The value of 2

2

2

22

yxt

¶

¶+

¶

¶=Ñ in rectangular coordinate system and

2

2

2

2 11

q¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶=Ñ

rrr

rrt in cylindrical coordinate system.

It is useful to separate the fields ),( 21 aaE and ),( 21 aaH into components parallel to

and transverse to z-axis. Thus

),(),(),( 212121 aaEaaEaaE zt += …(12)

and ),(),(),( 212121 aaHaaHaaH zt += …(13)

where the parallel components are given by

[ ]maaEmaaE z ),(.),( 2121 = …(14)

and [ ]maaHmaaH z ),(.),( 2121 = …(15)

the transverse components are given by

[ ] maaEmaaEt ´´= ),(),( 2121 …(16)

and [ ] maaHmaaH t ´´= ),(),( 2121 …(17)

where m is unit vector in the z-direction. Manipulation of the curl equations (5) and (6) and

the use of the z-dependence jkze lead to the determination of the transverse component in

terms of the axial components, thus

[ ]),(),()(

1),( 21212221 aaEmjaaHjk

kaaH ztztt Ñ´+Ñ

-= we

mew …(18)

and [ ]),(),()(

1),( 21212221 aaHmjaaEjk

kaaE ztztt Ñ´-Ñ

-= we

mew …(19)

where tÑ is the transverse gradient operator. It is ÷÷ø

öççè

æ

¶

¶+

¶

¶

yj

xi in rectangular coordinate

system. Equations (18) and (19) show that transverse components can be calculated from the

axial components. Therefore, it is sufficient to determine the axial components

),( 21 aaE z and ),( 21 aaH z as approximate solutions of equations

0),(),( 212

212 =¢+Ñ aaEkaaE zzt …(A)

and 0),(),( 212

212 =¢+Ñ aaHkaaH zzt …(B)

At the perfectly conducting bounding surface of interest, the field ),,,( 21 tzaaE and

),,,( 21 tzaaH satisfy the boundary conditions:

(i) Tangential component is continuous i.e.,



P a g e | 95

0),,,( 21 =´s

tzaaEn …(20)

(ii) Normal Component is continuous i.e.,

0),,,(. 21 =s

tzaaHn …(21)

Where n is a unit vector normal to the surface. The boundary condition (20) is equivalent to

0),( 21 =sz aaE …(C)

For normal component of ),,,( 21 tzaaH we have,

0),,(. 21 =-

s

tjezaaHn w

or 0),(. 21 =-

s

tjjkzeeaaHn w

or 0),(. 21 =s

aaHn

or 0),(),(. 2121 =+szt aaHaaHn

or 0),(.),(. 2121 =+szst aaHnaaHn

or [ ] 0),(),()(

1. 212122

=Ñ´++Ñ- sztszt aaEmjaaHjk

kn we

mew

because 0),(. 21 =sz aaHn

or 0),(. 21 =Ñszt aaHn

or 0),(. 21 =¶

¶

s

z aaHn

…(D)

where n¶¶ is the normal derivative at a point on the surface.

Because the boundary conditions ),( 21 aaE z and ),( 21 aaH z are different, they can not be

satisfied simultaneously. Consequently, the electromagnetic fields may be divided into

distinct categories.

(i) Transverse electric (TE) Mode:

From the boundary conditions it is clear that either ),( 21 aaE z is present

or ),( 21 aaH z or both. When ),( 21 aaH z is present and ),( 21 aaE z =0 everywhere, such waves

are termed as transverse electric waves(TE Mode). The boundary conditions as

0),( 21 =¶

¶aaH

nz …(22)

The transverse components are given by

),()(

),( 212221 aaHk

jkaaH ztt

®

Ñ-

=mew

…(23)



P a g e | 96

and ),()(

),( 212221 aaHmk

jaaE ztt

®

Ñ´-

=mew

we (24)

(ii) Transverse magnetic TM Mode:

In this case, ),( 21 aaE z is present and ),( 21 aaH z =0 i.e., there is component of E in the

direction of propagation but no component of H. the boundary condition is

0),( 21 =- aaE z …(25)

The transverse components are given by,

),()(

),( 212221 aaEmk

jaaH ztt

®

Ñ´-

=mew

we …(26)

and ),()(

),( 212221 aaEk

jkaaE ztt

®

Ñ-

=mew

…(27)

(iii) Transverse electromagnetic (TEM) Mode:

There is a degenerate mode called the transverse electromagnetic (TEM) mode in

which both ),( 21 aaE z is present and ),( 21 aaH z vanish. We shall find that in case of hollow

conducting pipes either TE or TM waves can propagate but TEM waves do not propagate.

In case of waveguide the only problem is to solve the equations A and B for different

co-ordinate systems by applying the boundary conditions C and D. Now we shall consider

rectangular wave guide.

4.8.Rectangular Waveguide:

When the cross section of the metal tube is rectangular, it is called a rectangular

waveguide. The rectangular wave guide is shown in figure 4.8. It has cross sectional lengths a

and b, and the z-axis parallel to the axis of the tube. The conductivity of the metal is assumed

infinite, and the interior filled with a loss free dielectric.

Fig.4.8.Rectangular waveguide. (a). TE wave in rectangular waveguide:

In this case

0),( =yxE z everywhere …(1)



P a g e | 97

),( yxH z satisfies the wave equation

( )22222 k where0),(),( kyxHkyxH zzt -=¢=¢+Ñ mew

0),(),( 2

2

2

2

2

=¢+÷÷ø

öççè

æ

¶

¶+

¶

¶yxHkyxH

yxzz …(2)

The boundary conditions for ),( yxH z are

úúúú

û

ù

êêêê

ë

é

===¶

¶

===¶

¶

bx

yxHand

axxx

yxH

z

z

y and 0yat 0),(

and 0at 0),(

…(3)

Now equation (2) is

0),(),(),( 2

2

2

2

2

=¢+¶

¶+

¶

¶yxHk

y

yxH

x

yxHz

zz …(4)

Let XYyYxXyxH z == )()(),( …(5)

Where X is a function of x and Y is a function of y,

Substituting equation(5) into equation(4) we have,

02

2

2

2

2

=¢++ XYkdy

YdX

dx

XdY

or 011 2

2

2

2

2

=¢++ kdy

Yd

Ydx

Xd

X

or 2

22

2

2 11

dy

Yd

Yk

dx

Xd

X-=¢+ …(6)

The left hand side is a function of x while right hand side is a function of y and both

are equal to each other. This is only possible when they are separately equal to a constant.

Hence,

2

2

22

2

2 11A

dy

Yd

Yk

dx

Xd

X=-=¢+ …(7)

Thus we should solve two equations

(i) 22

2

21Ak

dx

Xd

X=¢+

or 0)(1 22

2

2

=-¢+ Akdx

Xd

X

or 02

2

2

=+ XBdx

Xd …(8)



P a g e | 98

where )( 222 AkB -¢=

The solution of equation (8) is

BxCBxCX sincos 21 += …(9)

Where C1 and C2 are two constants.

(ii) 2

2

21A

dy

Yd

Y=-

or 02

2

2

=+ YAdy

Yd …(10)

The solution of equation(10) is given by

AyCAyCY sincos 43 += …(11)

Where C3 and C4 are two constants.

Now we shall apply the boundary conditions (3) to eqs (9) and (11). Conditions(3) are

equivalent to

by

Yaxx

x

X===

¶

¶===

¶

¶y and 0yat 0 and and 0at 0

Differentiating eq.(9) w.r.to x we get,

BxBCBxBCx

Xcossin 21 +-=

¶

¶

When x=0, C2B=0. This gives C2=0

When x=a, -C1BsinBa+0=0 i.e., C1BsinBa =0

( )alBlBaC pp ===\¹ or or 0sinBa 01

where l =0,1,2,3…

xa

lCX

pcos1= …(12a)

Again differentiating eq.(11) w.r.to y, we get,

AyACAyACy

Ycossin 43 +-=

¶

¶

Where y=0, C4A=0. This gives C4=0

When y=b, -C3AsinAb+0=0 i.e., C3AsinAb =0

( )blAlC pp ¢=¢==\¹ or Abor 0sinAb 03

where l ¢=0,1,2,3…

yb

lCY

p¢= cos3 …(12b)



P a g e | 99

Finally, the solution becomes

b

yl

a

xlCCyxH z

pp ¢= coscos),( 31

or b

yl

a

xlHyxH z

pp ¢= coscos),( 0 …(13)

hence, ll ¢, are two parameters and a set of given values ll ¢, defines a waveguide mode which

we call ll ¢, waveguide mode for TE wave designated as llTE ¢, mode.

Cut off frequency and Wavelength:

úû

ùêë

é ¢+=÷

ø

öçè

æ ¢+÷

ø

öçè

æ=+=¢

2

2

2

22

22

222

b

l

a

l

b

l

a

lABk p

pp …(14)

where the indices ll ¢ and specify the mode.

The cut off wavelength is given by

÷ø

öçè

æ=ú

û

ùêë

é ¢+=÷÷

ø

öççè

æ

¢l

p

l

2

2

11 21

2

2

2

2

kb

l

a

l

llc

Q …(15)

While cut off frequency will be

( ) ÷ø

öçè

æ=ú

û

ùêë

é ¢+=¢

l

pwpw

c

b

l

a

lcll

2

2

2

2

2

Q …(16)

The modes corresponding to ll ¢ and are designated as llTE ¢, mode. The case 0 =¢= ll gives a

static field which do not represent a wave propagation. So TE00 mode does not exist. If a<b,

the lowest cutoff frequency results for 1 and 0 =¢= ll i.e.,

( )b

kor 01

ppw =¢=

b

c ….(17)

The TE01 mode is called the principal or dominant mode.

Phase Velocity and Group velocity:

The phase velocity pv is given by

0

k

kc

kv p ==

w ÷

ø

öçè

æ=

ck

w0 Q

( ) ( )[ ]

1

c22

0222

0

0

kkkk

ck

¢-=

¢-= ( )22

02 kkk ¢-=Q

( )[ ]

1

2

0 c

p

cv

ll-=\ ÷

ø

öçè

æ=

l

p2 kQ ….(a)



P a g e | 100

where ¥== pc v,0 ll , i.e, phase velocity becomes infinite at cut off. The group velocity

with which energy is propagated along the axis of the guide will be given by

( )

( )

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ ¢-=

=¢+=

úûù

êëé ¢+

¶

¶=

¶

¶=

-

2

0

2

0

21

22

21

22

1

2 2

1

k

kc

ckkkkc

kkckk

vg

m

w

1

2

0

úú

û

ù

êê

ë

é

÷÷ø

öççè

æ-=\

c

g cvl

l …(b)

Transverse components of Electric field and Magnetic field intensity

The transverse components of electric field and magnetic field intensity may now be

calculated as follows:

coscos)(

),()(

),(

022

22

b

yl

a

xlH

yj

xim

k

j

yxHmk

jyxE ztt

pp

mew

wm

mew

wm

¢÷÷ø

öççè

æ

¶

¶+

¶

¶´

-

-=

Ñ´-

-=

®

úû

ùêë

é ¢+

¢¢

-

-=

¢÷÷ø

öççè

æ

¶

¶-

¶

¶

-

-=

b

xl

a

xl

a

lj

b

yl

a

xl

b

li

k

Hj

b

yl

a

xlH

yi

xj

k

j

pppppp

mew

wm

pp

mew

wm

cossinsincos)(

coscos)(

22

0

022

…(18)

From this equation, the components ),(E and ),( yxyxE yx can be calculated.

Similarly,

coscos)(

),( 022 b

yl

a

xlH

yj

xi

k

jkyxH t

pp

mew

¢÷÷ø

öççè

æ

¶

¶+

¶

¶

-=

úû

ùêë

é ¢÷ø

öçè

æ ¢+

¢÷ø

öçè

æ-

-=

b

yl

a

xl

b

lj

b

yl

a

xl

a

li

k

jkH pppppp

mewsincoscossin

)( 22

0 …(19)

with the help of equation(19) the components Hx ),(H and ),( yxyxH yz can be calculated.

(b) TM wave in rectangular wave guide:

In this case, 0),( =yxH z … (20)

),( yxE z satisfies the equation



P a g e | 101

0),(),( 2

2

2

2

2

=¢+÷÷ø

öççè

æ

¶

¶+

¶

¶yxEkyxE

yxzz …(21)

The boundary conditions for ),( yxE z are

úû

ùêë

é

===

===

byxE

axxyxE

z

z

y and 0yat 0),(

and 0or 0),( …(22)

This case can also be solved as the case of TE wave, hence we have,

b

yl

a

xlEyxE z

pp ¢= sinsin),( 0 …(23)

with úû

ùêë

é ¢+=¢

2

2

2

222

b

l

a

lk p

Cut off Wavelength: The cut off wavelength is given by

÷ø

öçè

æ=ú

û

ùêë

é ¢+=÷÷

ø

öççè

æ

¢l

p

l

2

2

11 21

2

2

2

2

kb

l

a

l

llc

Q …(24)

Cut off frequency: The cut off frequency is given by

( ) ÷ø

öçè

æ=ú

û

ùêë

é ¢+=¢

l

ppw

ck

b

l

a

lcll

2

2

2

2

2

Q …(25)

This implies that TE and TM modes of rectangular guide have the same set of cut off

frequencies. However, in this case 1or 0 and 1 =¢=¢= lll represents non-trivial solution since

this gives 0 ),( =yxE z and hence all the components of E and H will be zero; of course, this

mode i.e., 0 and 1 =¢= ll was dominant in TE mode. In TM mode the lowest mode has

1 =¢= ll and may be represented by TM11. The cut off frequency of lowest mode is given by

1a

11 2

1

2

221

2211 úû

ùêë

é+=úû

ùêë

é+=

b

ac

bac

ppw …(26)

Since a>b, the cut off frequency of lowest TM mode is greater than that of the lowest TE

mode by a factor ( ) 21221 ba+ .

4.9 Let us sum up



P a g e | 102

From this unit you have understood about boundary conditions at interfaces and

Reflection and Refraction at the surfaces which leads to Fresnel’s laws. Also we described

about the total internal reflection and critical angle. Finally you have learnt about the wave

guides with the particular reference to rectangular wave guide.

4.10 Lessons-end activities

(1) Write Fresnel’s equations

(2) What is meant by Brewster’s angle

(3) Define degree of Polarization

(4) Define reflection coefficient and Transmission coefficient between two non-

conducting media.


(i) Prove that for glass-air interface (n1= 1.5 and n2= 1.0 ) for no0rmal incidence

the reflection and Transmission coefficient are 0.04 and 0.96 respectively.

[Hint : R = 2

21

221

)(

)(

nn

nn

+

-; T=

221

21

)(

4

nn

nn

+]

(ii) A hollow rectangular waveguide has a= 6 cm and b= 4cm. The frequency of

the impressed signal is 3GHz. Compute for TE10 mode (a) cut-off

wavelength (b) guide wavelength.


1. Explain total internal reflection and critical angle

2. Derive an equation for rectangular wave guide

4.13 Sources/References

1. Electrodynamics by Gupta, Kumar, Singh

2. Electromagnetic fields and waves by R.N.Singh



P a g e | 103

UNIT – III

Lesson – 5

Scattering Electromagnetic wave


This unit deals with different types of scattering namely Thomson and

Rayleigh scattering and its importance in the interaction of electromagnetic

waves with matter. Also this unit deals with Normal and Anomalous

dispersion and its application to solids, liquids and gases.

5.1 Scattering and Scattering Parameters

If an electromagnetic wave is incident on a system of charged

particles, the electric and magnetic components of the wave will exert

Lorentz force on the charges and they will be set into motion. Since the

electromagnetic wave is periodic in time, so will be the motion of the

particles. Thus, there will be changes in the directions of motion and hence

there will be accelerations. The system will therefore radiate; that is energy

will be absorbed from the incident wave by the particles and will be re-

emitted into space in all directions. We describe such a process as

scattering of the electromagnetic wave by the system of charged particles. If

the energies of the incident and scattered radiations are equal the scattering

is called elastic otherwise inelastic.

Incident beam Electron Unpolarized Nucleus Scattered Radiation Fig. 5.1



P a g e | 104

Scattering is most conveniently characterized by following parameters

–

(A) Differential Scattering Cross-section: It is defined as the ratio of

the amount of energy scattered by the system per unit time per unit angle to

the energy flux density or intensity (i.e. energy per unit area per unit time in

a normal direction) of the incident radiations.

So if a solid angle d is subtended at the system of an area ds.

(fig 5.2 a) The mean power (i.e. energy per unit time) scattered by the system

will be given by

dPsr = Ssrds

where Ssr is the intensity of Scattered radiation. The mean energy scattered

per unit time per unit solid angle will therefore be

dPsr / d = Ssrds / d

or dPsr / d = Ssrr2 (as d = ds/ r2 ) -------(1)

If the incident energy flux density i.e. intensity (Poynting vector) is SIr, the

differential Scattering cross-section will be

d / d = dPsr/ d / SIr -------(A)

d / d = Ssrr2/ SIr

(Substituting the value dPsr/ d from Eq. 1)

From the above it is clear that the differential scattering cross section has

the dimensions of the area.

(B) Total Scattering Cross-section: We have defined the differential

scattering cross-section as

d /d = Ssrr2/ SIr

so d = Ssrr2 d / SIr

Therefore = (Ssr/ SIr) . ds (as d = ds/r2)

i.e. = Psr/ SIr [ as Psr = (Ssrds) ] --------(B)

is called the total scattering cross-section and is defined as the ratio

of the power scattered (total energy scattered per sec) to the intensity (energy

per unit are per unit time) of the incident radiations.

5.2 Scattering by a free electron (Thomson-Scattering)



P a g e | 105

Let there be an electron of mass m and change q in the path of a

plane polarized monochromatic wave in vaccum. Both the electric and

magnetic vectors E and B will exert a force on q given by the Lorentz formula

F = q [E + v xB] --------(1)

Where v is the velocity of the particle produced by the wave, and assumed

throughout this treatment to be non-relativistic so that v << c. Because in

a plane wave

i.e B = n x E /c

so expression (1) becomes

F = q[E+(v/c) x (nxE) ]

F = qE (as v /c << 1 ) ---------(2)

Thus only action of the electric field on the charge need to be

considered. And so the equation of motion

F = ma = m (d2r / dt2 ) --------(3)

Substituting for F in (2) becomes m (d2r / dt2 ) = qE. -------(4)

And because in a plane wave

E = E0 e- i(ωt – k.r)

Equation (4) becomes

(d2r / dt2 ) = (q E0/m) e- i(ωt – k.r) ---------(5)

Equation (5) implies that the acceleration, velocity and displacement of the

particle are all in the same direction as E0 which itself is constant and that

the charge is oscillating sinusoidally.

Now if the incident electromagnetic wave in which the electric vector is

along x –axis (as it is plane polarised) is moving along z-axis, then the

acceleration in the x – direction will be given by

(d2x/dt2) = (q E0/mω2) e- i(ωt – k.z) --------(6)

So that the displacement x at time t will be given by

x = - (qE0 /mω2) e- i(ωt – k.z) --------(7)

Now as an oscillating charge behaves like an oscillating dipole with dipole

moment

p = qx

It follows from equation (7) that



P a g e | 106

p = (q2 E0 /m ω2) e- i(ωt – k.z)

or p0 = (q2 E0 /m ω2) -------(8)

But as the average energy radiated per sec. per unit area in a normal

direction by an oscillating dipole is given by

Ssr = (1/4 0) (ω4p02/ c 3r2) sin2θ

or Ssr = (1/4 0) (ω4/8 c 3r2) [q2 E0 /m ω2]² sin2 θ

(putting the value of p0 from eqn. 8)

i.e. Ssr = (1/4 0) (q4 E02/8 m2c3r2) sin2 θ --------(9)

Further as for a plane wave

SIr = E x H

So the average value of SIr will be given by

SIr = (½ E0H0) (as E is to H )

i.e SIr = (1/2 0cE02 ) (as H0 c 0 E0) ------(10)

so the differential scattering cross – section

(d /d ) = (Ssr r2/ SIr)

= (1/4 0) (q4 E02 sin2 θ/8 m2c3) x (2 / 0cE02)

= (q2/4 0mc2)2 sin2 θ

or (d /d ) = r02 sin2 θ = r02 cos2φ ------(A)

Where r0 = (q2/4 0mc2) and is called the classical radius of the electron.

The incident radiation here has been taken to be plane polarised.

For unpolarised or randomly polarised waves an average must be taken over

all x

ds

+ x θ d Y p

φ E0 o - x

(a) (b)

Fig. 5.2

orientations of the plane of E. Suppose in Fig. 5.2(b) AB is the direction of

E in another wave incident on the particle of the fig. (a), the angle between

AB and the plane of fig. (a), containing field point is ψ. It is now preferable

to express the scattering in terms of the angle φ which is common to all

B θ O A



P a g e | 107

azimuths. In fig. (b) the plane POB is drawn perpendicular to the plane

containing AB so that the length BQ is given both by r cos θ and by r sin φ

cos ψ.

Hence

cos θ = sin φ cos ψ .

i.e. cos2 θ = sin2φ cos2 ψ

i.e sin2 θ = 1 - sin2φ cos2 ψ

i.e sin2 θ = 1 - cos2 ψ (1 - cos2 φ) ------------------(11)

[For a plane polarised light as ψ = 0 )

sin2 θ = 1 – (1- cos2 φ) = cos2 φ

which is also evident from fig. 5.2a in which θ = (( / 2) –φ).

2 2

(cos2 ψ ) av = (1/2 ) 0 ( cos2 ψ dψ) = (1/2 ) 1/2 0 (1+ cos 2ψ ) dψ 2

= (1/2 ) 1/2 ψ + (sin2ψ/2) = 1/2 ] 0

Averaging equation (11) over all ψ, we get

sin2 θ = 1 – ½ (1- cos2 φ ) [as (cos2 ψ )av = ½]

i.e sin2 θ = ½ (1+ cos2 φ ) --------(12)

Substituting the value of sin2 θ from, equation (12) in (A), we get

d /d = r02 ½ (1+ cos2 φ) ------------------(B)

This is called Thomson formula for scattering of radiation and is

appropriate for the scattering of X – rays by electrons or γ (gamma) – rays by

protons. In it angle φ is called the scattering angle and the factor

½ (1+ cos2 φ) is called degree of dipolarisation. From expression (B) it is

clear that:

(i) Scattering of electromagnetic waves is independent of the nature

of incident wave, (i.e.ω).

(ii) Scattering occurs in all directions and is maximum when φ = 0

or (i.e. in the forward and backward directions) while minimum when

(φ = ) or (3 /2) (i.e. in the side way directions).

(iii) Scattering depends on the nature of the charged particle i.e.

scatterer and is symmetrical about the line given by (φ = /2).



P a g e | 108

A plot of dirrerential scattering cross-section as a function of

scattering angle φ for plane polarized and unpolarised incident radiations is

shown below.

1

(1/r02) d /d .5

0 /2

The total scattering cross-section will be

= (d /d ) d

[If the incident radiation plane polarised then

/2

= r02 sin2 θ d - /2 r02 sin2 θ 2 cos θ dθ

/2 /2

i.e = (4 r02) 0 (sin2 θ cos θ dθ) = (4 r02) [ (sin3θ/3)]0 = ((4/3) x r02 )

or = 0 (r02 cos2φ d ) (as (sin2 θ = cos2φ )

i.e = 0 (cos2φ 2 sinφ dφ ) = 2 r02 0 (cos2φ sinφ dφ )

= 2 r02[( - cos3φ/3)]0 = (4/3 x r02) ]

= r 02 ½ (1+ cos2φ ) d

= r02 0 ½ (1+ cos2φ ) 2 sinφ dφ ( as = 2 (1-cosφ)

= r02 0 (sinφ + cos2φ sinφ ) dφ

= r02 [ (- cosφ) - (cos3φ/3) ]0

i.e = (8 /3) x r02 -------------------------(C)

Result (C) was first derived by Thomson and so after his name it is

called Thomson scattering cross-section.

Unpolarised

Plane polarised



P a g e | 109

A quantum mechanical calculation carried out by Klein and Nishina

shows that deviations from Thomson result become significant for incident

photon energy hv which is comparable with or larger than the rest energies

of the scattering electron mc2. According to them

KN = r02 { (8 /3) (1- 2hv/mc2 + ..........)} for hv<< mc2

and KN = r02{ ( mc 2/hv [loge (2hv/ mc2 + ½ ) ]} for hv >> mc2

These results are shown graphically in figure given below

1

Hv/

1/r02 d /d .5

0

/2

From these curves it is clear that:

(i) The scattering depends on the nature of the incident radiations.

(ii) Quantum mechanical result approaches the classical one on the long

wavelength side as the frequency υ = ω/2 goes to zero.

(iii) The scattering is not symmetrical and radiations more concentrated in

the forward direction i.e. φ = 0).

Apart from these there is another feature of Thomson scattering which

is modified by quantum considerations. Classically the scattered radiations

has the same frequency as the incoming waves but quantum mechanical

calculations shows that the frequency of the scattered radiation is lesser

than that of incoming wave and depends on the angle of scattering. The

relation between the wavelength of the scattered radiation at an angle φ and

the incident radiation is

λs = λi + h/mc(1-cosφ)

hv/mc2 = 0 hv/mc2 = 0.2 hv/mc2 = 1



P a g e | 110

5.3 Scattering by a Bound Electron (Rayleigh Scattering).

Considering now the same charge as in the previous section but

elastically bound so that a restoring force – mω02x is brought into play when

it is displaced. We also assume that there is a small amount of damping

proportional to dx/dt which may be produced in particle by Collisions or

radiation. The equation of motion now becomes

md2x/dt2 = qE – mγ dx/dt - mω02x

where γ is the damping constant per unit mass. Thus

d2x/dt2 + γ (dx/dt)ω02x = (q/m) x E0 e-i(ωt – kz) -----------(1)

The solution of this differential equation consists of two parts-

(a) The complementary function: It is obtained by solving equation

d2x/dt2 + γ (dx/dt) + ω02x = 0 ---------------(2)

Let the solution of equation (2) be

x = A t -----------------(3)

so that dx/dt = A e t

and d2x/dt2 = A 2 e t

Substituting the value of x, dx/dt and d2x/dt2 from above in equation

(2) we get

2 + γ + ω02 = 0

i.e. = γ/2 ± (γ2/4) – ω02 )

or = γ/2 ± i (ω02 - γ2/4) (as ω0 >> γ)

So that = A1 e[ - (γ/2) + i (ω02 – (γ 2/4)] t + A2 [ - (γ/2) - i (ω02 – γ 2/4)] t

i.e x = e ( - (γ/2) [ A1eiβt + A2e-iβt ] ------(a)

with β = (ω02 - γ2/4). The constant A1 and A2 can be determined by

applying initial conditions.

(b) The perpendicular integral : It is obtained by solving

(d2x/dt2) + γ (dx/dt) + ω02 x = (q/m) E0e-i(ωt – kz) ---------(4)

Let the solution of equation (4) be

x = Be-i (ωt – kz) --------------------(5)

dx/dt = -iω Be-i (ωt – kz)

d2x/dt2 = – ω2 Be-i (ωt – kz)

Substituting these values of x,dx/dt and d2x/dt2 in equation (4) we get



P a g e | 111

- ω2 B + γ ( - 2ωB) + ω02B = (q/m) E0

or B = qE0/m[ω02 - ω2 – iγω]

so that x = q E0 / m [ω02 - ω2 – iγω] x e-i (ωt – kz) -----------------(a)

= q E0/[(ω02 - ω2) + iγω] /m [ω02 - ω2)2 + γ2 ω2] x e-i (ωt – kz)

(q E0 e-i (ωt – kz) )/ (m [ω02 - ω2)2 + γ2 ω2]1/2 ) { (ω02 - ω2) /[(ω02 - ω2)2 + γ2 ω2]1/2 )

+ i (γω) / [(ω02 - ω2)2 + γ2 ω2]1/2 }

= (q E0 e-i (ωt – kz) ) / [(ω02 - ω2)2 + γ2 ω2]1/2 ] x [ cos δ + isin δ ]

x = (q E0 e-i (ωt – kz) ) / (m [ω02 - ω2) + γ2 ω2]1/2 ) ----------(b)

with δ = tan-1 γω/ (ω02 - ω2)

So from expressions (a) and (b) we conclude that the general solution

of equation (1) will be

x = e-(γ/2) t [ A1eiβt + A2eiβt ] + q E0 / m(ω02 - ω2)2 + γ2 ω2]1/2 ] e-i (ωt – kz)

-------------(c)

In this solution first term on the R.HS represents the free damped

vibrations of the charge. These vibrations die out soon on account of the

factor e-(γ/2) t and hence the frist term can be neglected in considering the

final motion. So equation (c) reduces to

x = q E0 / m(ω02 - ω2)2 + γ2 ω2]1/2 ] e-i (ωt – kz) -----(d)

Now as an oscillating charge is equivalent to an induced electric dipole of

moment

p = qx

It follows from equation (d)

p= q2 E0 / m(ω02 - ω2)2 + γ2 ω2]1/2 ] x e-i (ωt – kz

or p0 = q2 E0 / m(ω02 - ω2)2 + γ2 ω2]1/2] -----------(6)

But as average energy radiated per sec per unit area in a normal

direction by an oscillating dipole is given by

S = (1/4 0) (ω4 p02/8 c 3r2) x sin2 θ

So for the present situation

Ssr = (1/4 0) (ω4 p02/8 c3r2) q4 E02 sin2 θ / m2 [(ω02 - ω2)2 + γ2 ω2]

and as SIr ½ 0c E02

dσ/d = (Ssrr2 / SIr)

= (q2/(4 0mc2))2 (ω4 sin2 θ)/(ω02 - ω2)2 + γ2 ω2]

i.e. dσ/d = r02 (ω4 sin2 θ) / [(ω02 - ω2)2 + γ2 ω2]-------(A)



P a g e | 112

Now if is the angle of scattering and the incident radiations are

unpolarised then as in previous section

sin2θ = 1/2 ( 1 + cos2φ ) -----------(7)

So equation (A) in the light of (7) becomes

dσ /d = ½ ( 1 + cos2φ )r02ω4 / [(ω02 - ω2)2 + γ2 ω2] ------(B)

This is the required result. From this, it is clear that –

(i) Scattering depends on the nature of the incident radiations – i.e. ω

(ii) Scattering depends on the angle of scattering φ

(iii) Scattering depends on the nature of the scatterer i.e. ω0 and γ

The total scattering cross-section will be

σ = ( dσ /d ) d

= ½ ( 1 + cos 2φ )r02ω4 / [(ω02 - ω2)2 + γ2 ω2] d

= r02ω4/[(ω02 - ω2)2 + γ2 ω2] 0 ½ ( 1 + cos2φ) 2 sin φ dφ

σ = (8 /3) (r02ω4) / [[(ω02 - ω2)2 + γ2 ω2] --------------(C)

The expression (C) gives the total scattering cross-section for an

elastically bound electron. From this, it is evident that total scattering

cross-section is a function of the frequency of incident radiations. Fig. 5.3 is

a plot of σ against ω. From the curve of fig. 5.3 or from equation C it is

clear that

8 /3 r02(ω0/γ )2 Resonance

σ

Thomson

8 /3 r02

ω4 Rayleigh

0 ω0 ω

Fig. 5.3

(i) If ω >>ω0 then σ › (8 /3) r02 as γ>0



P a g e | 113

i.e σ › σT

i.e Thomson Scattering occurs and the charge behaves as if it were free.

This is likely to occur at the X-ray end of the electromagnetic spectrum,

(ii) if ω ~ ω0 then σ = (8 /3) r02 (ω0/ γ)2

which is generally very large compared to Thomson scattering cross-section.

The scattering is known as resonance scattering. One example of resonance

scattering occurs when sodium vapour is illuminated with the characteristic

yellow sodium radiation. The whole volume of the vapour then becomes

luminous because it is strongly scattering radiations of its natural

frequency.

(iii) if ω<< ω0 then σ › (8 r02/3) (ω/ ω0)4

i.e σR › K/λ4 (With K = (8 02/3) λ04)

i.e the amount of scattered light is proportional to 1/ λ4 where λ is the

wavelength of the incident radiation. This scattering is known as Rayleigh

scattering. This is likely to occur when ω corresponds to the frequencies of

visible light and ω0 to that of ultraviolet.

In case of Rayleigh scattering shorter the wave length the more

strongly is the incident light scattered. This is the result that is used to

account for the blue colour of sky; the light entering the eye-has been

strongly scattered away from its direct path from the sun by air molecules

and since shortest visible wavelength corresponds to blue and since shortest

visible wavelength corresponds to blue light-Similarly, the red colour of a

sun-set or sunrise arises because the light coming to the eye through a

thick layer of air has had virtually all the blue light scattered out of this

direct path leaving only the long wavelength (red) component of the light to

be seen. Further as red light has longest wavelength in the visible region it

is scattered least. So it can travel longer distances in atmosphere and can

be seen from larger distances in comparison to other wavelengths of visible

region. This is why red light is used for danger signals while the eye is most

sensitive to yellow green.



P a g e | 114

5.4 Coherence and incoherence in scattered light.

Scattering from real materials does not always exhibit the simple

features because the total mean intensity is the result of adding

contributions from large number of particles. The resultant intensity of a

number of waves is always found by adding the amplitudes taking due

account of the phases, and squaring the resultant. If the phase difference

between any two waves at a point in space is constant for all time, they are

said to be coherent, while if the phase difference varies in a completely

random way they are said to be incoherent. In practice a state somewhere

between the two extremes is common and the waves are then said to be

partially coherent

As we are concerned more with waves from different radiating

particles, all being excited by the same incident radiation, we might except

some degree of coherence. In general the resultant intensity of a set of

incoherent waves at any waves at any point is obtained by adding the

intensities. Thus if A1,A2 etc., are the amplitudes of the waves for a set of

sources then intensity I is

I = (A1 + A2+A3 + ……)2 …..(a1) for coherent sources

and I = A12 + A22 + A32 + ….(b1) for incoherent sources

The expressions (a1) and (b1) for N identical sources each of amplitude A

reduce to

I = N2A2 -------(a2) for coherent sources

and I = N A2 -------(b2) for incoherent sources

Let us now apply these ideas of coherence to the problem of scattering

by gas molecules. The phase of the scattered waves at the point P shown in

fig 5.4 arising from a molecule located at 1 will depend on the distance the

wave has travelled from the sources to 1 and from 1 of P. Since the

molecules of a gas have no permanent locations with respect to each other

their path lengths will be randomly related, their phases will be incoherent,

and we shall be justified in adding intensities as stated by equation (b1) or

(b2)



P a g e | 115

This procedure would not be permissible in discussing the scattering

of X-rays in crystals, where the various atoms have fixed positions in the

lattice so one observes coherent scattering (i.e equation a1or a2 apply)

P

2

1 p

2 3

1

(a) 3 (b)

Fig. 5.4

The use of equations (b1) and (b2) would also not be valid if we were looking

at the radiations which are scattered from a gas and are observed in a

direction close to the incident direction as shown in fig 5.4(b) If we are far

enough in front, the path lengths are all about the same, the phases are

approximately equal and we get coherent scattering (i.e. interference) rather

than incoherent scattering.

5.5 Normal and Anomalous, dispersion

If in a medium the index of refraction varies with frequency (i.e

wavelength) then the medium is said to be dispersive, the phenomenon itself

is called dispersion and the rate of change of refractive index with

wavelength i.e. dn/dλ is known as dispersive power.

Generally the variation of n is such that

(i) The index of refraction increases as the frequency increases.

(ii) The rate of increase dn/dω i.e. the slope of the n – ω curve is

greater at high frequencies.

However it is also found that over small frequency range there is often

decrease of index of refraction with-the increase in frequency.

104

N A N A

N ω

n = 1



P a g e | 116

n = 96

A – Regions in which the dispersion is anomalous N – Regions in which the dispersion is normal

Fig. 5.5

In these narrow spectral regions due to its abnormal behaviours the

dispersion is called anomalous. Normal and anomalous dispersion are

shown graphically in fig. 5.5.

5.6 Dispersion in Gases (Lorentz Theory)

In order to investigate the frequency dependence of refractive index n

or dielectric constant r and to discuss dispersive effects, Lorentz assumed

that in case of gases –

(i) There is no appreciable interaction between the atoms in case of

atomic gases or between the molecules in case of molecular gases.

(ii) As an electromagnetic wave passes through a gas the electric field

induces dipole moment in the gas molecules.

(iii) In polarisation the positions of the electrons are altered from their

equilibrium value while nuclei remain stationary.

(iv) The electrons are bound to the nucleus in an atom by linear

restoring force.

(v) There is a damping proportional to the velocity of the electron.

(vi) Over an atom or a molecule E is constant in space i.e.

E = E0e-i(ωt – k.r) E0e-iωt

In the light of above assumptions the equation of motion of an electron will

be

m (d2r/dt2) +( mγ0 dr/dt) + (mω02r) = eE

or (d2r/dt2) +γ0 (dr/dt) + (ω02r) = (e/m)E --------(1)

This equation is already discussed in the previous hint and its solution was

found to be

r = ((e/m) E0e-iωt )/ (ω02 – ω2) - i γ0ω -----------(2)



P a g e | 117

So the dipole moment which results from the displacement of the

electron under consideration

P = er = [(e2/m) E0e-iωt ] / (ω02 – ω2) - i γ0ω ----------(3)

Now if there are N electrons per unit volume in the gas, the

polarisation vector

P = Np

In the light of equation (3) is given by

P = N(e2/m) E /(ω02 – ω2) - i γ0ω --------(4)

In deriving the above equation we have assumed that there is only one

type of charge which is characterised by the constants ω0 and γ0. It is quite

reasonable to expect that the electrons are not all in identical situations

within molecules and that there should be different pairs of characteristic

frequencies ω0j and associated damping factors γ0j each pair reflecting that

particular environment in which the given type of electron is found. So if we

define fj as the probability that an electron has characteristic frequency ω0j

and damping coefficient γ0j then the generalization of equation (4) yields

P= (Ne2E/m) (Σj) fj / (ω0j2 – ω2) – iγ0jω

So the electrical polarisability

= (e2/m) Σj fj / (ω0j2 – ω2) – iγ0jω ( as = Pm/Em=NPm/NEm = P /NE)

i.e r = 1 + (Ne2/ 0m) Σj fj / (ω0j2 – ω2) – iγ0jω

(as r = 1+x0 = 1+P// 0E = 1+N / 0)

n2 = 1 + 1/(4 0) .(4 Ne 2/m) Σj fj / (ω0j2 – ω2) – iγ0jω ------(A)

(as n = r)

This is the required result which expresses the frequency dependence of r

or n. To study what this equation implies we consider the following

situations –

(A) Static Case: If ω>i.e , the frequency of the incident wave is very small in

comparison to the natural frequency of the electrons, equation (A) reduces to

n2 = 1 + 1/(4 0) .(4 Ne 2/m) Σj fj / (ω0j2 ----(a)

Equation (a) clearly shows that the index of refraction is a constant greater

than unity and depends on the nature of the medium i.e. ω0j



P a g e | 118

(B) Normal dispersion : if γ0j› 0 and ω < ω0j i.e. the region is remote from

the natural frequencies of the electrons (which is true for optical

frequencies) equation (A) reduces to

n2 = [ 1+1/(4 0) .(4 Ne 2/m) Σ (fj / ω0j2 – ω2] --------(b)

From equation (b) it is clear that the refractive index is real and increase

with frequency of the incident waves i.e. for a given medium red light has

the lowest while violet largest index of refraction in the optical range of

frequencies.

(c) Anomalous dispersion : If ω ω0j i.e. in the extremely narrow spectral

region in which the impressed frequencies include one of the so many

natural frequencies of the electron. For simplicity we assume that there is

one natural frequency i.e., ω0j = ω0 so that equation (A) becomes

n*2 = 1 + [1/(4 0) .(4 Ne2/m) ] x [1/((ω02 – ω2) – iγ0ω]

Now as the index of refraction of gases under normal conditions is

approximately unity so that expression (1 + y)1/2 = 1+ ½ y for y < 1 may be

employed to obtain

n* = [1 + 1/(4 0) ] x[ (2 Ne2/m) ] x 1/((ω02 – ω2) – iγ0ω

Multiplying numerator and denominator of the second term on RHS by its

complex conjugate

n* = 1 + [1/(4 0)] x [ (2 Ne2/m) ] [ (ω02 – ω2)+ iγ0ω] /[ω02 – ω2)2+ γ02ω2]

or n* = n + ik ---------(1)

with n = [1 + 1/(4 0)] x [ (2 Ne 2/m) ]x [ (ω02 – ω2) /( ω02 – ω2)2+ γ02ω2]

------–(2)

and k = [1/(4 0) ] x (2 Ne 2/m) x [ γ0 ω] /( ω0 – ω2)2 + γ02ω2] ------------

(3)

(Here Eq. (1), (2), (3) = c )

Equation (c-1) shows that the index of refraction is a complex function of

frequency of the electromagnetic waves propagating through the gas.

The real part of equation (c-1) i.e.., n is plotted as function of

frequency ω in fig. 5.6. At very low frequencies n is slightly greater than

unity, n increases with increasing ω, reaching a maximum at ω1 (ω0 - γ0/2)

falling rapidly to unity at (ω - ω0), n continues to decrease rapidly until (ω0



P a g e | 119

+ γ0/2) where upon it increases again and approaches unity asymptotically

for large values of ω.

γ0

n

96

k

ωj ω0 ω ω Fig. 5.6

The imaginary part of n* i.e., k is also plotted in fig. 5.6. It

corresponds to absorption of the electromagnetic waves propagating through

the gas. The imaginary part of n* has a typical resonance shape. k is

maximum at (ω = ω0), where n is unity and a width at half maximum

approximately equal to γ0. Therefore in regions where n changes rapidly, the

gas is relatively high absorbing.

For any real gas there exist many resonant frequencies ω0j and

corresponding damping coefficients γ0j so that

n* 1+[1/(4 0) ] x[ (2 Ne 2/m) ] Σ fr/(ω0j2 – ω2)2 + iγ0j2ω2 -----(d)

The behaviour of the real and imaginary parts of the equation (C) is

illustrated in fig. 5.6 (a)

n

k

ω0 1 ω0 2 ω0 3



P a g e | 120

ω

Fig. 5.6 (a)

It is worthy to note here that this classical theory cannot

predict the values of the resonance frequencies ω0j ; only a correlation of

observations relating to optical properties of matter can be attempted.

Quantum theory must be used for complete description and even though a

quantum calculation can be in principle yield values for the resonance

frequencies, the computations can be carried out exactly for the most simple

cases.

5.7 Dispersion in Liquids and Solids.

In the materials the molecules are sufficiently close to each other so

the effect of interactions among the molecules can no longer be neglected.

Since the material is polarised, we expect that the actual electric field on a

given charge will have a contribution from the polarisation and hence it will

be different from the applied field. The usual way of approximating this is to

imagine a small sphere centered at the position of the electron in question to

be cut out of the material as shown in fig 5.7 . The sphere is to be large

enough microscopically and small enough macroscopically so that the

material outside it can be described in terms of the continuous polarisation

vector P. Then because of the discontinuity in P there are bound surface

charges on the surface of spherical volume with density

σ' = P.n = P cos θ

So the contribution to the local field E' by the change on an area ds

d E' = σ'ds / 4 0R2

The components normal to the direction of P cancel, so integrating over all

the surface we find that local field E' parallel to P is equal to

2

E'= dE' cos θ = (P/ 4 0 ) 0 0 (cos2 θ R2 sinθ dθ dφ/ R2) = (P/3 0) -----(1)



P a g e | 121

E

Fig. 5.7

The result is the contribution to the local field from the material

outside the sphere. We still have to calculate the contribution due to the

molecules within the small sphere. It can be shown that this contribution

averages to zero in an isotropic material such as a liquid or when the

molecules are arranged in an cubic lattice. Therefore if E is the applied

electric field, the total electric field acting on the charge is

ET = E + E' = (E + P/3 0) (using equ. 1)-------(2)

So the equation of motion of an electron in this case will be

m(d2r/dt2) + mγ0 (dr/dt) +(m 02r)= e [E + P/3 0]

m(d2r/dt2) + mγ0 (dr/dt) +(m 02r)= e [E + (Ner/3 0] (as P = Ner)

or (d2r/dt2) + γ0 (dr/dt) + (ω02 – (Ne2/3 0m) r = (e/m). E -------(3)

The solution of equation (3)

r = (e/m). E / [(ω02 – (Ne2/3 0m) - ω2 –iγ0ω]

So the dipole moment which results from the displacement of the

electron under consideration will be

þ = er = (e2/m). E / [(ω02 – (Ne2/3 0m) - ω2 –iγ0ω]

Now as there are N electron per unit volume

P = Np = N (e2/m). E/[(ω02 – (Ne2/3 0m) - ω2 –iγ0ω] ----------(4)

However as electrons are no all in identical situations with the

molecules, there should be different pairs of characteristic frequencies ω0j

P

+ P – + – R + θ – + θ – dE` + + –



P a g e | 122

and associated damping factors γ0j, each pair reflecting the particular

environment in which the given type of electron is found. So if we define fj

as the probability that an electron has characteristic frequency ω0j, and

damping coefficient γ0j, then the generalisation of equation (4) yields

P =(Ne2E/m) Σ fj / [(ω0j2 – (Ne2 fj /3 0m) - ω2 –iγ0ω]

i.e = (p/E) =(Np/NE =(P/NE)= (e2/m) Σj fj /[(ω0j2 – (Ne2 fj /3 0m) - ω2 –iγ0ω

5.8 Let us sum up:

This lesson described various methods adopted in scattering of electro

magnetic waves with particular reference to Thomson scattering (free

electron) and Rayeigh scattering (bound electron). Also you have understood

about normal and anomalous dispersion in gases, liquids and solids.


(i) Write a note on basic concepts about scattering and scattering parameters. (ii) What is meant by total scattering cross – section

(iii) Write the equation for the force in a plane EM waves


(i) What is resonance scattering? Explain

(ii) The intensity of X-rays scattered by an electron at distance r

from it and making an angle θ write the original direction is

given by

2

Ie = e2/(4 0mc2) I ½(1+cos2 θ) / r2 where I is the intensity

of the incident beam. Calculate (a) the intensity of scattered

radiation Is interms of Ic for n electrons (i) when they scatter

independently (ii) when all of them act as a single scattering

centre.


(i) Write short notes on

(a) Normal and anomalous dispersion

(b) Dispersion in liquids and solids

(ii) Explain in detail



P a g e | 123

(a) Scattering by a free electron

(b) Scattering by a bound electron.

5.12 References

(i) Electrodynamics by Gupta, Kumar, Singh

(ii) Electromagnetic fields and waves by R.N. Singh



P a g e | 124

UNIT – III

Lesson - 6

COHERENCE AND INTERFERENCE


We have so far assumed that light sources emit perfect harmonic

waves. In an ideal harmonic wave there exists a definite relationship

between the phase of the wave at a given time and at a certain time later;

and also at given point and at certain distance away. In reality, light

sources do not emit perfectly harmonic waves. Even a very best practical

monochromatic source emits a finite range of wavelengths and the light

waves are quasi-monochromatic. If it were not so, the light waves would

have been ideally coherent and interference would be observed at all times.

In practice, light is emitted form a light source when excited atoms

pass from the upper excited state to lower energy state. The process of

transition from upper state to a lower state lasts for brief time of about

10-8s. It means that an atom starts emitting a light wave as it leaves the

excited state and ceases emission as soon as it reaches the lower energy

state. Therefore, an emission event produces a light burst. Each light burst

occurs over a period of 10 -8s only, during which period a train of finite

length having a certain limited number of wave oscillations is generated.

Such a light burst is known as wave train or a wave packet. After some time

the atom again receives energy and jumps into excited state and

subsequently emits another burst of light. These emission events occur

quite randomly. Each atom in the source acts independently and different

atoms emit wave trains at different instants and their combination in

millions and millions constitutes the light from a light source. In order to

appreciate some of the peculiarities of natural light, the following fact is to

be well understood. The light emitted by an ordinary light source is not an

infinitely long, simple harmonic wave but is composed of jumble of finite

wave trains. We therefore call a real monochromatic source as a quasi-



P a g e | 125

monochromatic source. The wave trains issuing out of a quasi-

monochromatic source are as shown in Fig. 6.0

E (t)

t

Fig. 6.0

6.1 WAVE TRAIN

Fig. 6.1 shows a wave train generated by an atom

x

l

Fig. 6.1

If such a wave train lasts for a time interval t, then the length of the wave

train in a vacuum is

l = c t ------(1)

Where c is the velocity of light in a vacuum.

For example, if t = 10-8s, and c = 3x108 m/s,

then l = (3x10 8 m/s)( 10-8s) = 3 m.

The number of oscillations present in the wave train is

N = l/λ ----(2)

Where λ is the wavelength. If we assume λ = 5000Å = 5x10-7m, then

N = 3m/5x10-7m = 6x106

Thus, a wave train contain about a million wave oscillations in it.



P a g e | 126

Adding together the wave packets generated by all atoms in the light

source, one finds a succession of wave trains, as shown in Fig. 6.1(a). In

passing from one wave train to the next, there is an abrupt change in phase

and also in plane of polarization. It is not possible to relate the phase at a

point in wave train Q to a point in wave train P.

P Phase break Q

T (or x)

tcoh tcoh

(l coh) (l coh)

Fig. 6.1 (a)

Consequently there is no correlation between the phase of different

wave trains. Each wave train has a sustained phase for only about 10-8s,

after which a new wave train is emitted with a totally random phase which

also lasts only for about 10-8s. The phase of the wave train from one atom

will remain constant with respect to the phase of the wave train from

another atom for utmost 10-8s. It means that the wave trains can be

coherent for a maximum of 10-8s only. If two light waves overlap, sustained

interference is not observed since the phase relationship between the waves

changes rapidly, nearly at the rate of 108s times per second.

6.2 COHERENCE LENGTH AND COHERENCE TIME

The wave train, shown Fig. 6.1(a) appears fairy sinusoidal for some

number of oscillations between abrupt changes of frequency and phase.

The length of the wave train over which it may be assumed to have a fairly

sinusoidal character and predictable phase is known as coherence length.

We denote it by l coh. We may consider coherence length as approximately

equal to the length of the wave train, c t, over which its phase is

predictable. The time interval during which the phase of the wave train can

be predicted reliably is called coherence time. It is the time, t, during

which the phase of the wave train does not become randomized but



P a g e | 127

undergoes change in a regular systematic way. Coherence time is denoted

as tcoh. We can therefore write.

l coh = c t -------(3)

tcoh = t --------(4)

l coh = c tcoh ----------(5)

A wave train consists of a group of waves, which has a continuous

spread of wavelengths over a finite range λ0 centred on a wavelength λ0.

According to Fourier analysis the frequency bandwidth υ is given by

υ = 1/ t

where t is the average lifetime of the excited state of the atom. However, t

is the time during which a wave train is radiated by the atom and

corresponds to the coherence time, tcoh, of the wave train

Therefore υ = 1 / t = 1/ tcoh ---------(6)

Using the relation (5) into Eq. (6) , we get

υ = c/ tcoh -------------(7)

6.3 BANDWIDTH.

A wave packet is not a harmonic wave. Therefore, it cannot be

represented mathematically by simple sine l max

functions. The mathematical representation

of a wave packet is done in terms of Fourier l max/2 λ

integrals. If light emitted from a source is

analyzed with the help of a spectrograph,

it is known to be made up of discrete spectral λ(nm)

lines. Wave packets emitted by atoms form λ/2 λ 0 λ/2

these spectral lines. Therefore, spectral line

and a wave packet are equivalent descriptions. Fig 6.3

The wavelength of a wave packet or a spectral line

is not precisely defined There is a continuous spreads of wavelengths over

a finite range, λ, centered on a wavelength λ0. The maximum intensity of

the wave packet occurs at λ0 and the intensity falls off rapidly on either side

of λ 0, as shown in Fig. 6.3. The spread of wavelengths is called the

bandwidth.



P a g e | 128

The bandwidth is the wavelength interval from λ0 – ( λ/2) to λ0 +

( λ/2) which contains the major portion of the energy of the wave packet.

In practice a source, which is said to produce line spectrum, produces a

number of sharp wavelength distributions.

6.4 RELATION BETWEEN COHERENCE LENGTH AND BANDWIDTH.

The frequency and wavelength of a light wave are related through the

equation

υ = c / λ --------(8)

Where λ0 is the vacuum wavelength.

Differentiating Eq. (8) on both sides, we get

u= - c/λ2 x λ --------(9)

Using the relation (7) into eau. (9) , we obtain

c/l coh = -( c/ λ2) * λ ------(10)

Rearranging the terms, we get

l coh = λ2/ λ --------(11)

The minus sign has no significance and hence is ignored. Eq. (11)

means that the coherence length ( the length of the wave packet) and the

bandwidth of the wave packet are related to each other. The longer the wave

packet, the narrower will be the bandwidth (see Fig. 6.4(a). In the limiting

case, when the wave is infinitely long, we obtain monochromatic radiation of

frequency v0 (wavelength λ0).

λ

λ

λ - +

λ0

Fig. 6.4 (a)

From the equ.(2), the coherence length may be designated as the

product of the number of wave oscillations N contained in the wave train

and of the wavelength, λ. Thus



P a g e | 129

l coh = N λ ---------(12)

Equating (11) and (12), we get

N = λ / λ

λ / λ = 1/N --------(13)

Eq. (13) shows that the larger the number of wave oscillations in a wave

packet, the smaller is the bandwidth. In the limiting case, when N is

infinitely large, that is when the wave packet is infinitely long, the wave will

be monochromatic having a precisely defined wavelength. The dependence

of bandwidth on the length of the wave packer is schematically shown in

Fig. 6.4 (a)

6.5 COHERENCE

Coherence is an important property of light. It refers to the

connection between the phase of light waves at one point and time, and the

phase of the light waves at another point and time. Coherence effects are

mainly divided into two categories: temporal and spatial. The temporal

coherence is related directly to the finite bandwidth of the source, whereas

the spatial coherence is related to the finite size of the source.

TEMPORAL COHERENCE

Temporal coherence is also known as longitudinal coherence. Let a

point source of quasi-monochromatic light S (Fig. 6.5) emit light in all

directions. Let us consider light traveling along the line SP1P2. The phase

relationship between the points P1 and P2 depends on the distance P1P2 and

the coherence length of the light beam.

The electric fields at P1 and P2 will be

correlated in phase when a single wave

train extends over greater length than P3

the distance P1P2; that is if the distance P2

P1P2 is less than the coherence length l coh. S P1

Then, the waves are correlated in their Fig. 6.5

rising and falling and they will preserve a

constant phase difference.



P a g e | 130

The points P1 and P2 would not have any phase relationship if the

longitudinal distance P1P2 is greater than l coh since in such a case many

wave trains would span the distance. It means that different independent

wave trains would be at P1 and P2 at any instant and therefore the phase at

the two points would be independent of each other. The degree to which a

correlation exists is known as the amount of longitudinal coherence.

Monochromaticity

From Eq.(11) and Fig(6.4 (a)) we conclude that temporal coherence is

indicative of monochromaticity of source. An ideally monochromatic source

is an absolute coherence source. The degree of monochromaticity of a

source is given by

ξ = v /v0 -------(14)

when the ratio v /v0 = 0, the light wave is ideally monochromatic.

Purity of spectral line.

The width of a spectral line is given by λ (see Fig. 6.3). It is seen

from equ.(11) that it is related to the temporal coherence. Thus,

λ = λ2 / l coh ---------(15)

6.6 SPATIAL COHERENCE

Spatial coherence refers to the continuity and uniformity of a wave in

a direction perpendicular to the direction of propagation. If the phase

difference for any two fixed points in a plane normal to the wave propagation

does not vary with time, then the wave is said to exhibit spatial coherence.

It is also known as lateral coherence. Again looking at the point source S

(Fig. 6.5), SP1 = SP3 and therefore, the fields at points P1 and P3 would have

the same phase. Thus, an ideal point source exhibits spatial coherence, as

the waves produced by it are likely to have the same phase at points in

space, which are equidistant from the source. On the other hand, an

extended source is bound to exhibit lesser lateral spatial coherence. Two

points on the source separated by a lateral distance greater than one

wavelength will behave quite independently. Therefore, correlation is absent

between the phase of the waves emitted by them. The degree of contrast of

the interference fringes produced by a source is a measure of the degree of



P a g e | 131

the spatial coherence of its waves. The higher the contrast, the better is the

spatial coherence.

Fig. 6.6 Spatial coherence

6.7 DETERMINATION OF COHERENCE LENGTH

The coherence length can be measured by means of Michelson

interferometer. In a Michelson interferometer, a light beam from the source

S is incident on a semi-silvered glass plate G (see Fig.6.7) and gets divided

into two components; one component is reflected (1) and the other (2) is

transmitted. These two beams, 1 and 2, are reflected back at mirrors M1

and M2 respectively and are received by the telescope where interference

fringes are produced. It is obvious that the beams produce stationary

interference only if they are coherent.

Let M2' be the image of M2 formed by G. If the reflecting surfaces M1

and M2' (the image of M2) are separated by a distance d, then 2d will be the

path difference between the interfering waves. The condition of fixed phase

relationship between the two waves, 1 and 2, will be satisfied if

2d<<l coh

In such a case distinct interference fringes will be seen. If, however,

2d>>l coh

then phase of the two waves are not correlated and interference fringes will

not be seen. To determine the coherence length of wave emitted by a light

source, the distance d between the mirrors M1 and M2' (the image of M2) is

varied by moving one of the mirrors. As the distance varies, the contrast of

the fringes decreases and ultimately they disappear. The path difference 2d

at the particular stage where the fringes disappear gives us the coherence

length.



P a g e | 132

The light from a sodium lamp has coherence length of the order of 1

mm, that of green mercury line is about 1 cm, neon red line 3 cm, red

cadmium line 30 cm, orange krypton line 80 cm, and that of a commercial

He – Ne laser is about 15m. The coherence length of light from some of the

lasers goes up to a few km.

M1

M'2

1

S 2

G 1 M2

2

T

Fig. 6.7

6.8 CONDITION FOR SPATIAL COHERENCE

The degree of spatial coherence of a beam of light can be deduced

from the contrast of the fringes produced by it. The broader the source of

the light, the lesser is the degree of coherence. In Young’s` double slit

experiment, if the slits S1 and S2 are directly illuminated by a source,

interference fringes are not observed. Instead the screen is uniformly

illuminated. The absence of fringes indicates that the light issuing from the

slits do not posses spatial coherence. If a narrow slit is introduced before

the double slit, light passing through the narrow slit S illuminates S1 and

S2. The waves emerging from them, having been derived through wave front

division, are coherent and stationary interference pattern will be observed

on the screen. If the width of the slit S is gradually increased, the contrast

of fringe pattern decreases and fringes disappear. When the slit S is wider,

S1 and S2 receive waves from different parts of S which do not maintain

coherence. When S is narrow it ensures that the wave trains incident on



P a g e | 133

slits S1 and S2 originate from a small region of the source and hence they

have spatial coherence.

A S1

B'

S 22 θ

B 2α 2θ d A'

S2

B

Fig. 6.8

Formation of distinct fringe pattern depends on two parameters in the

double slit experiment. One is the size of the slit S and second is the

separation, d, between the two slits S1 and S2 . From Fig. 6.8(a), it is seen

that the path difference between the two waves passing at the edges A and B

of the slit S is

= b sin a

Distinct fringes will be obtained when

<<(1/2 ) λ

Therefore, it requires that

b sin a = <<(1/2)* λ ---------(16)

This equation determines the size of the source, which is spatially

coherence to produce fringes of satisfactory contrast. Equ.(16) shows that

the smaller is the size of the source, the better is the spatial coherence.

Spatial coherence of waves decrease with increasing size of the light source.

From 6.8 it is seen that the edges A and B of the slit S subtend an

angle 2θ at the centre of the double slit and the double slit subtends an

angle 2 a at the centre of S. Waves from A produce interference pattern

which has its centre at A'. Similarly waves from B produce fringe Pattern

with its centre at B'. The patterns cancel each other if the maximum of one

falls on the minimum of the other. The condition for no fringes is that

2θ = (λ / 2d) --------(17)

when the slit S is made wider, the edges A and B move away from each

other. As a result the patterns at A’ and B' also move away. When the



P a g e | 134

maxima of one pattern fall on the maxima of the other, the fringes appear.

Therefore, the condition for fringes is that

2θ = λ / d

or d = λ /2θ ------(18)

In other words, there is restriction imposed on the separation of the slits

S1 and S2. d must be very small compared to λ /2θ in order to maintain

spatial coherence and obtain fringe pattern on the screen.

Light produced by lasers is highly coherent. Therefore, when a laser

is used as the source, interference fringes can be obtained without the aid of

slit S.

P

A

α

b

B

Fig. 6.8 (a)

6.9 THE LINE WIDTH

In the Michelson interferometer experiment discussed in the previous

section, the decrease in the contrast of the fringes can also be interrupted as

being due to the fact that the source is not emitting at a single frequency

but over a narrow band of frequencies. When the path difference between

the two interfering beams is zero or very small, the different wavelength

components produce fringes superimposed on one another and the fringe

contrast is good. On the other hand, when the path difference is increased,

different wavelength components produce fringe patterns which are slightly

displaced with respect to one another, and the fringe contrast becomes

poorer. One can say that the poor fringe visibility for a large optical path

difference is due to then non-monochromaticity of the light source.

The equivalence of the above two approaches can be easily understood

if we consider the Michelson interferometer experiment using two closely

spaced wavelengths λ1 and λ2 we had shown that for two closely spaced



P a g e | 135

wavelengths λ1 and λ2 (Like D1 and D2 lines of sodium), the interference

pattern will disappear if

(2d/ λ2) – (2d/ λ1) = 1/2 --------(1)

where 2d represents the path difference between the two beams. Thus

2d = (λ1 λ2)/2(λ1 - λ2) λ2 / 2(λ1 - λ2) ------(2)

Instead of two discrete wavelengths, if we assume that the beam consists of

all wavelengths lying between λ and λ + λ, then the interference pattern

produced by the wavelengths λ and λ +1/2 λ will disappear if

2d = λ2/2 ( 1/2 λ ) = λ2 / λ -------(3)

Further for each wavelength lying between λ and λ +1/2 ( λ), there will be

a corresponding wavelength (lying between λ and λ + (1/ λ) and λ+ λ)

such that the minima of one falls on the maxima of the other, making the

fringes disappear. Thus for

2d λ 2 / λ ---------(4)

the contrast of the interference fringes will be extremely poor. We may

rewrite the above equation in the form

λ λ2/2d -------(5)

implying that if the contrast of the interference fringes becomes very poor

when the path difference is ~d, then the spectral width of the source would

be ~ λ2/2d.

Now we had observed that if the path difference exceeds the coherence

length L, the fringes are not observed. From the above discussion therefore

follows that the spectral width of the source

λ, will be given by

λ ~ (λ2 / L) = λ2 / c c -------(6)

Thus the temporal coherence c of the beam is directly related to the spectral

width λ. For example, for the red cadmium line, λ = 6438 Å,

L 30 cm ( c 10-9 sec)

( λ - λ2) / c c = (6.438x10-5)2 / 3x 1010 x 10-9

~ 0.01 Å

for the sodium line, λ 5890 Å, L 3 cm ( c 10-10 sec) and λ ~ 0.1 Å.

Further since u = c/λ, the frequency spread v of a line would be

u ~ (c/ λ2 ) λ ~ (c/L) ---------(7)



P a g e | 136

where we have disregarded the sign. Since c = L/c, we obtain

u ~ 1 / c ------------(8)

Thus the frequency spread of spectral line is of the order of the inverse of

the coherence time. For example, for the yellow line of sodium

(λ = 5890 Å),

c ~ 10-10 s › v ~ 10 10 Hz

u= c/ λ = 3x1010 / 5.89 x 10-5 5x1014 Hz

we get

u /u~ 1010 / 5 x1014 = 2 x 10-5

The quantity u /u represents the monochromaticity (or the spectral

purity) of the source and one can see that even for an ordinary light source

it is very small. For a commercially available laser beam, c 1~ 50 nsec

implying v /v ~ 4 x10 -8. The fact that the finite coherence time directly

related to the spectral width of the source can also be seen using Fourier

analysis.

6.10 THE SPATIAL COHERENCE (Experiment)

Till now we have considered the coherence of two fields arriving at

particular point in space from a point source through two different optical

paths. In this section we will discuss the coherence properties of the field

associated with the finite dimension of the source.

We consider the Young’s double-hole experiment with the point source

S being equidistant from S1 and S2 (see fig.6.10 (a)). We assume S to be

monochromatic so that it produces interference fringes of good contrast on

the screen PP'. The point O on the screen is such that S1O = S2O. Clearly

the point source S will produce an intensity maximum around the point O.

We next consider another similar source S' at a distance l from S. We

assume that



P a g e | 137

P

S' S1

l α d

α Q Q

S2

a1 a2

P'

(a)

S1

l S O' O

θ S2

a

(b)

Fig. 6.10 (a) Young’s double – hole interference experiment with two independent point sources S and S'. (b) The same experiment with an extended source.

Waves from S and S' have no definite phase relationship. Thus the

interference pattern observed on the screen PP ' will be a superposition of

the intensity distributions of the interference patterns formed due to S and

S'. If the separation l is slowly increased from zero, the contrast of the

fringes on the screen PP ' becomes poorer because of the fact that the

interference pattern produced by S' is slightly shifted from that produced by

S. Clearly, if

S'S2 - S'S1 = λ/2 --------(1)

the minima of the interference pattern produced by S will fall on the maxima

of the interference pattern produced by S' and no fringe pattern will be

observed. It can be easily seen that



P a g e | 138

½

S'S2 = a2 + (d /2 + l)2 a + (1 / 2a) (d/2 + l)2

and ½

S'S1 = a2 + (d/2 - l)2 a + (1/2a) (d/2 - l)2

Where a = a1 + a2

And we have assumed a >> d, l, Thus

S'S2 - S'S1 (l d)/a

Thus for the fringes to disappear, we must have

λ/2 = S'S2 - S'S1 (l d)/a

or

l (λa) /2d

Now, if we have an extended incoherent source whose linear dimensions is ~

λa /d then for every point on the source, there is a point at distance of (λa)

/2d which produces fringes which are shifted by half a fringe width.

Therefore the interference pattern will not be observed. Thus for an

extended incoherence source, interference fringes of good contrast will be

observed only when

l << λa /d --------(2)

Now, if θ is the angle subtended by the source at the slits

(see fig. 6.10 (b)] then θ /a and the above condition for obtaining fringes of

good contrast takes the form

d <<λ /θ --------(3)

On the other hand, if

d ~ λ/θ -------(4)

the fringes will be of very poor contrast. Indeed, a more rigorous diffraction

theory tells us that the interference pattern disappears when

d = 1.22 λ/θ, (2.25 λ)/θ, (3.24 λ)/θ ----(5)

Thus as the separation of the pinholes is increased from zero, the

interference fringes disappear when d = (1.22 λ)/θ; if d is further increased



P a g e | 139

the fringes reappear with relatively poor contrast and they are washed out

again when d = 2.25 λ/θ, and so on. The distance

l w = λ/θ (Lateral coherence width)

gives the distance over which the beam may be assumed to be spatially

coherent and is referred to as the lateral coherence width.

6.11 (I) MICHELSON STELLAR INTERFEROMETER - Theory

Using the concept of spatial coherence, `Michelson developed an ingenious

method for determining the angular diameter of stars. The method is based

on the result that for a distant circular source, the interference fringes will

disappear if the distance between the pinholes S1 and S2 (fig. 6.11 (a) is

given by (see equ. 5)

d = (1.22 λ)/θ --------(1)

Where θ is the angle subtended by circular source. For a star whose

angular diameter is 10-7 radians the distance d for which the fringes will

disappear would be

d ~ (1.22 x 5 x 10-5) / 10-7 600 cm

where we have assumed λ 5000Å. Obviously, for such a large value of d,

the fringe width will becomes extremely small. Further one has to use a big

lens, which is not only difficult to make, but only a small portion of which

will be used. In order overcome this difficulty, Michelson used

L

S1

S d

S2

f

Fig. 6.11 (a) S is source of certain spatial extend; S1 and S2 are two slits separated by a distance d which can be varied. The fringes are observed on the focal plane of the lens L



P a g e | 140

two movable mirrors M1 and M2 as shown in Fig. 6.11(b) thus he effectively

got a large value of d. The apparatus is known as Michelson’s stellar

interferometer. In a typical experiment the first disappearance occurred

when the distance M1M2 was about 24 feet, which gave

θ (1.22 x 5x 10-5) / (24 x 12 x 254) radians 0.02”

for the angular diameter of the star. This star is known as Arctures. From

the known distance of the star, one can estimate that the diameter of the

star is about 27 times that of the sun.

M1

M2 L

Fig. 6.11(b) Michelson’s stellar interferometer

We should point out that a laser beam is spatially coherent across

the entire beam. Thus, if laser beam is allowed to fall directly on double-slit

arrangement ( Fig. 3.2.11 (c) ), then as long as the beam falls on both the

slits, a clear interference pattern is observed on the screen. This shows that

the laser beam is spatially coherent across the entire wave front.

Laser beam

Screen

Fig. 611(c)- if a laser beam falls on a double – slit arrangement, interference are observed on the screen.

This shows that the laser beam is spatially coherent across



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the entire wave front.

Fig. 6.11(b) shows the interference pattern obtained by Nelson and Collins

by placing a pair of slit of width 7.5µm separated by distance 54.1µm on the

end of the ruby rod in a ruby laser the interference pattern aggress with the

theoretical calculation to with in 20%. To show that the spatial coherencies

is indeed due to laser action, they showed that below, threshold (of the laser)

no regular interference pattern was observed; only a uniform darkening of

the photographic plate was obtained.

6.11 (II) Michelson Stellar Interferometer (Another Method)

Using the concept of spatial coherence, Michelson developed an

ingenious method for determining the angular diameters of stars. In this

interferometer, the separation of effective slits for receiving start light may

be increased considerably while the separation of slits for producing fringes

was small and fixed. He accomplished this by mounting four mirrors M1, M2

, M3, M4 on a long iron girder in front of the telescope objective (fig. 6.11(c) )

S1 and S2 are slits in front of the telescope objective L. Light from the star is

incident on the mirrors M1 and M2 which reflect into M3 and M4 and which

reflect into M3and M4 and which, in their turn, pass light into telescope

thorough the apertures of double slit. When the mirrors have been adjusted

so that M1 is parallel to M3 and M2 to M4 and the optical paths M1M3S1 and

M2M4S2 are equal, the beams overlap in the focal plane of the objective and

the Airy`s disc is seen crossed by a number of fine fringes. Airy`s disc is

elongated and much larger than would be formed by the telescope alone

because of its restricted aperture in the present case. Starting with M1 and

M2 close to the fixed mirrors M3 and M4, they moved out symmetrically

maintaining optical paths M1M3S1 and M2M4S2 equal until the fringes

disappeared. Let a be the least separation of the mirrors M1 and M2 for

disappearance of fringes. Then, the angular diameter of the star is

Ф = (1.22 λ) / a

λ is the effective wavelength for the star.

The fringe spacing, which depends on the separation b of S1 and S2

remains constant as the separation a of the movable apertures M1 and M2 is

varied.



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Ф

B A M1

Ф S1

Light from star M3

A b

M4 F

B P S2 A

Ф M2

Airy`s

Fig. 6.11 (c) Michelson stellar Interferometer

6.12 Let us sum up

In this lesson we described the theory of partial coherence, coherence time and

coherence length and derivation of each. Also we have described about the spatial

coherence for extended sources.


(i) What is partial coherence?

(ii) Explain coherence time and coherence length

(iii) What is spatial coherence?


(i) Derive an expression for coherence length and time.

(ii) Write a note on Multi layer films


(i) Write a neat diagram explaining the measurement of stellar diameter using

Michelson interferometer.

6.16 References

(i) Text book of optics by Brijlal & Subramaniyam

(ii) Optics by Ajoy Ghatak



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UNIT – IV

Lesson – 7

Optics of Solids


This lesson deals with optics of solids, in which the wave equation in a

conducting medium in derived. Also you study about to theory of optical

activity with special reference to Quartz. Finally you study the principle

involved in Faraday rotation.

7.1 THE WAVE EQUATION IN A CONDUCTING MEDIUM For a conducting medium

J = E -----------(1)

Where represents the conductivity of the medium. Thus, Maxwell’s

equation become

div E = 0 -------(2)

div H = 0 ----------(3)

curl E = - µ ( H / t) -------(4)

curl H = E + ( E / t) ------(5)

Taking the curl of Eq. (4), we get

Curl curl E = - µ ( / t) curl H

or

grad div E - 2E = - µ ( E / t) - µ ( 2E / t2)

Using Eq. (2), we get 2 E - µ ( E / t) - µ ( 2E / t2) ------(6)

Which is the wave equation for a conducting medium. For a plane wave of

the type

E = E0 exp [ i (kz – ωt ] -------(7)

We obtain

-k2 + iω΅ +ω2 µ = 0 -------(8)

Which shows that k must be a complex number. If we write

K = + i β ------------(9)



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Then

- ( 2 + 2i β) + iω΅ + ω2 µ = 0

Equating real and imaginary parts, we get

2 – β2 = ω2 µ ---------(10)

and

β = ω /2 --------(11)

Substituting for β in Eq. (10) and solving for , we get

= Ö ω µ ½ ± ½ (1 + ( 2 / ω2 2)1/2 1/2 -----(12)

We must choose the positive sign, the negative sign, would make complex.

Thus

= ω Ö µ ½ ± ½ (1 + ( 2 / ω2 2)1/2 1/2

β = ω / 2 --------(13)

Now, when k is complex, Eq. (7) becomes

E = E0 exp ( - βz) exp [ i ( z – ωt) -------(14)

Which represents an attenuated wave. The attenuation is due to the Joule-

loss. For a good conductor

/ ω >>1 ----------(15) and one obtains

β (ω΅ /2 )1/2 -------(16)

Indeed of / ω <<1, the medium can be classified as s dielectric and if /

ω >>1 (say 100), the medium can be classified as a conductor. For

0.01 / ω 100,

the medium is said to be a quasi-conductor. Thus depending on the

frequency, a particular material can behave as a dielectric or a conductor.

For example, for fresh water / 0 80 and 10-3 mhos/m. (Both and

can be assumed to be constants at low frequencies.) Thus

/ 10-3 / (80 x 8.55 x 10-2) 1.4 x 10 6 sec-1

For ω = 2 x 1010 sec-1

/ω 2 x 10-5

Thus fresh water behaves as a good conductor for υ 103 sec-1 and as

dielectric for υ 107 sec-1. On the other hand, for copper one may assume

0 and



P a g e | 145

5.8 x 107 mhos/m and for ω 2 x 1010 sec-1

/ω (5.8 x 107) / (2 x 1010 x 8.9 x 10-12) 108

Thus even for such frequencies it behaves as an excellent conductor.

From Eq. (14) it can be easily seen that the field decrease by a factor e

in traversing a distance

= 1 / β

which is known as the penetration depth. For copper, µ µ 0 = 4 x 10-7 N / amp2

and [ (2/ω΅ ) ]½ ( 2 / 2 v x 4 x 10-7 x 5.8 x 107 )½

0.065 / v x m

Thus for v 100 sec-1, 0.0065 m = 0.65 cm whereas for v 10 8 sec-1,

6.5 x 10-6 m showing that the penetration decreases with increase in

frequency.

7.2 FARADY ROTATION

Consider linearly polarized light propagating through a medium. If a

magnetic field is applied along the direction of propagation of the polarized

wave, then the plane of polarization gets rotated – this rotation is usually

referred to as Faraday rotation after the famous Physicist Michel Faraday

who discovered this phenomenon in 1845. In the presence of (longitudinal)

magnetic field, the modes of propagation are the left circularly polarized

(LCP) wave and the right circularly polarized (RCP) wave. Thus the situation

is somewhat similar to the phenomenon of optical activity. The angle θ by

which the plane of polarization rotates is given by the empirical formula

θ = VH

where H is the applied magnetic field, l is the length of the medium and V is

the constant. For silica V 2.64 x 10-4 deg / Ampere 4.6 x 10-6 radians /

Ampere.

The Faraday rotation has a very important application in measuring

large currents using single mode optical fibers. We consider a large length

of a single mode fiber wound in many turns in the form of a loop around a

current carrying conductor (see Fig.7.2 ). If a current I is passing through

the conductor then by ampere’s law



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H .dI = NI

where N represents the number of loops of the fiber around the conductor.

Thus if a linearly polarized light is incident on the fiber, then its plane of

polarization will get rotated by the angle.

Fig. 7.2 A single mode fiber wound helically around a current

carrying conductor. The rotation of the plane of polarization

is detected by passing the light through a Wollaston prism

and then an electronic processor

------

θ = VNI

The rotation θ does not depend on the shape of the loop. As an

example, for I = 200 Amperes and N = 50, θ 0.26 degree. The light from

the fiber is allowed to fall on a Wollaston prism and the outputs are

measured separately; the Faraday rotation θ is given by

θ = constant (I1 – I2) / (I1 + I2)

where I1 and I2 are the currents in the electronic processor due to the two

beams coming out of the Wollaston prism. We can determine the actual

variation of the output with the current passing through the conductor.

Such a set up can be used to measure very high currents (10000 Amperes).

7.2 (a) General idea of Polarization

A simple method for eliminating one of the beams is through selective

absorption; this property of selective absorption is known as dichroism. A

crystal like tourmaline has different coefficient of absorption for the two



P a g e | 147

linearly polarized beams into which the incident beams splits up.

Consequently, one of the beams gets absorbed quickly and the other

component passes through without much attenuation. Thus if an

unpolarized beam is passed through a tourmaline crystal, the emergent

beam will be linearly polarized

Y Incident unpolarized

z

Linearly polarized wave x

Tourmaline crystal

Fig. 7.2 (a) when an unpolarized beam enters a diachroic

crystal like tourmaline, it splits up into two linearly polarized

components. One of the components gets absorbed quickly and

the other component passes through without much attenuation.

------- Another method for eliminating one of the polarized beam is through

total internal reflection. We will show that the two beams have different ray

velocities and as such the corresponding refractive indices will be different.

If one can sandwich a layer of material whose refractive index lies between

the two, then for one of the beams, the incidence will be at a rarer medium

and for the other it will be at a denser medium. This principle is used in a

Nicol prism which consist of a calcite crystal cut in such a way that for the

beam, for which the sandwiched materials is a rarer medium, the angle of

incidence is greater than the critical angle. Thus this particular beam will

be eliminated by total internal reflection. Figure (7.2 (b) ) shows a properly

cut calcite crystal in which layer of Canada Balsam has been introduced so



P a g e | 148

that the ordinary ray undergoes total internal reflection. The extra-ordinary

component passes through and the beam emerging from the crystal is

linearly polarized.

Calcite 90º

68º

Fig 7.2 (b) The Nicol prism. The dashed outline corresponds

to the natural crystal which is cut in such a way that the

ordinary ray undergoes total internal reflection at the Canada

Balsam layer

7.3 THEORY OF OPTICAL ACTIVITY

As mentioned earlier, in an isotropic dielectric, the D vector is in the

same direction as E and we have

D = E = 0n2E ---------(1)

Where 0 ( = 8.854 x 10-12 MKS units) is the permittivity of free space and

n = / 0 is the refractive index of the medium. Now if we dissolve cane

sugar in water, the medium is still isotropic; however because of the spiral

like structure of sugar molecules, the relation between D and E is given by

the following relation

D = 0 n2E + ig k x E

0 n2 (E + i k x E ) ----------(2)

where

= g / 0 n2

and k is the vector along the direction of propagation of the wave. The

parameter can be positive or negative but it is usually an extremely small

48º e -ray

Axis Optic o -ray axis



P a g e | 149

number (<<1). Without any loss of the generality, we may assume

propagation along the z- axis so that kx = ky = 0 and k1 = 1 giving

x y z

K x E 0 0 1 = x Ey + y Ex

Ex Ey Ez

Thus Dx 0n2 -ig 0 Ex = Dy ig 0n2 0 Ey -------(3)

Dz 0 0 0n2 Ez

The matrix is still Hermitian but there is small off diagonal imaginary

element. The presence of these off - diagonal terms give rise to optical

activity.

n2w / c2 µ0 [ E – ( k . E) K = D

We write the x and y components of the above equation and since kx = 0 = ky

and kz =1, we get

(n2w / c2 µ0) Ex = Dx = 0n2 Ex - igEy and

(n2w / c2 µ0 ) Ey = Dy = igEx + 0n2 Ey

Thus

( (n2w / n2) – 1) ) Ex = -i Ey and

( (n2w / n2) – 1) ) Ey = -i Ex

where, we have used that fact that c = (1 / 0µ0.) For nontrivial solutions.

(n2w / n2 – 1)2 = 2 giving

nw = n 1 ± --------------(4) and

Ey = ± i Ex ------------(5)

We write the two solutions as nr ( = n 1 + and n1 ( = n 1 - ;

the corresponding propagation constants will be given by



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k = kr = (ω/c) nr = (ω/c) n 1 + ---------(6)

and

k = kl = (ω/c) nl = (ω/c) n 1 - ---------(7)

For nw = nr, if

Ex = E0ei(krz – ωt)

Then

Ey = + iEx = E0eikrz – (ωt + /2)

Which would represent an RCP (Right Circularly Polarized) wave and hence

the subscript r. Similarly For nw = nl , if Ex = E0ei(klz – ωt)

Then Ey = -iEx = E0 eiklz – (ωt - /2)

Which would represent an LCP (Left Circularly Polarized ) wave and hence

the subscript l. The RCP and LCP waves are the two “modes” of the

“optically active” substance and for an arbitrary incident state of

polarization, we must write it as a superposition of the two modes and

study the independent propagation of the two modes. Hence,

nr - nl = n [ 1+ - 1 - ] -------(8)

n

7.4 OPTICAL ACTIVITY IN QUARTZ

One observes optical activity for a plane polarised wave propagation

along the optic axis of quartz crystal. The general theory of propagation of

electromagnetic wave in such crystals is quite difficult; if the propagation is

not along the optic axis the “modes” are very nearly linearly polarized and

one may use the analysis discussed already. If the propagation is along the

z-axis then we may write (Eq.3)

Dx 0n20 -ig 0 Ex

Dy = ig 0n20 0 Ey

Dz 0 0 0n2e Ez

Where n0 and ne are constant of the crystal. Carrying out an identical

analysis, we would get

nr n0 [ 1 + ½ ]



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and

nl n0 [ 1 – ½ ] giving

nr - nl n0 and

ρ = / λ0 (nl - nr ) ( n0 ) / λ0

where λ0 is measured centimetres. For quartz

ρ ± 8.54 radians /cm at λ0 = 4046.56 Å

± 3.79 radians /cm at λ0 = 5892.90 Å

± 2.43 radians / cm at λ0 = 7281.35 Å

In quartz, we can have nr > nl or nr < nl . For λ0 = 4046.56 Å, we readily get

- n l - nr - 1.1 x 10-4

We may compare this with the value ne – n0 0.9 x 10-2. At higher

wavelengths, the value of - n r - nl - is much less.

7.5 THEORY OF FARADAY ROTATION

The equation of motion for the electron, in the presence of an external

electric field E is given by

(d2r / dt2) +(ω2 r) = - q/m E --------(1)

In the presence of static magnetic field B, we would have an additional (VxB)

term

. (d2r / dt2 )+ω2 r + (q/m) r x B = - (q/m) E -----(2)

where r = x x + y y + z z represents the position vector of the electron, x, y,

and z, are the unit vectors and q ( = + 1.6 x 10-19 C) is the magnitude of the

electronic charge. We assume the magnetic field to be in the z- direction.

Bx = 0 = By and Bz = B0 -------(3)



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Thus

. x y z r x B dx/dt dy/dt dz/dt = [ x (dy/dt) – y (dx/dt) ] B0

0 0 B0

Now for a circular polarized light wave propagating along the z – direction

E± = ( x ± i y) E0ei( kz –ωt ) ---------(5)

Where the upper and lower signs correspond to RCP and LCP respectively.

If we now write the x and y components of (Equ. (2), we would get

(d2x / dt2) + (ω02x) + (qB0 / m) ( dy/dt) = ( - q/m) E0ei( kz –ωt) ------(6)

and

(d2y / dt2) + (ω02y ) - (qB0 / m) ( dx/dt) = ± (i q/m) E0ei( kz –ωt) -----(7)

where the upper and lower signs corresponding to RCP ( Right Circularly

Polarization) and LCP ( Left Circularly Polarization) respectively. Writing

x = x0 e i( kz –ωt) and y = y0 e i( kz –ωt)

we get x (ω2 – ω02) (ω2 – ω20) x0 + iωc iω y0 = + (q/m) E0

(ω2 – ω20) y0 + iωc iω x0 = ± (q/m) E0

x - iωcω

--------------(8 , 9)

where

ωc = (qB0 / m)

is the electron frequency. If we multiply Eq. (8) by (ω2 – ω02)and Eq. (9) -

iωcω and add the two equation we should get

[ (ω2 – ω02)2 – ω2cω2 ]x0 = (q/m)E0 [ (ω2 – ω02)2 ± ωcω

giving

x0 = q E0 / m [(ω2 – ω02)2 ± ωcω ]

similarly

y0 = ±i q E0 / m [(ω2 – ω02)2 ± ωcω ] = ±ix0

Thus the polarization is given by



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P = - Nqr

= -Nq . q E0 ( x ± i y ) / m [(ω2 – ω02)2 ± ωcω] e i(kz –ωt)

= Χ = E±

where the susceptibility Χ is given by

Χ =( Nq2 /m) . 1 / [(ω02 – ω02)2 ± ωcω]

Thus the modes are circularly polarized and the corresponding refractive

indices are given

n2 ± = 1 + ( Nq2 /m 0) . 1 / [(ω2 – ω02)2 ± ωcω]

Where the upper and lower signs correspond to RCP and LCP respectively.

7.6 Let us sum up

From this lesson, you have learnt the propagation of light in conducting media

and also the propagation of light in crystals. Further you have learnt about the optical

activity, Faraday rotation in solids.


(i) What is Faraday rotation?

(ii) What is optical activity?


(i) Write a note on reflection by a conducting medium?

(ii) Write a note on Magneto and electro optic effects.?


(i) Explain the theory of optical activity?

(ii) Explain the theory of Faraday rotation?

7.10 References

(i) Text book of optics by Brijlal & Subramaniyam

(ii) Optics by Ajoy Ghatak



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UNIT – IV

Lesson – 8

Optical Fibres


The refractive index `n` or index of refraction is related to the speed of

light (c) in free space and the speed of light in the medium (v) i.e.., n = c/v

and the value of C = 3x108 m/s. The value of `n` for air is 1.00, for water

1.33 and for glasses 1.50.

When light ray travelling in medium of refractive index (n1) is incident

on a dielectric material of lower refractive index (n2) a part of the light is

reflected and a part is refracted at the loundary. According to Snell’s law

n1 Sinφ1 = n2 Sinφ2

Incident ray

Reflected ray

Lower index n1 φ1

(air)

Higher index n2

(glass) φ2 Refracted ray.

Fig. 8.0 (a) Incident light ray on glass – air interface

Here φ1 is the angle of incidence and φ2 is the angle of refraction. As the

angle of incidence φ1 increases, the angle of refraction φ2 also increases and

approaches 90º. When the angle refraction is 90º, the angle of incidence φ 1

is called the critical angle (φc). When the angle of incidence is greater than

the critical angle (φc), the ray is completely reflected back into the

originating medium fig. 8.0 (b). This phenomenon is known as Total

Internal Reflection.



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n2

n1

φ φ

Fig. 8.0 (b) Total internal reflection

8.1 OPTICAL FIBRES

An optical fibre is a dielectric waveguide and is normally cylindrical in

form. It consists of bundle of very thin individual fibres. A beam of light

enters at one end of the bundle of fibres and reaches the other end with less

amount of energy loss in a direction parallel to its direction, as shown is fig.

8.1 (c). Optical fibres can also transmit an image from one end to the other

end.

φ Core φ

8.1 (c) Fig. Total internal reflection

8.1.1 Propagation of Light in Optical Fibres

The light rays propagating into the fibre core are of two types namely;

meridional rays and skew rays. The light rays which pass through the axis

of the core are known as meridional rays. The passage of such rays in a step

index fibre is shown in fig. 8.0(a)

Cladding

Cladding



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a) Meridional (b) Skew rays

Fig. (8.1d) Propagation optical fibre

The rays which never pass through the axis of the core are known as

Skew rays. A typical passage of skew rays in a step index fibre is shown in

fig 8.0 (b). The skew rays will not utilise the full area of the core and they

travel farther than meridional rays and undergo higher attenuation.

8.1.2 Acceptance Angle

The fibre core will propagate the incident light, which are incident at

an angle less than the critical angle (φc) will continue to be propagated down

its length. The geometry of the launching of light rays into an optical fibre is

shown in fig 8.1.2.

A Eventually lost by

radiations Φc Acceptance cone Φc Core

B Cladding

8.1.2 Acceptance angle

Let a meridional ray be incident at an angle θa in the core – cladding

interface of the fibre. Any ray A which is incident at an angle greater than θa

will be transmitted into the core cladding interface at an angle less than the

critical angle θc and will not undergo total internal reflection.

A ray B which is incident at an angle greater than θa is eventually lost.

Incident rays which are incident on fibre core within conical half angle φa

will be refracted into fibre core and propagate into the core by total internal

reflection. φa is called the angle of acceptance or minimum acceptance angle

or total acceptance angle.



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8.1.3 Structure of optical fibres

An optical fibre consists of an inner glass cylinder of high refractive

index (n1) which is known as the core. The core is surrounded by a

cylindrical shell of glass or plastic of lower refractive index (n2) called

cladding. Since n1 > n2 total internal reflection takes place.

The thickness of the core is 50µm and that of the cladding is 100µm

to 200µm. The overall thickness of an optical fibre is nearly 125µm to

200µm. Thus an optical cable is small in size and of light weight.

Φc core n1 Φmax

Φ

Incident ray

8.1.3 Configuration of optical fibre

8.1.4 Numerical Aperture:

Numerical aperture (NA) is a measure of the light rays that can be

accepted by the fibre. Consider a ray of light entering the optical fibre at an

angle φ1. Let φ1 be less than the angle of acceptance φa. The ray enters from

air of refractive index n0 to the core of refractive index n1 . n1 > n2, the

refractive index of the cladding. By Snell’s law

n0 Sin φ1 = n 1 φ2 ------(1)

Air (n0)

n2

φ1 A C n1

φ2 φ φ

B n2

Fig. 8.1.4 Numerical aperture

From the above fig, in triangle ABC

φ2 = ( /2 – φ) -----(2)

Cladding n2



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Substituting in (1)

n0 Sin φ 1 = n 1 Sin( /2 – φ )

n 0 Sin φ1 = n 1cos φ -----(3)

n0 Sin φ 1 = n1(1-sin2 φ)1/2 -----(4)

When total internal reflection takes place φ = φc and φ1 = φa Equ. (4)

becomes

n0 Sin φ a = (n 12 - n 22)1/2

i.e., NA = n0Sin φ a = (n 12 - n 22) 1/2 ------(5)

Refractive index of air n0 = 1, NA is equal to sin φa.

The NA is terms of refractive index difference ( ) between the core and

cladding in given by the relation

NA = (n 12 - n 22)1/2 = n1 (2 ) 1/2 where = (n 21 - n 22 ) / 2n12 = (n1 – n2) / n1

NA is the Numerical Aperture passing through an optical fibre is given by

NA is the Numerical Aperture passing through the number of modes optical

fibre is given by

N 4.9 (N Ad/λ)2

Where d is he diameter of the core and λ is the wavelength of light used.

When d 0.76 λ/NA the fibre propagates only a single mode.

Fibre optic materials and their properties. Optical fibres are made up

of silica (glass) and plastic. The optical fibre materials have the following

properties.

i) Low scattering cross section

ii) Low optical (absorption, attenuation, dispersion) energy loss

iii) Efficient guide for light waves.

8.2 Classification of optical fibres

Optical fibres are classified based on

i) Number of modes and

ii) Refractive index profiles.

Based on number of modes

Depending on the number of modes, optical fibres are classified as

single mode fibre and multimode fibres.



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Single mode fibre: If only one mode s transmitted through a fibre, then it

is a single –mode fibre. It can be excited with LASER diodes. The core

diameter is of the order of a few wavelengths. The difference between the

refractive indices of core and cladding is small.. A typical single mode fibre

may have a core radius of 3µm and numerical aperture of 0.1 at a

wavelength of 0.8µm.

Multimode fibre: If more than one mode is transmitted through a fibre,

then it is a multimode fibre.

Advantages of multimode fibre: The larger core radii of multimode fibres

make it easier to launch optical power into the fibre and facilitates end to

end connection of similar fibres. Another advantage is that light can be

launched into multimode fibre using a light emitting diode (LED) source.

The disadvantage of multimode fibres is that they suffer from intermodal

dispersion.

Based on Refractive index profile:

The optical fibres are classified in the following manner:

1) Step index fibre

2) Graded index fibre

Step index fibre: It is based on the comparison of the refractive indices of

air, cladding and core. The refractive index (n) profile with respect to the

radial distance (r) from the fibre axis is given by

r r

a

Refractive Index (nr)

n1 Core

n2 Cladding

(a) Multimode

r



P a g e | 160

Refractive index a Core

(nr) n1 Cladding

n2

(b) Single mode

Fig. 8.2 Step index fibre

For core n(r) = n 1 ; r < a For cladding n (r) = n 2 ; r a

The step index fibres propagate both single and multimode signals

within the core. A typical step index fibre which propagates multimode

signals is shown is fig. 8.2(a). The diameter of the core is 50µm or greater.

A typical single mode or monomode step index fibre is shown is fig. 8.2(b).

The core diameter of the single mode fibre is of the order of 2 to 10µm.

A typical structure of the single mode step index fibre is shown in fig.

8.2(c) The core and cladding are made of glasses with high purity to avoid

losses due to absorption. The outer layer is of less refractive index and

made of glass of lesser purity. The thickness of the cladding is at least 10

times as that of the core radius. The same can be increased for easy

handling and for reducing the damage due to micro bending.

Buffer jacket

Primary coating Refractive Index

Cladding n1

n2

Core

Fig. 8.2 (c) Single mode fibre waveguide structure

In practical step index fibres the core (of radius a ) refractive index

n1 = 1.48. The refractive index of the cladding n 2 is slightly less than η1.

n 2 = n 1(1- )

The parameter is the core –cladding index difference or simply the

core index difference. is of the order of 0.01 since the core refractive index

.



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is larger than the cladding refractive index., electromagnetic energy at

optical frequency is made to propagate along the fibre wave guide by total

internal reflection at the core-cladding interface.

ii) Graded index fibre: In graded index fibre, the refractive index of the core

varies with the radial distance from fibre axis. The refractive index at the

fibre axis is maximum at the core – cladding interface. The index (n) profile

of the radial distance (r) from the fibre axis is given as

for core n (r) = n 1[1-2 (r/a)α]1/2 ; r < a

for cladding n (r) = n 1[1-2 ) 1/2 = n 2 ; r a

Where α is the profile parameter which gives the characteristic refractive

index profile of the optic fibre core. The index difference for the graded –

indeed fibre is given by

= (n 12 - n 22)/2 n 1 (n 1 - n 2) / n 1

The approximation on the right –hand side of the above expression gives the

index difference of the step –index fibre.

When α is infinite the fibre will be step index. When α = 2 the fibre

will be parabolic and when α = 1 the fibre will be triangular as shown in the

figure Refractive index

α2 α1

n2 n2

-a Core axis b Radial distance

Fig. 8.2(d) Refractive index profile against values

A parabolic refractive index profile core with α=2 is used for

multimode propagation as shown 8.2 (e) . In the graded-index fibre, the

refractive indeed of the core is maximum at the core axis and a gradual

decrease from the centre of the core results in the creation of many

refractions of the incident rays as shown in Fig. 8.2 (e). The above

mechanism can be easily explained by considering the Fig. 8.2 (f). The



P a g e | 162

expanded diagram shows the various refractive index values from high to

low at different interfaces within a graded-index fibre. When the ray is

incident at the air core medium, the refracted ray shows a gradual curve

due to the ever increasing a angle of incidence until the condition for total

internal reflection at met and then the ray travel back to the core axis

gradually be continuous refraction.

r

r

n2 a

Refractive Index n1

core

Cladding

Fig. 8.2 (e) Ray transmission multimode graded fibre

The intermodal dispersion less in the case of multimode graded-index fibre

than multimode step index fibre due to their refractive index profile. The

multimode graded index fibre with relatively high numerical aperture finds

application in high bandwidth and medium half.

r Refraction Total int.reflection

n6 n5

n4 n3 n2 Core n1 Cladding

Fig 8.2(f) Curved ray path in a graded fibre

A typical multimode graded indexed wave guide structure is shown

Fig. 8.2 (g). Similar to single mode wave guide, in order to avoid the

absorption loses high purity core and cladding martial were selected. The

features of outer layer ae to provided and improved rigidity to resist bending.

The outer layer may be of same material with different refractive index. The



P a g e | 163

thickness of the cladding will be normally very thin as it is used only as a

barrier to prevent the impurity ions from outer layer to migrate into the core

region. For higher refractive index of the outer layer, the thickness of the

cladding will be of 10 wavelengths. The diameter of the core will be in the

order of 50 -60µm with NA of 0.2 – 0.3

Buffer jacket

Primary coating Refractive Index

Cladding

n1

Core nn

Fig. 8.2 (g) Multimode grated index wave guide structure

8.2.1 TRANMISSION CHARACTERISTICS OF OPTICAL FIBRES

The study of the transmission characterises of the optical fibres helps

in selecting the fibre for optical communication. Following are the two

factors which affect the transmission of light waves in optical fibres.

(i) Attenuation

(ii) Dispersion

(i) Attenuation:

The attenuation or transmission loss of light signals when it

propagates through an optical fibre is an important factor to be considered

in the design of an optical fibre communication. The main sources, which

are responsible for attenuation are namely electron absorption. Raleigh

scattering, material absorption and impurity absorption. The first three

absorption are known as intrinsic absorption mechanism since these

depend on the glass fibre. The absorption due to impurity is known as

extrinsic absorption mechanism. The attenuation/wavelength

characteristics of a glass fibre is shown in Fig. (4.3.1)

.



P a g e | 164

Attenuation (dB/Km)

Impurity absorption

100-

10-

1- Rayleigh scattering

Electron Material absorption absorption

0.1

0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8

Wavelength (µm)

Fig.8.2.1 Attenuation/wavelength of a silica based glass fibre

(i) Intrinsic absorption

In intrinsic absorption attenuation is due to Rayleigh scattering

material absorption and electron absorption. In the case of Rayleigh

scattering, the scattering of light takes place due to the small irregularities

in the structure of the core. The irregularities in the core are due to the

density fluctuation during the glass manufacturing. The fundamental

Rayleigh scattering is significant only when the wavelength of the incident

light is in the same order as the dimension of the scattering mechanism.

The above losses can be reduced by having the operation at longer

wavelength.

It is clear from Rayleigh scattering that the operational long

wavelength will produce low losses. However, the atomic bonds associated

with the core material will absorb long wavelength light and is known as

material absorption. Therefore, the operation at wavelength greater than

1.55 mm will result in a significant drop in attenuation.

The absorption of photons by exciting the electrons from core of the

atom to higher energy states due to incidence of flight in the ultraviolet



P a g e | 165

region is known as electron absorption. This type of absorption takes place

only in the lower wavelength region and is shown in the figure.

(ii) Dispersion

In optical fibres communication due to digital modulation, broadening

of the transmitted light pulse as they travel for a long distance take space.

This to the dispersion mechanisms involved within the fibre during

transmission. The broadening of light pulse using the digital bit pattern

1011 as they are transmitted along a fibre is shown in Fig. 8.2.1 The initial

pulses in the fibre input are shown in Fig 8.2.1 (a). It is clear from the Fig.

8.2.1(b) and Fig. 8.2.1(c) each pulse broadens and overlaps with is

neighbours and finally becomes indistinguishable at the receiver input. This

effect is known as intersymbol interference (ISI). The ISI is more effective in

digital communication than analog.

1 0 1 1

Amplitude Amplitude 1 0 1 1

Time Time

8.2.1 (a) Fibre input 8.2.1 (b) Fibre output at a distance L1

No zero level Composite pattern

Amplitude

ISI

Time

8.2.1 (c) Fibre output at a distance L1 > L2

Fig. – Broadening of light pulse

8.3 FABRICATION OF OPTICAL FIBRES:

Generally, fibres are prepared by using glass and plastic materials. Glass

fibres are fabricated mainly on the basis of Sio2 and doping is done with

oxides such as Geo2, PbO, B2O3 etc., Similarly the plastic fibres are



P a g e | 166

commonly made with polystyrene as core and methyl-methacrylate as

cladding materials. The different methods used for the fabrication of optical

fibres are the following.

I) modified chemical vapour deposition (MCVD)

ii) Fibre drawing from preform

iii) Fibre drawing from a double crucible.

8.3 (a) Modified chemical vapour deposition (MCVD)

A block diagram of the experimental setup used for the formation of

silica cladding tube and deposition of core glasses is shown in figure 8.3 (a)

a) Formation of Silica Vitrifying flame

CLADDING TUBE Sio2 Glass

Deposition flame

Si,O2 Soot

Mandrel

Sicl4 O2

Fig. 8.3 (a) Formation of silica cladding tube

(b) DEPOSITION OF CORE GLASSES:

Silica tube

Core Reaction area Exhaust gases

material glass

Rotating joint Rotating joint

Reaction flame

Fig:8.3 (b) Schematic diagram used in MCVD



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The MCVD is the method used to produce a glass preform. A glass

preform is a final fibre which has refractive index profile as that of final

fibre. The optical fibres are drawn from the glass preform by heating and

pulling a thin stand. The first step in the preparation of glass perform is the

process of producing a dio2 tube or substrate. This forms the cladding of

the final fibre and hence it needs a dopant to maintain a difference in

refractive index between core and cladding. The tube or substrate is made

by depositing a layer of d io2 particles and dopants known as soot on rotating

ceramic former or mandrel. When the soot reaches the required

temperature say 1400ºC it is vitrified into a clear glass. Then, the rotating

ceramic former is withdrawn. In other words the complete glass preform

can be made by depositing the glass first and then the cladding using the

deposition flame as shown in fig (a). After this process, the rotating ceramic

former can be withdrawn. Then the tube is collapsed to get the required

glass preform. This process is known as outside vapour phase oxidation.

In MCVD process, the tube is then fitted with the lathe and the core

particle as gaseous constituents is passed into it as shown in Fig. 8.3(b).

A clear glass obtained by sintering of core particles. The vapour is shut off

after achieving the required core depth. Then, the necessary heating is

given to the tube get the tube to collapse resulting in the required glass

preform. The glass preform is then placed in a pulling tower where the fibre

is drawn out.

8.4 DISPERSION:

The term dispersion describes the pulse broadening effect in fibres.

The dispersions produced by the properties of the core material and the line

width of the light signal passing through the fibre. As shown in Fig. 8.4 (a)

the pulse that appears at the output of a fibre is wider than the input pulse.

Dispersion may, therefore be defined as the output light pulse width

produced by an input pulse of zero width i.e.., a monochromatic light which

has single wavelength.. In that case, Pulse width available at the output

would be totally a result of fibre dispersion. Let us take a practical case of



P a g e | 168

an input pulse of with W1 and the output pulse of width with W2 > W1. The

fibre dispersion dT is defined as

dT = W22 – W21

W1 Output pulse

(a)

Input pulse W2

Fibre

Fig. 8.4 (a)

Dispersions measured in units of time either in nanoseconds (10-9s)

or picosecs. (10-12). The total dispersion produced by an fibre depends

directly on its length. Here, manufactures give the dispersion per unit

length of the fibre either in nanoseconds per kilometre (ns/km) or

picoseconds per kilometre(ps/Km). This permits the users to calculate the

expected total dispersion for a given length of fibre. If L is the length of the

fibre in kilometre, then the total fibre dispersion is given by

dT = L x (dispersion /Km)

The dispersion can be divided into two categories

i) Intermodal dispersion

ii) ) Intramodal dispersion

8.4 (a) Intermodal dispersion:

--- ---------

This type of dispersion result from the fact that the light ware

propagates through a fibre in different modes as explained in figure. There

is difference in the propagation times for the different modes which

produces dispersion between these different modes. For full understanding

of this dispersion, consider two extreme modes as shown in figure.

--------------------------------

n2

------θc------------------------------------- n1 Zero mode



P a g e | 169

The critical mode propagating at angle θc and another mode where the

propagation angle is zero i.e., zero mode. A pulse of light launched into the

fibre will propagate along the fibre in both modes.

For zero modes, travel time will be minimum and is given by

Tdo = L/V1 = L/C/n1 = L x n1/C

Where V1 is the velocity of light through the core material, C is the velocity of

light in vacuum and L is the length of the fibre. For the light rays travelling

at the critical angle θc, the delay would be maximum and is given by

Idc = (L/Cos θc )/ V1 = (L/Cos θc)/ (C/ n1)

= (L x n 1) / (Cos θcxC)

The time difference (Tdc - Td0) represents the pulse width dT at the output

dT = Tdc - Td0

= (L x n1)/Cos θcxC – (L x n1/C)

Using the relation Cos θc = n 2 / n 1; we get

dT = (Lx n1/C) x { (n1 – n 2) / n 2}

Where n1 and n2 are the refractive indices of the core and cladding

respectively. The expression (n 1 – n 2) /η2 = . Hence, the pulse width can

be expressed as

dT = = (Lx n1/C) x

Substituting the value of in terms of numerical aperture we get

dT = [Lx(N.A)2] /(2x n1xC) (or)

dT/L = (N.A)2 /(2x n1xC)

Where dT is the amount of pulse broadening and dT/L is the broadening per

unit length of the fibre.

8.4. (b) Intramodal dispersion:

It is due to the fact that he light signal propagating in fiber does not

consist of a single frequency but a group of frequency. It is also called

chromatic dispersion. It is related to the line width of the light pulse. It is

often expressed in terms of Pico seconds per Kilometre per nanometre of line

width.



P a g e | 170

8.5 Fibre losses:

The three major causes of light losses in an fibre are as under:

i) Material loss: It is due to the absorption of light by the fibre material. It

includes absorption due to light interacting with the molecular structure of

the material as well as loss due to material impurities. The loss due to the

atomic structure of the material itself is relatively small. Losses due to

impurities can be reduced to by better manufacturing.

ii) Light Scattering:

Cladding

Light is scattered by the molecules of the material due to structural

imperfections and impurities. The scattered light does not propagate down

the fibre, it is lost.

As shown in figure when light is scattered by an obstruction, it

produces power loss. In optical fibres, obstruction refer to density

variations in the material which produce changes in the refractive index.

The index variations behave like point sources scattering light in all

directions. Scattering can also be due to micobends i.e.., fibre deformation.

This type of loss has been made negligible with the help of improved

manufacturing techniques.

iii) Waveguide and Bend losses:

These are caused by imperfections and deformations of the fibre

structure which cause radiation of light away from the fibre. Fig. Shows

radiation loss caused by a change in diameter. As seen in the absence of

deformation, the light ray would have been confined inside the fibre. Very

small changes in the core size (i.e.., micobends) also cause radiation of light

as well as back scatter shown in figure, obviously, in both instances, the

loss results from scattering of light.

Core



P a g e | 171

------------------------ ------------------------ Ray leaves fibre

Back Scattered Radiated light

All the above losses are dependant on the wavelength of light used. By

carefully choosing the operating wavelength, the above losses can be

minimized.

Calculation of losses:

Cladding

Pi P0

As shown in figure, if Pi is the power input to the fibre and P0 the

power output the fibre loss is given by the ration P0/ Pi . The decibel loss is

given by

Loss = 10log P0/ Pi decibel

Since the loss increase with fibre length, the loss is usually given by decibels

per Kilometre (dB/Km).

8.6 APPLICATIONS

1) The small size and large information-carrying capacity of optic fibres is

much higher as compared to copper twisted-par cables in telephone

systems.

Fiber Core



P a g e | 172

2) Continuous passive links (no repeaters) more than 100Km long have been

produced.

3) The low weight of fibre cable a compared to coaxial lines gives them

distinct advantages for submerged cable application because of the relative

ease of transporting and laying the fibres.

4) Optic fibre communication are compatible with electrified railways

because they are not effected by EMI

5) Fibres are used in application that ae primarily video i.e.., broadcast

television and cable T.V remote monitoring and surveillance.

6) Fibre systems are particularly suited for transmission of digital data such

as that generated by computers

7) Military application of fibre optics include communication, command and

control links on ships and aircraft, data links for satellite earth stations and

transmission lines for tactical command post communication

8.7 Fibre Endoscope

In medicine, a fibre endoscope is used to study the interior of the

lungs and the other parts of the human body that cannot be viewed directly.

It can also be used to study the tissues and blood vessels far below the skin.

Te fibre endoscope uses bundle pf flexible fibres.

The schematic representation of flexible fibre endoscope is shown in

Fig. 8.7 It consists of two fibres namely inner and outer fibre. The inner

fibre illuminates the inner structure of the object under study, while the

outer fibre is used to collect the reflected light from that area. Thus, one can

view the inner structure of the object.

The optical light source is attached at the viewing end of the

endoscope. The optical arrangements, which are used to collect and view

the object, are arranged at the respective fibre ends. One can also achieve a

better imagine of the object by interfacing a telescope system at the internal

part of the endoscope.



P a g e | 173

Outer fibre conduct light to object

`

``

Object

Inner fibre conduct image to observer

`

Fig. 8.7 Flexible fibre endoscope

8.7.1 Advantages of optical fibres in communication

(i) A large number of telephone signals, nearly 15,000 signals can be passed

through the optical fibres in particular time without any interface, where as

in an ordinary copper cable, only 48 signals can be passed in particular

time.

(ii) The signal leakage is nil due to total internal reflection and hence there is

no cross-talk.

(iii) It is one of the most ideal means for communication in explosive

environments.

(iv) It gives a nearly foolproof communication during the wartime

(v) The cost of the cables is very low compared to metal cables.

(vi) Optical fibres have immunity to adverse temperature, moisture and

chemical reactions.

8.7.2 Other Applications

(a) In computers

(i) To exchange the information between different terminals in a network

(ii) The optical fibres are used to exchange information in cable television,

space vehicle submarine, etc.,

(b) In industry

The optical fibres are used industrial automation, security alarm system and

process control.

(c) Optical applications.



P a g e | 174

(i) Fibre optic delay lines the fibre optic cables are widely used interethnic

fields to produce required delay

(ii) Fibre optic sensor it is used as a sensor for electric field, magnetic field,

temperature, mechanical force, etc.,

8.8 Waveguide Dispersion Equations:

We had discussed material dispersion which results from the

dependence of the refractive index of the fibre on wavelength. Even if we

assume that the refractive indices n1 and n2 to be independent of λ0; the

group velocity of each mode does depend on the wavelength; physically, this

is due to the fact that the spot – size of the mode depends on the wavelength

this leads to what is known as waveguide dispersion. The detailed theory is

rather involved* but a convenient empirical formula for a step-index single-

mode fiber is given by

w = L/c n2 [0.080 + 0.549(2.834 – v)2 λ0/ λ0

for 1.4<v<2.6

If we assume L = 1 Km = 103m, λ 0 = 1nm and c=3x10-4 m/ps we get

Dw = - n2 /3 λ0 x 107 [0.080+0.549(2.834 –v)2] ps/Km.nm

Where λ0 is measured in manometers. As before, the negative sign indicates

that longer wavelengths travel faster. The total dispersions given by the

sum of material and waveguide dispersions:

Dtot = Dm+Dw

Let us consider the two single – mode fibers discussed earlier.



P a g e | 175

30

20

10

0

-10

-20

-30

1.1 1.2 1.3 1.4 1.5 1.6 λ0 (µm)

The wavelength dependence of Dm, Dw and Dtot

for a typical conventional single-mode fiber (CSF) with parameters as given in example 24.6. The total dispersion passes through zero around λ0 1300nm which is known as zero dispersion wavelength.

8.9 Let us sum up

On going through this lesson you would have understood about the

propagation of light in optical fibre. Also about the meaning of Acceptance angle and

Numerical aperture. Finally about the fabrication technique and the types of optical fibre

and their application.


(i) What is optical fibre?

(ii) Explain Acceptance angle?

(iii) What is meant by Numerical aperature?

(iv) What is intermodal dispersion


(i) Explain the wave propagation through optical fibre?

(ii) In an optical fibre, the core material has refractive index 1.6 and refractive

index of clad material is 1.3. What is the value of critical angle? Also calculate

the value of angle of acceptance.

CSF Dm

Dtot

Dw

Dis

per

sion (

ps/

nm

-Km

)



P a g e | 176

[ Hint: Critical angle sin fc =1

2

n

n ; Acceptance angle qa = sin -1 Ö n1

2-n22]

(iii) The numerical aperture of an optical fibre is 0.5 and the core

refractive index is 1.54. Find the refractive index of the cladding.

[Hint: n2 = Ö n12-(NA)2 ]

(iv) A step index fibre cable has an acceptance angle of 30° and a core index of

refraction is 1.4. Calculate the refractive index of cladding.

[Hint : sin2 qa = n12-n2

2 find n2 ]


(i) Write a note on Fibre losses

(ii) Explain different types of Fabrication technique of optical fibre

(iii) Write any two applications of optical fibres.

8.13 References

(i) Optics and Spectroscopy by R. Murugesan

(ii) Text book of optics by Brijlal and Subramaniyam and M.N. Avadhanulu.



P a g e | 177

UNIT – V

Lesson – 9

LASERS


this lesson provides a concise account of principles and operation of LASER. It covers the

characteristics of a LASER beam which make it distinct from the light beam.

Also it deals with the type of lasers. Some of the laser commonly used in practical

applications are described in detail. Semiconductor diode laser is given special attention and

the relevant theory and principles are discussed at length.

INTRODUCTION

It is one of the most important inventions of the 20th century. It is a light source but

differs vastly from common light sources and is not used in illumination purposes. Lasers are

useful in radio and microwave transmitters which produce coherent, monochromatic and

polarized beam of electromagnetic radiation. It occupies a unique place in modern

technology. It has wide applications in metalworking, printing, entertainment,

communications, medical diagnosis, surgery, ophthalmology, weapon guidance etc.

9.1 Characteristics of LASER

The discovery of laser made enormous impact in the scientific world and showed that the

function of optics is very much alive.

(i)Directionality

The light beam can travel as a parallel beam up to a distance d2/l, where d is the

diameter of the aperture through which the light is passing and l the wavelength of the

light used. After traveling a distance d2/l, the light beam spreads radially. In ordinary

light the angular spread is given by Dq = l/d. For a typical laser beam, the angular spread



P a g e | 178

is 1mm per1m, but for ordinary source of light, the angular spread of light is1m per1m.

This shows the directionality of the laser beam. For example the laser beam can be

focused to moon from the earth with an angular spread of a few kilometers.

(ii) Intensity

The intensity of the laser beam is very high. If a person is allowed to observe the

laser beam from the same distance, the entire laser beam penetrates through his eye and

will damage the eye of the observer. This shows the high intensity of the laser beam.

(iii) Monochromaticity

The band width (Dn=0) of the laser beam is narrow, while ordinary light spreads

over wide range of frequencies.

(iv) Coherence

The degree of coherence of the laser beam is very high than the other sources. The

coherence of laser emission results in extremely high power (5´1012 WCm-2). The height

from the laser source consists of wave trains that are identical in phase and directions of

propagation. Also it can be focused to a very small area of 0.7 mm thickness.

9.2 Principle pf LASER (Atomic basis for LASER action)

We can get an idea from the theory of interaction of radiation with matter

regarding the working of laser. Consider an atom that has only two energy levels E1 and

E2.when it is exposed to radiation having stream of photons, each with energy hn , three

distinct processes takes place:

(i) Absorption.

(ii) Spontaneous emission and

(iii) Stimulated emission.

(i) Absorption

An atom or molecule in the ground state E1 can absorbs a photon of energy hn

and go to the higher energy state E2.This process is known as absorption and is shown in fig

9.2



P a g e | 179

Fig. 9.2

The rate of upward transition R12 from ground state E1 to excited state E2 is proportional

to the population of lower energy level N1 (number of atoms per unit volume) and to the

energy density of radiation rn

i.e. R12µ rn

µN1

Thus R12= B12rn N1 ……………………. (1)

Where B12 is the probability of absorption per unit time.

Normally, the higher energy state is an unstable state and hence, the atoms will make a

transition back to the lower energy state with the emission of photon. Such a emission can

take place by one of the following two methods.

(ii) Spontaneous emission

In this the atoms or molecule in the higher energy state E2 return to the ground

state by emitting their excess energy spontaneously. This process is independent of external

radiation .the rate of the spontaneous emission is directly proportional to the population of the

energy level E2.

i.e. R21µN2

R21 = A21 N2……………… (2)

Where A21 is the probability per unit time that the atoms will

spontaneously fall to the ground state and N2 the number of atoms per unit volume in the

state E2.This process is shown in the Fig. 9.2 (i)



P a g e | 180

Fig. 9.2 (i)

Stimulated emission

In this a photon having energy E1 equal to the difference in energy between the two

levels E2 and E1, stimulate an atom in the higher state to make transition to the lower

state with the creation of second photon as shown in figure 9.2(ii)

Fig. 9.2 (ii)

The rate of stimulated emission R21 is given as

R21 = B21rn N2…………. (3)

Where B21 is the probability per unit time that the atoms undergo transition from higher

energy state to lower energy state by stimulated emission. Under conditions of thermal

equilibrium, the population of the energy levels obey the Boltzmann’s distribution

formulae.



P a g e | 181

9.3 EINSTEIN’S THEORY OF STIMULATED EMISSION

In 1917, Einstein proposed a mathematical expression for the existence of stimulated

emission of light. It is known as Einstein’s expression.

Consider two level energy systems (E1 and E2). Let N1 and N2 be the number of atoms

in the ground state and excited state. Assume that only the spontaneous emission is present

and there is no stimulated emission of light.

At thermal equilibrium condition,

The rate of absorption= the rate of emission of light

From the equation (1) and (2)

rn´B12 N1 = A21N2

112

221

NB

NA=nr (4)

according to Boltzmann’s distribution function, the population of atoms in the upper and

lower energy levels are related by

kTE

kTE

e

e

N

N/

/

1

2

1

2

-

-

= (5)

where k is the Boltzmann’s constant and T is the absolute temperature.

Substituting 1

2

N

N in equation (4)

( ) kTEEeB

A /

21

21 12 --=n

r (6)

kTheB

A/

21

21 1nnr

-=

according to blackbody radiation, the energy density



P a g e | 182

1

18/3

3

-=

- kThec

hnn

npr (7)

Where h is the Plank’s constant and c is the velocity of light. Comparing (6) and (7) one can

observe that they are not in agreement. To rectify this discrepancy, Einstein proposed another

kind of emission known as stimulated emission of radiation. Therefore, the total emission is

the sum of the spontaneous and stimulated radiation.

At thermal equilibrium condition

The rate of absorption= the rate of emission

From the equation (1), (2) and (3) we get

B21N1rn = A2N2 + B21N2rn

221112

221

NBNB

NA

-=nr

(8)

dividing each and every term on the R.H.S of equation (8) by N2, we get

21

2

112

21

BN

NB

A

-÷øö

çèæ

=nr

substituting for 2

1

N

Nfrom the equation (5)

( )( )21

/

12

21

21 BeB

AkTEE -

=-nr

we know that E2-E1= hn. Hence

( )21

/

12

21

BeB

AkTh -

=nnr

(9)

the coefficient A21,B12,B21 are known as Einstein’s coefficients.

Comparing equation (9) with (7) we get B12=B21 and

3

3

21

21 8

c

h

B

A np=

(10)



P a g e | 183

Equation (7) and (9), the ratio of the stimulated emission to spontaneous emission is

given by

( )( ) 1

1/

221

221

21

21

-==

- kTheNA

NB

SPR

STRn

nr

(11)

SP = spontaneous emission

ST = stimulated emission

From the equation (11)

Einstein proved the existence of the stimulated emission of radiation.

Note: the spontaneous emission produces incoherent light, while the stimulated emission

produces coherent light. In ordinary light source the spontaneous emission is dominant. For

laser action, stimulated emission should be predominant over spontaneous emission and

absorption. To achieve this, an artificial condition known as population inversion is required.

9.4 POPULATION INVERSION

Consider a two level energy system (E1 and E2). A photon of energy equal to the

energy difference between the two levels is incident on the system; Einstein showed that

under normal circumstances both the process absorption and stimulated emission are equally

probable.

In a system containing a very large number of atoms, the dominant process will

depend on the virtual number of atoms in the upper and lower state. A large population in the

upper level (N1< N 2) will result in stimulated emission dominating over the absorption. If

there are more number of atoms in the lower level i.e. ground state (N1> N2) there will be

more absorption than stimulated emission.

Under conditions of thermal equilibrium, Boltzmann’s distribution function relating

N1 and N2 given by equation (5) is obeyed

( ) kTEEe

N

N /

1

2 12 --= (12)

where k is the Boltzmann’s constant and T is the absolute temperature. For stimulated

emission to be dominant, it is necessary to increase the population of the upper energy level,

so that it is greater than that of the lower energy level. This is known as population inversion.



P a g e | 184

In the above equation, if T is negative, then stimulated emission would dominate over

absorption. Such a condition is not possible. Different methods to increase population

inversion are given below

9.4.1 METHODS OF ACHIEVING POPULATION INVERSION

(i) Optical pumping

In this method, an external source like xenon flash lamp is employed to produce a

high population in the higher energy level of the laser medium. This is illustrated in

fig9.4.1(a). This method is used in solid state laser.

(ii) Direct electron excitation

The direct electron excitation in a gaseous discharge may be used to produce the

desired inversion. This method is used in some gaseous ion laser. In this type shown

in fig9.4.1(b), laser medium itself carries the discharge current under suitable

conditions of pressure and temperature. In this method the electrons directly excite the

active atoms to achieve higher population in certain higher energy levels compared to

the lower energy levels.

(iii) Inelastic atom-atom collisions

Here, electric discharge method is employed to cause collision and excitation of

the atom. In this method, a combination of two types of gasses are used say A and B, both

having the same excited state A* and B* that coincide. In the first step, during electric

discharge, A gets excited to A* (metastable) due to collision with electrons

A + e ® A* +e1 (13)

A* + B ®A + B*

The excited A* atoms now collide with B atoms so that the latter atoms gets excited to

higher energy B*. This type of transition is used in He-Ne laser.



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(iv) Chemical reactions

In this method, the molecules undergo chemical changes in which one of the product of

the reaction is a molecule or an atom that is left in an excited state under appropriate

conditions. Now population can occur. An example of this type of laser, in which

hydrogen fluoride molecule in the excited state result from the following chemical

reaction

H2 + F2 ® 2HF

9.4.2 METASTABLE STATES

Normally excited states have short life times and release their excess energyin a

matter of nanoseconds (10-9sec) by spontaneous emission. Atoms do not stay at such

excited states long enough to be stimulated to emit their energy. Though, the pumping

agent continuously raises the atoms to the excited level, many of them rapidly undergo

spontaneous transitions to the lower energy level. Population inversion can not be

therefore established. For population inversion the excited atom are required to wait at

upper level till a large number of atoms accumulate at that level. Thus, what is needed is

an excited state with a longer life time. Such a longer- lived upper levels from where an

excited atom does not return to lower level at once, but remains excited for an appropriate

time, are known as metastable states. Phosphors are an example of materials having

metastable states. They emit persistent- light called phosphorescence because of

metastable states excistent in them. Atoms stay in metastable states for about 10-6- 10-3s.

This is 103 to 106 times longer than the time of stay of atoms at excited levels. Therefore,

Fig. 9.4.1



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it is possible for a large atoms to accumulate at metastable level. The metastable state

population can exceed the population of the lower level and lead to the state of population

inversion.

If the metastable states do not exists there could be no population inversion, no

stimulated emission and hence no laser action.

9.4.3 OPTICAL RESONATOR

In laser, feedback is obtained by placing the active medium between pair of mirrors

which are facing each other. The photons triggered many stimulated emissions and these

photons at the end are reflected back into the medium by the mirror. The photon reverse

their direction and travel through the medium till they reach the mirror at the other end.

Hence the pair of mirrors contributes an optical resonator. This structure is actually an

open resonator as the sides open.

As the light bounces back and forth in the optical resonator, it undergoes

amplification as well as it suffer various losses. These losses occur mainly due to the

scattering and diffraction of light within the active medium. For the proper build up of

oscillations, it is essential that the amplification between two consecutive reflections of

light from rear end mirror can balance the losses. The mirrors could be either plane or

curved and are designed such that one of them reflects all the light that reaches it while

the other one reflects most of the light incident on it. A small portion of light is

transmitted through this mirror as laser output beam. The basic components of laser are

shown in fig. 9.4.3

Fig. 9.4.3

(a) Closed cavity used at microwave frequencies

(a) (b) (c)



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(b) Open cavity used in optical frequencies

A laser consist of an active medium (AM) placed between two mirrors (M and OM)

and is pumped by pumping agent (P). An additional element (IE) may be located within

the cavity to achieve certain desired characteristics in the laser outputs.

9.4.4 ACTION OF OPTICAL RESONATOR (LASER ACTION)

In fig.9.4.4 shows the action of an optical resonator. In this the active centres in the

laser medium are in ground state (Fig.9.4.4 (a)). Through a suitable pumping mechanism,

the medium is taken into the state of population inversion (Fig.9.4.4 (b)). Here the

possibility is that some of the large number of excited atoms decays spontaneously. They

emit photons in various directions. Each spontaneous photon can trigger many stimulated

transitions along the direction of its propagation and stimulated photon will travel in

different directions. In the absence of end mirrors, the net effect would have been that

incoherent light is produced without a specific direction. In the presence of end mirrors a

specific direction is imposed on the photons. Photons propagating along the optic axis of

the pair of mirrors are restricted whereas photons emitted in the other direction will pass

out of the sides of the resonator and are lost, shown in (Fig.9.4.4 (c)).

A majority of photons travelling along the axis are reflected back on reaching the end

mirror. They propagate towards the opposite mirror and on their way they stimulate more

and more atoms and build up their strength as shown in the fig.9.4.4(d). Thus the mirror

provide positive feedback of light into the medium so that stimulated emission acts are

sustained and the laser operates as an oscillator.

The photons that strike the opposite mirror are reflected once more in the medium fig

9.4.4(e). The photons will travel along the axis generating more photons and cause more

and more amplification. Hence stimulated emission increased. At each reflection at the

front end mirror, light in partially transmitted through it. The transmitted component

constitutes a loss of energy from the resonator. When the losses at the mirrors and within

the medium balance the gain, the laser oscillations build up and laser output beam will

become bright as in fig 9.4.4(f). A highly collimated intense beam emerges from the laser.



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As the end mirrors reflected strongly into laser medium, the light travels are higher

within the resonator. Consequently, the radiation density r(n) is large within the medium

and sustains the stimulated emission. It is desirable that the laser oscillations at one

frequency produce monochromatic light. This is achieved by coating the mirror with

suitable material which is highly reflect photons of the desired frequency and at the same

time absorb the photons of the undesired frequency.

Fig. 9.4.4 Light amplification and oscillation due to action of optical inversion



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Threshold condition

For the proper build up of oscillations, it is essential that the amplification between

two consecutive reflections of the light from rear end mirror can balance the losses.

Threshold gain can be calculated by considering the change in intensity of a beam of light

undergoing a round trip within the resonator.

Let us assume that the laser medium fills the space between the mirrors M1 and M2

which have reflectivity r1 and r2. Let L be the distance between mirrors. Let the intensity

of the light beam be the I0 at M1. Then in travelling from M1 to mirror M2, the beam

intensity increases from I0 to I (L), which is given by

I (L) = I0 e (g-as)L (1)

Where g(=-a) is referred to as the gain coefficient per unit length and is a positive

quantity. After reflection at M2, the beam intensity will be

I (2L) = r1r2I0e (g-as) 2L (2)

The amplification obtained during the round trip is

( ) ( ) LserrI

LIG 2

21

0

2 ag -== (3)

The product r1r2 represents to losses at the mirrors whereas as includes all the distributed

losses such as scattering, diffraction and absorption occurring in the medium. These

losses are balanced by gain, when 1³G or I(2L)=I0. it leads to

( )

12

21 ³- Lsrr

ag

( )

21

2 1

rre Ls ³-ag

taking log on both sides, we get

( ) 21ln2 rrL s -³-ag



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( ) 21ln2

1rr

Ls -³-ag

21ln2

1rr

Ls -³ ag

21

1ln

2

1

rrLs +³ ag

The above equation is the condition for the lasing. It shows that the initial gain must

exceed the sum of the losses in the cavity. This condition is used to determine the

threshold value of pumping energy necessary for the laser action g, the amplification of

the laser will be dependent on how hard the laser medium is pumped. As the pump power

is slowly increased, a value of gth called threshold value will be reached and the laser

starts oscillating. The threshold value is

21

1ln

2

1

rrLsth += ag (5)

Therefore for the laser to oscillate,

thgg ³

the above equation is the threshold condition for lasing and states the criterion when the

net gain would be able to counteract the effect of losses in the cavity.

9.5 SOLID STATE LASER.

The term solid state has different meanings in the field of electronics and laser. In this

type the active centres are fixed in a crystal or glassy material. They are electrically non-

conducting. Also called doped insulator laser.

The basic principles that underline the operation of solid state laser is shown in

Fig 9.5.5. The active centres are dispersed in a dielectric crystal or a piece of glass. The

crystal atoms do not participate directly in the lasing action but acting as a host lattice to

the active centres which are present in small concentrations. The crystal is usually shaped

into a rod with reflecting mirrors placed at each end. Light from an external source such

as flash lamp excites the active centres in the rod and linear in shape. The linear lamp and

laser rod are placed close to each other in a reflective cylinder, which focusses the pump



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light on the rod. The mirrors on the sides of the laser rod form a resonant cavity and

provide necessary feedback to generate laser beam.

It is required tha the host material should be transparent to the pump light and should not

absorb light at the laser wavelength. Most of the excitation energy ends up as heat rather

than light. Excess heat damages the laser crystal and hence has good thermal

conductivity. Normally, cooling can be done by circulating water.

Fig. 9.5 Solid State Laser

The active centres are ions of the metallic elements like Chromium, Neodymium (Nd),

erbium (Er), cerium (Ce) and titanium (Ti). These lasers are simple to maintain and

capable of generating high peak powers.

9.5.1 SEMICONDUCTOR LASER.

A semiconductor diode laser is a specially fabricated pn junction device which emits

light when it is forward biased. It is made from gallium arsenide (Ga-As) which operated

at low temperatures and emitted light is in the near IR region.

It is a pn junction diode with a p-type and n-type regions heavily doped. Under large

applied forward bias, electrons and holes are injected into and across the transition region

in considerable concentration. As a result, the region around the junction contains a large

number of electrons within the conduction band and a large number of holes within the

valence band. When the population density is high enough, a condition of population

inversion is achieved and recombination may be stimulated resulting in laser action.



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In the case of semiconductor laser there is no need of external mirrors. In a

germanium or silicon semi-conductor, due to such a recombination, only heat is

generated.

A typical Ga-As laser is shown in fig.9.5.1

9.5.1 Ga-As laser schematic representation

The Ga-As laser convert electricity into light. The efficiency can go up to 100% when the

temperature is reduced to 100K. The operating current is supplied from a pulse generator

upto 5-20ms. The energy separation between the conduction band and valence band is 1.4

eV and hence, the wavelength of the light emitted is 8874A° at room temperature.

Advantages

1) The efficiency is more and can be increased by decreasing the temperature

alone.

2) It can have a continues wave output or pulsed output.

3) The modulation of the output is possible.

4) It is highly economical and the arrangement is compact.

Disadvantage

The spatial and temporal coherence are very poor.

9.5.2 GAS LASER- HELIUM-NEON LASER



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The advantage of gas laser is that they can be operated continuously. The gas laser

show exceptionally high monochromaticity and high stability of frequency. The output of the

laser can be tuned to a certain available wavelength. Hence the gas lasers are widely used in

the industries.

Helium-Neon LASER

Fig. 9.5.2 He-Ne laser

The two important parts of the laser are as follows.

Active medium:

The active medium used in this type of laser is a mixture of helium and neon gas.

These two gases are mixed under pressure of 1mm Hg of helium and 0.1mm Hg of neon in

the ratio of 10:1 the mixture of these gases is filled within the discharge tube for laser action.

Gas discharge tube

It is made up of fused quartz tube with a diameter of 11.5 cm and length 80 cm. The

end faces of the discharge tube is tilted at the Brewster angle known as Brewster window,

since the window are transparent to the preferred direction of polarisation. A fully reflecting



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concave mirror is placed at one end of the discharge tube and a partially reflecting concave

mirror at the other end.

Working Principle

The population inversion for laser action was achieved by inelastic atom-atom

collision. The collisions of the atoms are made using any one of the following methods.

(i) Direct current discharge

(ii) Alternate current discharge

(iii) Electrodeless high frequency discharge

(iv) High voltage pulses

Due to electric discharge in the gas, an energetic electron interacts with the ground state

helium atoms. The impact of the electron results in exchange of some of its energy to the

helium atom. As a result, helium atoms are excited to higher levels1S0 and 3S1 known as

metastable state. The life time of these levels are relatively low. The collision of the first kind

is represented as

He + e 1 ® He* +e2 (1)

These two energy levels are very close to 2s and 3s levels of the neon atom and the collision

of the second kind takes place between the He and the Ne atoms, the neon atom goes to the

excited state,

He* +Ne ® He +Ne* (2)

Explanation

Therefore there are three types of transitions one from 2S to 2P the other from 3S to

3P and another from 3S to 2P levels of the laser action. These transitions constitute laser

beam in the infrared region (11,523 A° , 33912 A° ) and 6328 A° in the visible region. The

transition from 3S to 3P at 3.39mm will have adverse effect on the laser emission. The line is

suppressed in order to get a maxim um power output at 6328 A° .



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In the fig 9.5.2(a), the energy levels are represented in terms of Paschen’s notation

and their corresponding electronic configuration are given in parentheses. The neon atom in

the terminal level 2P decay very rapidly to the 1S metastable state in 10-8 s, much faster than

spontaneous rate of decay from 2S to 2P level. Thus, the lower is relatively kept empty and

population inversion achieved between 3S and 2P.

9.5.3 MOLECULAR GAS LASERS-CORBON DIOXIDE LASERS(CO2)

This laser was first developed by C.K.N.Patel

Principle: To understand this, one has to recall the rotational and vibration spectrum of CO2

molecules. The three atoms can be considered as a ring over a straight line, the outer atom

being O with a carbon atom C at the centre. There are three models of vibrations and in each

mode the centre of gravity remains fixed.

Fig. 9.5.3 Co2 molecules- different modes of vibration

1) As shown in above fig 9.5.3(a), the carbon atom is fixed in its position and each oxygen

atom vibrate in the opposite direction symmetrical to the carbon atom with each other along a

straight line and is known as symmetric mode of vibration. The corresponding frequency is

called symmetric stretching frequency.

2) as shown in fig 9.5.3(b), the oxygen atom and the carbon atom may vibrate at right angles

to the line passing through centre of gravity. This is known as the bending mode and the

corresponding frequency is bending frequency.

3) as shown in fig 9.5.3(c), in asymmetric mode of operation, two oxygen atoms may vibrate

about the central C atom asymmetrically, and at the same time the carbon atom also vibrates



P a g e | 196

from its mean position. The corresponding frequency is called asymmetric stretching

frequency.

In addition to the above three vibrational modes the molecule can rotate and the quantized

rotational energy levels are also possible. A series rotational levels is associated with each

vibrational levels are denoted by J values.

Construction of CO2 laser

This laser depends on radiation power, i.e. output-power on the diameter of the tube.

It can be raised by increasing the diameter. In a powerful CO2 laser, the length of the

discharge tube is several metres and its diameter will be several centimeters as shown in the

fig 9.5.3(d)

Fig. 9.5.3(d) Co2 laser

The laser is powered with an a.c supply of frequency 50 cycles or d.c supply. In order to get a

high output power, a metallic mirror of gold is employed for proper reflection, the laser are

either water cooled or air cooled. The efficiency of CO2 laser is about 30%. The gas mixture

can be pumped either longitudinally or transversely into the gas discharge tube.



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Working principle

The CO2 laser uses two additional gasses, N2 and He. The nitrogen plays a similar

role as that of the He in the He-Ne laser. The N2 molecules go into an excited state by

collision of first kind with the electrons.

N2 +e1 ® N2* + e2 (1)

The excited atom undergoes a collision of the second kind and makes the CO2 molecule to be

excited.

N2* + CO2 ® CO2* + N2 (2)

The frequency of the CO2 laser in an energy level diagram is shown in fig 9.5.3(e)

Due to various factors, the most powerful transition in CO2 laser at normal operating

temperature occurs at 10.6mm. The operating temperature plays an important role in

determining the output power of the laser. The contamination of carbon monoxide and

oxygen will also have some effect in the laser action. The unused gasses can be

Fig. 9.5.3(e) Co2 laser- Energy level diagram

pumped out and fresh Co2 must be pumped. The temperature can be reduced by restricting the

tube diameter and also the addition of helium to the mixture of N2 and He. The helium serves

not only to improve the conductivity of heat to the walls of the tube, but also decreasing the

population in the lower levels. The power output coming from this laser is 10KW. It is a four-

level molecular laser and operates at 10.6 mm in far IR region.



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9.5.4 Ion gas lasers à Argon laser

Helium, Neon, Argon, Xenon and Krypton are noble gases and they have electronic states

capable of laser transitions. If these noble gases are first ionized by electron collisions then

they are easy to pump.

Argon- ion laser:

General description: The argon laser belongs to the group of ion lasers. It is a four- level laser

which operates in visible region over a wavelength range from 3510 to 5200 . It is used in

laser light shows. The Argon-ion laser can provide 25 visible wavelengths ranging from

4089 to 6861 and more than 10 UV wavelengths ranging from 2750 to 3538 .

Structure: In Argon lasers, the active medium is argon gas and the active centres are ionized

Argon atoms. It is shown in fig 9.5.4.

Fig. 9.5.4 ion laser tube

It consists of a narrow-water-cooled ceramic tube in which an arc discharge takes

place. The electrodes are arranged at ends of the capillary tube. The anode and cathode

spaces communicated through a return gas path which ensures free circulation of the gas. A

magnet surrounds the discharge tube. Its function is to reduce the discharge area and increase

the concentration of ions along the axis of tube. This increases the output power and

efficiency.

Working: The initial high voltage pulse ionizes the gas so that it conducts current Electrons in

the current transfer energy directly to Argon atoms, ionizes them and raises the ions to a

group high energy levels, shown in the fig 9.5.4(i). It is about 35eV above the ground state of



P a g e | 199

neutral Argon atom. Different processes populate the metastable upper laser level. They are

(a) electron collisions with Ar positive ions in the ground state, (b) collisions with ions in

metastable states and (c) radiative transitions from higher states. The life time of upper level

is 10-8sec while that of the lower level is of the order of 10-9sec. Therefore, the condition for

population inversion is satisfied. Transitions can occur between many pairs of upper and

lower lasing levels and therefore many laser wavelengths are emitted.

But of the large number of wavelengths emitted, the most important and the most

intense lines are 4881 (blue) and 5145 (green). Argons ions quickly drop from the

Fig. 9.5.4(i) Energy level for an argon atom

lower level to the ground state of the ion by emitting UV light at 740 . The ground state ion

either recaptures an electron and becomes a neutral atom or it is again excited to the upper

levels.

During the operation the positive ions tend to collect at cathode where they are

neutralized and diffuse slowly back into the discharge. However it leads to a pressure

gradient. A gas return path is provided between anode and cathode to equalize the pressure.

Without such return path, the discharge may eventually be extinguished.



P a g e | 200

This laser needs active cooling. Any desired wavelength can be selected through

proper cavity optics. For example, a small prism may be inserted into the optical cavity and

the position of the end mirror is rotated such that it comes into a position normal to the path

of the light of desired wavelength. Then, only the desired wavelength is sustained in to and

fro reflections while the other wavelengths are lost from the cavity after a few reflections.

9.5.5. SOLID STATE LASERS

RUBY LASER:

In solid state laser, the active medium is a crystalline substance. It is the first

successful laser achieved by Maiman in 1960. The experimental setup is shown in figure

9.5.5. It has three parts

Fig. 9.5.5 Ruby Laser

(i) Ruby rod: Ruby is a crystalline substance of aluminium oxide doped with 0.05% by

weight of chromium oxide. The resultant pink colour is due to the presence of Cr3+ ions in the

appropriate concentration which replace aluminium atoms in the crystal lattice.

(ii) Resonating cavity: For this, a pink rod of 4 cm length and 0.5 diameter is used. The end

faces of the rod are grounded so that they are parallel and polished to a high degree. Further,

the end faces are silvered in such a way that one end face becomes fully reflecting while the

other end partially reflecting. In some cases, separate glass pieces are attached at the end

faces.



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(iii) Xenon flash tube: The laser action is achieved by using optical pumping. Helical xenon

flash tube that surrounds the ruby rod provides the pumping light to raise the chromium ions

to the upper energy levels. The flash of the xenon tube lasts for several milliseconds and the

tube consumes several thousand joules of energy. Only a part of the energy is used to excite

the Cr3+ ions, while the rest heats up the apparatus. A separate cooling arrangement is used to

reduce the temperature.

Ruby Laser-Energy level diagram

Working principle: The energy level diagram illustrating the operation of ruby laser is

shown above. The pumping light from the flash tube is absorbed by the Cr3+ ions raising them

from the ground state E0 to the excited state E1 or E2. From these levels, a rapid radiationless

transition to the level E, which is a metastable state, takes place. Decay from E is relatively

slow so that with sufficient excitation, population inversion between E and the ground state

E0 can occur. The photons are allowed to pass back and forth millions of times in the active

medium with the help of mirrors at the ends. When the condition for laser action is satisfied,

an intense pulse of light of the wavelength 6934 corresponding to the transition E to E0 is

obtained.



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9.6 Q-SWITCHING

There are basically three different techniques to obtain high pulse from a laser. They

are known as Q-switching, cavity dumping and mode locking. For many of the practical

applications, we require large powers even for a short time. Q-switching and cavity dumping

make large peak powers available for short time.

The method of controlling the laser output power is called Q-switching. Evaluation of

giant- laser pulse through Q-switching is shown in fig.9.6. Q-switching is widely used, but not

all laser can be Q-switched.



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Fig. 9.6 Evolution of a giant pulse through Q-switching

Method of Q-Switching –Electro Optic Shutter:

It can serve as voltage controlled gate which rapidly switches the cavity from high

loss to a low loss condition. It consists of a crystal that becomes double refracting when an

electric field is applied across the crystal. It is shown in fig. 9.6(i)



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9.6 (i) Electro-optic cell used as Q-switch

A voltage applied to the crystal during the pumping of the laser by the light from the

flash lamp. The magnitude of the voltage is chosen such that it transforms the electro-optic

crystal into a quarter wave plot. The light emitted by the laser becomes linearly polarized

light on passing through the polarizer. The linearly polarized light incident on the crystal

splits into mutually orthogonal components. As the two component travel through the electro-

optic crystal, a phase retardation of 900 is produced between them. On emerging from the

crystal, the two components combine to produce circularly polarized light. The light beams

reflect from the mirror and returns to the cavity travelling in the opposite direction. On

reflection the sense of rotation of circularly polarized light reverses in and on re-passing

through the crystal, the two components of circularly polarized light experience a further

retardation of 900 with respect to each other. Coming out of the crystal at the other end, the

components recombine to produce linearly polarized light. The direction of polarization of

light is now at 900 with respect to its original direction of polarization and transmission axis

of the polarizer. This light is not allowed to pass through the polarizer. Hence, light does not

comeback to the laser rod and the cavity is switched off. Thus cavity Q is reduced to a low

value. When the voltage applied to the crystal turned off, double refraction is absent in the

crystal and the state of polarization of the light passing through the crystal is unaffected.

Light now freely travels in both direction and returns to the laser rod to be further amplified.

Now the cavity is switched on and the Q regains its high value. The Q-switching is

synchronized with pumping mechanism such that the voltage applied to the crystal drops to

zero value at the time when the population inversion in the laser medium attains its peak

value. Two types of electro-optic shutters known as Kerr cell and Pockels cell are available



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for Q-switching. In this Pockel cell is preferred over the Kerr cell because of the lower

voltage needed to produce desired effect.

9.6.1 MODE LOCKING FOR ULTRA SHORT PULSES

Q-switching produces giant pulses of duration of about 10-7 to 10-8 sec. The technique

of mode locking allows the generation of pulses of lesser duration of about 10-11 to10-12 sec.

The matching of phase of different modes are called mode locking. An essential requirement

for mode locking is existence of multimode operation in the active medium. The broadened

laser cavity in general supports oscillation in many axial modes simultaneously. The axial

modes are uniformly spread out on the frequency scale and are separated by ν given by

c/2mL.The resultant output of an ordinary laser (laser without mode locking) depends on the

relative frequencies, phases, phases and amplitudes of these modes. All these parameters are

time-varying, because each of the modes oscillates independently on the other mode.

The following figures illustrates

(i) output of a non-mode locked laser consisting of random fluctuations in intensity

(ii) Output of a mode locked laser, consisting of a narrow intensive pulse.

(iii) All modes are in phase

(iv) Time spacing and duration of pulses are produced by a mode- locked laser.

Thus we find the output of a mode- locked laser consist of a sequence of short pulses

at time intervals of 2µL/c; which is equal to the round trip transits time for light within

the cavity. Each pulse has a peak power equal to N times the average power. The

mode-locked condition can be viewed as a condition in which a pulse of light is

bouncing back and forth inside the cavity and every time it hits the mirror, a fraction

of it is transmitted as the output pulse



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Output of a non-mode laser consisting of a random fluctuations in intensity

Output of a mode locked laser consisting of a narrow intense pulses



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Time spacing and duration of pulses produced by a mode- locked laser

9.6.2 TECHNIQUES FOR MODE LOCKING

The most common methods are

1. Mode locking by modulating the loss of the cavity externally

2. Mode locking by means of a saturable absorber placed inside the laser cavity.

In the first an acoustic or electro-optic modulator is driven by an external signal and hence it

is called active mode locking. In the second, employs a saturable absorber whose absorption

coefficient varies non-linearly with light intensity and is called passive model locking



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1. Active mode locking:

Kerr cell and Pockels cells or the acousto–optic shutter can be used as modulators for

achieving mode locking. If we consider electro-optic modulator (EOM) is kept inside the

cavity. Let the frequency of one of the axial modes be νm. Let the frequency of modulation be

equal to ∆ν, the intermode frequency separation. Since, the signal with frequency ∆ν is

applied to the electro-optic modulator, the loss of the laser activity is modulated at the same

frequency ∆ν. Consequently, the amplitude of the mode corresponding to ν m is also

modulated at ∆ν. The amplitude modulated mode at a frequency ν m generates two side bands

at frequencies ν m+ ∆ν and νm- ∆ν. Since ∆ν is chosen such that it is equal to intermode

spacing, the waves having new frequencies νm+ ∆ν and νm- ∆ν coincide with the two modes

located on either side of νm. Three modes become locked in phase. Since, the amplitudes of

these new modes are also modulated at the frequency ∆ν, they generate new side bands which

in turn correspond to some other axial modes. The process continues and at the end all modes

are forced to oscillate with a definite phase relationship. As a result, mode locking is

achieved.

2. Passive mode locking

Here a saturable absorber (an organic dye) placed within the laser activity. The solution has

the property of becoming more and more transparent as the intensity of the light falling on it

increases. The saturable absorber placed adjacent to one of the resorant mirrors. Initially, the

laser medium emits spontaneous radiation which forms within the cavity a radiation field

consisting of fluctuations of intensity having a noise like structure. Some of the intensity

peeks within the fluctuations bleach the saturable absorber more than the other components

and pass through. They get amplified in its round trip through the active medium. If the cell

recovery time is too short, the cell becomes opaque (absorbing) immediately after this pulse

passed through. As other fluctuations are weak, they are attenuated in the cell. Thus, the

strongest pulse will grow faster while the other less intense fluctuations are suppressed. At

regular intervals of time T, this pulse arrives at the output mirror and it is partially emitted.

Therefore, the laser output consists of a regular sequence of pulses with a pulse repetition

time of T.



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9.7 APPLICATION

LASER is bringing revolutionary changes in our lives. They are applied in

entertainment electronics, communications, surgery and related medical fields, information

processing and even in warefare. The list is unending and continue to grow. Laser

applications are broadly divided into two groups, one involving beams of high power and the

other involving beams of low power. High power, gas and solid laser are used in material

processing, nuclear fusion, medical field, defence etc. the low power semi-conductor laser are

used in CD players, optical floppy discs, optical memory cards, data processing and

information processing devices, hologram preparation and viewing, range finders, optical

communications etc.

9.7.1 LASER IN MECHANICAL ENGINEERING

It is used in cutting, drilling welding etc. jobs to be done on both metals and non-

metals. It requires transfer of energy from the laser beam to workpiece. It can happen only if

the material has high absorption at the wavelength corresponding to the laser beam. Once the

surface of the material absorbs energy, the material starts to melt and then vaporize. At high

intensity of radiation the vapour will be ionized to produce plasma. The plasma layer formed

between the laser and the workpiece absorbs light and prevents the laser beam from reaching

the workpiece. Therefore it is essential that the plasma should be removed to increase energy

coupling.

(i) Drilling: Drilling holes by a laser beam is based on the intense evaporation of materials

heated by a series of powerful light pulses of short duration of 10 -4 to 10 -3 sec. The energy

supplied for drilling should be such that rapid evaporation of material takes place before

radial distribution of heat into the work piece occurs. Use of short pulses minimises the

energy diffused laterally into the work piece and assists in controlling the size and shape of

the hole. Laser drilling is a non-contact process and does not require a physical drill bit. The

problems of wear and broken drills do not arise. The process becomes faster and drilling

operation can be done with extremely high precision and in any desired direction. CO2 laser

and Nd-YAG laser are used, the former laser is suitable for drilling in metallic and non-

metallic materials, the latter is used for drilling holes in metals only.



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(ii) Cutting: A wide range of materials can be cut by CO2 lasers. The materials include

paper, wood, cloth, glass, quartz, ceramics, steel etc. Laser cutting is done with the assistance

of air, oxygen or dry nitrogen gas. The advantage of laser cutting is that it is fine and precise.

The process does not introduce any contamination. It is automatised and high production

rates can be achieved.

(iii) Welding: It is joining of two or more pieces into a single unit. The metal plates are held

in contact at their edges and a laser beam is made to move along the line of contact of the

plates. The laser beam heats the edges of the two plates to their melting points and causes

them to fuse together where they are in contact. The advantage in this is that there is no

possibility of introduction of impurities into the joint. The work pieces do not get distorted as

the total amount of power input is very small compared to the conventional welding

processes. The most common laser used in welding is the CO2 laser. Gases like He, Ar, N2

are often used with laser welding for protection against oxidation of metal surfaces.

9.7.2 Lasers in Defence

Here it involves mainly ranging , guiding weapons to the intended target and laser

beam itself acting as a weapon. The optical radars, using laser beams, are employed in

detecting distant objects and collecting information about them. This is called range finding.

Nowadays, low flying aircrafts are used in ground attacks. They are fitted with laser

instrumentation (CO2 laser) for measuring the range of the target and guiding the bomb. Also

used in intercontinental Ballistic Missiles, which can hit the target within a few minutes. It is

required that the missile should be detected and destroyed before it can reach the target.

9.7.3 Lasers in Medicine

(i) They are used in destroying stones and gallstones. An optical fibre is threaded through

until it faces the stone directly. Laser pulses launched through the fibre shatters the stone into

small pieces that can pass through without pain.

(2) At times Tumors develop in brain and spinal cord. They cannot be operated in

conventional way because of their delicate nature. Laser surgery in such areas is possible and

safer.



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(3) Laser welding of human blood vessels is more advantageous than conventional suturing.

Argon laser with 1mm spot size is used for this purpose.

(4) It is possible with laser to selectively destroy cancer. In the treatment, a dye called

hemotopophyrin derivative (HPD) is injected into the patient body. The healthy cells flush

out the dye and the dye in the concentrated in cancer cells. The suspect areas are illuminated

with red light (628-632nm) and cells having HPD absorb light strongly. As a result, the dye

reacts chemically with oxygen in cancer cells and produce a highly reactive form of oxygen

that kill the cancer cells.

9.8 Let us sum up

From this unit you have learnt about the principles and operation of lasers. It dealt

with different types of lasers, its relevant theory and principles. Also it dealt with the

characteristics of laser beam which makes distinct from the light beam generated by

common light sources. Finally Q-switching and mode locking and their advantages

are discussed which will be useful. Finally laser application in mechanical, defence

and medical field are discussed.


(i) What is population inversion and spontaneous emission?

(ii) What are the characteristics of LASER light?

(iii) What is metastable state and mention its importance?

(iv) What is the working principle of He-Ne Laser?

(v)What is the reason for monochromaticity of laser beam?


(i) Find the ratio of population of the two states in He-Ne laser that produces light

of wavelength 6328 at 27°C .(Ans: 1.1*10-33 )

[Hint: N2 / N1 = e- (E2-E

1)/KT

E2-E1 = 12400/6328 eV = 1.96 eV]



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(ii) The half-width of the gain profile of a He-Ne laser material is about 2 * 10-3 nm

(∆λ). What should be the length of the cavity in order to have a single longitudinal

mode oscillation? [λ = 6328*10-10m and µ = 1] (Ans: 10cm)

[Hint: L = λ2/2µ∆λ]


(i) What do you understand by an optical resonant cavity? Explain.

(ii) What is a semi-conductor laser? Discuss its main features and condition for action.

What are its merits and de-merits.

(iii) (a) Describe the working of solid state laser.

(b) Explain the principle and working of a He-Ne laser.

(iv) Write short notes on (a) Q-Switching

(b) Mode locking

(c) CO2 laser

9.12 References

(i) An introduction to LASERS

by M. N. Avadhanulu (S. Chand & Co)

(ii) Optics

by Brijlal & Subramaniam and M. N. Avadhanulu



Electromagnetism (B U study material)

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