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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 1
1. Describe Michelson and Morley experiment with the help of
neat diagram and discuss its negative result. Answer: In 1887
Michelson and Morley performed an experiment to find the velocity
of earth relative to ether which was assumed to be stationary or an
absolute frame of reference in which the speed of light is same (=
c) in all direction
Fig.1. Experimental arrangement of Michelson-Morley
experiment
Where, S = Monochromatic light source, P = partially silvered
glass plate which inclined at 45 with the vertical, m1 and m2 =
Perfectly reflected glass mirror, T = Telescope.
Fig.2. Theory of the Michelson- Morley experiment
Where, lMOPM;ctPMMP;vtMMPOOP 111111
In 11MPM ,
2
2
2
22
22
22
222222
2222222222
2222
11
22
1
c
v1c
lt
c
v1c
lt
vc
ltlvct
lvctctvltc
ctlctMMPMMP 1
---------- (1) Now the total time of travel of light from the
plate P to the mirror M1 and the back from M1 to the plate P is
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 2
2
221
2
2
2
21
c2
v1
c
l2
c
v1
c
l2
c
v1c
l2t2t {By binomial expansion} ------ (2)
And the total time of travel of light from the plate P to the
mirror M2 and the back from M2 to the plate P is
c2vc
lvcvc
vc
l)vc(vc
vc
l
vc
l
vc
lttt
222222bf2
2
21
2
2
2
2
2
22
222 c
v1
c
l2
c
v1
c
l2
c
v1c
l2
c
v1c
lc2
vc
lc2t ;
{By binomial expansion} ------ (3) The time difference in the
time of travel of the light in the two mutually perpendicular
directions
3
2
2
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
12
c
lv
c2
v
c
l2
c2
vv2
c
l2t
c2
v
c
v
c
l2
c2
v1
c
v1
c
l2
c2
v1
c
v1
c
l2
c2
v1
c
l2
c
v1
c
l2ttt
------------- (4) Hence, the path difference for the light rays
travelling in the two mutually perpendicular directions is
2
2
3
2
c
lv
c
lv.ctcx -------- (5)
Now the entire apparatus is rotated by 90. Therefore the path
difference is
2
2
c
lvx ------- (6)
Thus, the change in path difference between the light waves
coming from the two directions due to rotation of the apparatus by
90 is
2
2
2
2
2
2
2
2
2
2
c
lv2
c
lv
c
lv
c
lv
c
lvxxx
---------- (7)
In the Michelson-Morley experiment .Sec/m103c.;Sec/m103v;m11l
84
m1022
103
10322
103
103112
c
lv2x 8
16
8
28
24
2
2
--------- (8)
For the yellow light ( mean wavelength =5500=5.510-7m ) the
number of fringe shifted is
4.055
22
105.5
1022xn
7
8
----------- (9)
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 3
This result is called negative result of Michelson-Morley
experiment because no fringe shift was observed by Michelson and
Morley when the experiment was repeated after six month.
Conclusions from the negative result- 1. The velocity of earth is
zero relative to the ether. 2. The speed of light does not depend
on the motion of source or on the motion of observer. 3. The
concept of ether to be stationary is found to be wrong. 2. State
the basic postulates of Theory of Relativity. Answer: The following
are the two basic postulates of special theory of relativity
proposed by Einstein- 1. All the frames of reference in relative
motion with a constant velocity are equivalent for the description
of the laws of physics. In other words, the laws of physics are the
same in all inertial frames of reference moving with a constant
velocity with respect to one another. 2. The speed of light is
equal in all the inertial frames of reference. In other words, the
speed of light in free space has the same value in all inertial
frames of reference. The speed of light is 2.998 108 m/Sec. 3.
Deduce the Lorentz transformation equation for space and time.
Answer: The equations relating the coordinates of a particle in the
two inertial frames on the basis of special theory of relativity
are called the Lorentz Transformations.
Fig.1. Two inertial frame in relative motion
In fig.1, S and S are the two inertial frames. Frame S is moving
relative to the frame S with velocity v along the positive X-axis.
Let there be a point source of light at the origin O of the frame
S. the wavefront of the light emitted at t=0, when reaches a point
P, the position and time observed by the observers at O and O are
(x,y,z,t) and (x,y,z,t) respectively. According to the special
theory of relativity, the speed of light is same (= c) in both the
frames, hence the time taken by the light to reach from O to P as
observed in the frame S is
222222
2222
222
zyxctc
zyxt
c
zyx
c
OPt
0ctzyx 22222 ----------- (1)
And the time taken by the light to reach from O to P as observed
in the frame S is
22222
2
222
2
222
zyxctc
zyxt
c
zyx
c
POt
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 4
0ctzyx 22222 ----------- (2)
From eq.(1) and (2), we get
2222222222 ctzyxctzyx ------------------ (3)
Since the frame S is moving relative to the frame S along
S-axis, the lengths in direction perpendicular to the direction of
motion are un affected i.e., zzandyy , ----------------- (4)
therefore from eq.(3),
222222 ctxctx ----------- (5) The transformation between x and x
coordinates is given by simple relation
vtxkx ------------ (6) Where k is constant, independent of x and
t.
And tvxkx ------------- (7) Where k is constant, independent of
x and t. Substituting the value x from eq. (6) in eq. (7), we
get
kvtkxk
xtvtvkvtkx
k
xtvvtxk
k
xtvvtxkkx
kk
11
v
xtkt
kk
11xvtk
k
xkxkvttv ------------- (8)
Putting the value of x from eq. (6) and the value of t from eq.
(8) in eq. (9) , we get
2
2222222
kk
11
v
xtkcvtxktcx
22
22222222222
kk
11
v
x
kk
11
v
xt2tkctvvt2xktcx
2
------- (9)
Equating the coefficients of t2 on both sides of the eq. (9), we
have
2
22
2
22222222222222 kvc
cvckcvkkcckcvkc
2
2
2
2
2
2
2
2
2
2
2
c
v1
1
c
v1c
ck
c
v1
1
c
v1c
ck --------------- (10)
And similarly, equating the coefficient of 2xt on both sides of
eq. (9), we get
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 5
kk
11cv
kk
11
v
kcvk
kk
11
v
kcvk0 22
22
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
c
v1
c
v1
c
v1
c
v1k
k
1
c
v1
kk
1
kk
11
c
v
2
2
c
v1
1k -------- (11)
Substituting the value of k and k in eq. (6) and (8), we get
2
2
c
v1
vtxx
and
2
2
2
c
v1
cxvt
t
------------ (12)
Thus the Lorentz transformations are
zz;yy;
c
v1
vtxx
2
2
and
2
2
2
c
v1
cxvt
t
------------ (13)
4. Show that the mass of a body depends on its velocity. Deduce
an expression for the variation of mass with velocity. Answer:
According to the Newtonian mechanics, mass is an invariant
quantity. If a stationary body of mass m is acted upon by a finite
and constant force F for a time t due to which it acquires a
velocity v, then the gain in momentum of the body is tFmvP
------------ (1) The maximum momentum that can be gained by a body
is
mcPmax ----------- (2)
Where c is the velocity of light. According to the special
theory of relativity the mass of body is not constant, but it
depends on the velocity of the body. When the body is at rest ( v=0
), the mass of the body is m0 and when the velocity of the body
becomes equal to the velocity of light ( v=c ), the mass of the
body becomes infinite. Consider a frame S relative to which a
particle is moving with the velocity v along the X-axis. The moving
mass of the particle in frame S is m. Consider another frame S in
which the particle is stationary. The rest mass of the particle in
frame S is m0. If the displacement of the particle relative to the
frame S in time t is y along Y-axis, then the velocity of particle
along Y-axis in the frame S is
t
yvy
and its momentum is
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 6
t
ymP 0y
------------- (3)
Similarly, t
yvy
and
t
ymP 0y
------------- (4)
From the Lorentz transformations yy , hence yy and according to
time dilation
2
2
c
v1
tt
t
y.
c
v1m
cv1t
ym
t
ymP
2
2
22
y
------------- (5)
Since the momentum of the particle is invariant, therefore Py=Py
From eq. (3) and eq. (4), we get
2
2
0
2
2
02
2
0
c
v1
mm
c
v1mm
t
y.
c
v1m
t
ym
------------ (6)
The above expression gives the variation of mass with velocity.
5. Establish Einstein mass-energy relation. Answer: According to
the Einstein, the mass and velocity are equivalent to each other.
If a mass m is converted into energy, the energy produced is E=mc2,
where c is the speed of light. This is called the mass energy
equivalence relation. Let a particle of mass m moving with a
velocity v is displaced by a distance dx by applying a force F on
it. The work done by the force is
dx.dt
dmvdx
dt
dvmdx.
dt
dmv
dt
dvmdx.mv
dt
ddkdx.Fdwdk
dmvmvdvdk 2 ;
vdt
dx ----------- (7)
But 22022222
02
2
2
2
2
0cmvmcmm
c
v1m
c
v1
mm
------------- (8)
On partial differentiation of eq. (8), we get
vdvm2dmmv2dmmc20vdvm2dmmv2dmmc2 222222
mvdvdmvdmc 22 -------------- (9)
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 7
From eq. (7) and (9), we get
dmcdk 2 ------------ (10) Hence gain in kinetic energy in
acquiring the velocity v from rest is
202
0
2
m
m
2 cmmcmmcdmck
0
------------ (11)
The eq. (11) gives the expression for the kinetic energy of the
particle at relativistic velocity.
But at v=0, the rest energy of the particle = 20cm ------------
(12)
Total energy of the particle E= kinetic energy + rest energy
2
0
2
0
2 cmcmmcE 2mcE ---------- (13)
The above equation is called the Einstein equation of
mass-energy equivalence. 6. What is chain reaction? Answer: When a
neutron fissions a uranium nucleus then, besides the fission
fragments a few fast neutrons are also emitted. If one or more of
the emitted neutrons are used to fission other nuclei further
neutrons are produced and the process is repeated. The reaction
thus becomes self-propagating and is known as a chain reaction. The
chain reactions may be two types:
I. Uncontrolled Chain Reaction and II. Controlled Chain
Reaction.
Fig.1: Chain Reaction
1. Uncontrolled Chain Reaction: In this type of chain reaction,
the number of neutrons is allowed
to multiply indefinitely and the entire energy is released all
at once as a violent explosion. Such a chain reaction takes place
in nuclear bomb.
2. Controlled Chain Reaction: In this type of chain reaction,
the reaction is first acceleration so that the neutrons are built
up to a certain level and thereafter the number of fission
producing neutrons is kept constant by some means. Such a
controlled chain reaction is used in nuclear reactors.
7. Explain the construction and working of Nuclear Reactor. Give
well-labeled diagram of Nuclear Reactor. Answer: Nuclear reactors
are the device or furnace to produce a large amount of energy
through initiate and control a sustained nuclear chain reaction.
The energy produced during nuclear reaction is also
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 8
termed as nuclear power and generally used for the production of
electricity and for the propulsion of ships. The main components of
any nuclear reactor are
1. Core: This is the main part of nuclear reactor, which
contains the nuclear fuel and where nuclear reaction takes place.
There are hundreds of thousands of individual fuel pins, which
assembled in core of nuclear reactor.
2. Moderator: The moderator used to slow down the speed neutrons
quickly as the fuel has high fission cross-section for low energy
neutrons. Depends on the type of nuclear reactor, many compounds
can be used as moderator like water, graphite rods, beryllium,
heavy water as well as some organic compounds.
3. Reflector: Reflector is used to reduce the leakage of
neutrons. Reflector reflects back the neutrons, which are escaping
from the core. Generally, moderator can be used as reflectors in
the case of thermal reactors. However, in the fast reactors nickel,
stainless steel and molybdenum are used as reflectors.
4. Cooling System: Since nuclear fission occurs in nuclear
reactor, which produces a large amount of heat, hence there must be
some coolant to absorb this excess heat. Coolant absorbs the heat
from the core, which consists of pipes through which the coolant is
pumped. The heat absorbed by coolant can be transferred to another
working medium through a heat exchanger and then returns to the
reactor. Some common coolants are heavy and light water, and liquid
metals like sodium, lithium, potassium etc.
5. Control System: This system is designed to control the number
of neutrons involve in nuclear chain reaction which indirectly
control the rate of the chain reaction as well as power level.
Control system includes sensing elements, which can measure the
number of neutrons in the reactor, and control rods, which are able
to absorbed strong neutron like cadmium or boron, and other
devices, which can regulate the position of the control rods. The
rate of nuclear chain reaction can be controlled by inserting and
ejecting these rods in reactor. In fact, the size of control rods
also affects the rate of reaction.
6. Protective shield: With a large amount of energy, nuclear
reactions are also a good source of radioactive rays like alpha,
beta and gamma rays. These radioactive rays are very dangerous, so
the exposure to these radiation must be prevent as it can affect
the person working near the reactor. The nuclear reactor is cover
by a protective shield, which made up of steel and concrete and the
arrangement of protection is known as Radiation shielding.
7. Steam Generator Component: This component used in Pressurized
water reactors to produce steam from water by absorption of heat
produced in nuclear reactor core. Some typical nuclear reactors are
as follows.
Fig.2: Nuclear Reactor
I. Boiling water reactor (BWR): It is a thermal reactor in which
steam produced in the core,
which can be fed directly to the turbine and produce
electricity. The fuel used in such type of reactor is uranium oxide
and light water is used as the coolant and moderator. Example of
boiling water reactor is nuclear power station of Tarapur, which
used for power generation of 210 MW.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 9
II. Pressurized water reactor (PWR): It is same as boiling water
reactors but here water keeps at high pressure so it does not boil
after absorbing heat also. It prevents the mixing of water from the
reactor and the water in the steam generator; hence, most of the
radioactivity cannot move out of the reactor area.
III. Fast Breeder Test Reactor: In this reactor, plutonium acts
as nuclear fuel. This reactor involves artificial transmutation of
Uranium-238 to Plutonium-239 by absorption of neutrons. The product
of transmutation that is plutonium consumed in reaction and again
creates or breeds from the uranium. Here liquid sodium acts as
coolant with normal steam generator. Because of absence of
moderator to slow neutrons, this reactor called as Fast Breeder
Test Reactor.
8. Write a short note on Fusion Reaction in Stars. Answer:
Nuclear fusion is the process by which two or more atomic nuclei
join together, or "fuse", to form a single heavier nucleus. During
this process, matter is not conserved because some of the mass of
the fusing nuclei is converted to energy, which is released. Fusion
is the process that powers active stars. There are extreme
astrophysical events that can lead to short periods of fusion with
heavier nuclei. This is the process that gives rise to
nucleosynthesis, the creation of the heavy elements during events
such as supernovae. Mark Oliphant first accomplished building upon
the nuclear transmutation experiments by Ernest Rutherford, carried
out several years earlier, the laboratory fusion of heavy hydrogen
isotopes in 1932. During the remainder of that decade, Hans Bethe
worked out the steps of the main cycle of nuclear fusion in stars.
There are two types:
1. The protonproton cycle and 2. The carbonnitrogenoxygen
cycle
1. The proton-proton cycle: Bethe, in 1939 suggested that the
production of stellar energy is by thermonuclear reaction in which
protons are continuously transformed into helium nuclei. In
proton-proton cycle is given by-
1
1
4
2
3
2
3
2
3
2
1
1
2
1
0
1
2
1
1
1
1
1
HHeHeHe:3Step
HeHH:2Step
eHHH:1Step
So, that we have
MeV7.26114
2
1
1 H2HeH4
Fig.1: The proton-proton chain dominates in stars Fig.2: The CNO
cycle in stars 2. The carbonnitrogenoxygen cycle: In this cycle,
the synthesis of hydrogen nuclei with carbon takes place. The
carbon nuclei absorb the protons in succession and ultimately
discharge a particle becoming carbon nuclei again. The reactions
are given below:
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 10
4
2
12
6
1
1
15
7
0
1
15
7
15
8
15
8
1
1
14
7
14
7
1
1
13
6
0
1
13
6
13
7
13
7
1
1
12
6
HeCHN:6StepeNO:5Step
OHN:4StepNHC:3Step
eCN:2StepNHC:1Step
The overall process may be written as
MeV27eHeHe40
1
4
2
4
2
9. What is mass spectrograph? Answer: A mass spectrograph is an
instrument that measures the nuclear masses. Mass spectrographs
were designed by F.W. Astons by A.J. Dempster and by K.T.
Bainbridge usingdifferent properties. The initial mass
spectrographs measured atomic masses with a precision of 1 part in
103 whereas the modern instruments can yield a precision of 1 part
in 107. 10.Explain construction and working of Bainbridge Mass
Spectrograph. How to relative proportion of isotopes are
determined. Answer: Bainbridge mass spectrograph is the recent and
advanced type mass spectrograph and having high resolving power,
precise symmetric image and linear mass scale which could not be
obtained in Astons or Dempsters spectrographs. This is possible
because it uses velocity selector and power electromagnet.
Principle: Whatever be the velocities of the positive ions in the
process of their generation, which are to be investigated, they are
made perfectly homogeneous in velocity by the use of a special
device called velocity selector. They are subjected to an extensive
transverse magnetic field, which deflects the positive ions along
circular path of different radius. The deflected ions are then
brought to focus on a photographic plate. The radius of the
semi-circular path is directly proportional to the mass of ion
describing the path. Construction & working: Fig.1. represent
the schematic diagram of Bainbridge mass spectrograph. A vacuum
chamber is placed in a acting perpendicular to its large surface.
The given beam of ions is collimated by two narrow parallel slits
S1 and S2. These ions are having different velocities. It is then
allowed to pass through the velocity selector. The velocity
selector is a device in which the electric and magnetic fields are
applied in the same region but perpendicular to each other. These
two fields are arranged in such a way that deflections produced by
them are equal and opposite.
Fig.1. Bainbridge Mass-spectrographs
The ion beam produced by discharge tube or a spark is collimated
by slits S1 and S2 and allowed to enter the velocity selector. The
electric field strength E and magnetic field strength B are
adjusted such that only ions of given velocity v pass un-deflected
experiencing equal and opposite electric and magnetic forces-
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 11
(1)B
EvqvBqE
Only the ions havening velocity B
E enter the evaluated chamber D through to the slit S3. The
positive
ions entering into D are subjected to a strong uniform magnetic
field of strength B perpendicular to their path. The force acting
each ion is 'qvB and traverse circular paths of radius r given
by
(2)qB'
mvrqvB'
r
mv 2
Thus is ions different masses m, but having same charge q,
describe semi-circles of different radii
(3)rB'
v
m
q
Using value ofB
Ev , we get
rBB'
E
m
q
As E, B and B are constants,
r
1
m
q, if q is same for all ions, then mass rm . The ions
different
masses strike the photographic plate of different points and we
obtain a typical mass spectrum on the photographic plate, the mass
scale being linear. 11. Explain construction and working of Aston
Mass Spectrograph. How to relative proportion of isotopes are
determined. Answer: In 1919, F.W. Aston designed a mass
spectrograph. This spectrograph enables the measurement of the mass
of single atomic ions and is useful for the investigation of
isotopes. Principle: The deflection of the beam of positive rays in
the electrostatic field depends on the velocity of ions and hence
they are dispersed. This dispersed beam is then subjected to a
magnetic field, which is perpendicular to electrostatic field.
Construction, Working & Theory: The different parts of the
apparatus are shown in fig.1. AO is direction of positive rays
before entering the electrostatic field. S1 and S2 are slits which
provide a fine pencil of positive rays. Plates P1 and P2 and the
direction of the field being from P1 and P2 maintain the
electrostatic field. The beam is deflected and dispersed downwards.
Let and d be the angles of deviation and dispersion. Using a
diaphragm D some of the rays are selected and are allowed to pass
between the poles of an electromagnet. The magnetic field being
perpendicular to the plane of the paper and inward. According to
the Flemings left hand rule, the beam will be deflected upwards.
This magnetic field nullifies the dispersion produced by electric
field and recombines the particles, which are brought to focus in
the form of sharp lines on a photographic plate CD. These lines are
same as those of spectral lines.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 12
Fig. 1. Aston Mass Spectrograph
Say, q= charge on positive ray particle; m'= mass of each
particle; E= electrostatic field; B= magnetic field; = velocity of
each particle; = angle of deviation produced by magnetic field; d=
angle of dispersion produced by magnetic field; = angle of
deviation produced by electric field; d= angle of dispersion
produced by electric field. Taking into account that the deflection
in electrostatic field is small, the curve near the vertex may be
considered as circular of a radius r, we have
2
2
v'm
'Eq
r
1
r
v'm'Eq
Hence, the deflection , which is proportional to, r
1is given by
where C1 = CE, because E = constant.
)1(2
d
v
dv
v2
v'm
'qC2
dv
ddispersion
21
If r is the radius of curvature in magnetic field then
)2(v'm
'qC
v'm
'Bq'C
v'm
'Bq
'r
1
'r
v'mv'Bq 2
2
(Since B is constant)
)3(d
v
dv
vv'm
'qC
dv
ddispersion
22
From eqns. (1) and (2) we have,
)4(2
d
dd2
dd
2
dd
2
d
v
dv
Hence, if a deflection is given, the dispersion due to the
electric field is twice that due to magnetic field. The small
changes d and d refer to the particles with identical mass and
charge but processing velocity differ by dv. In the absence of
magnetic field, the dispersion produced in the beam for a distance
(a+b) is given by
)5(d)ba(
212 v'm
'qC
v'm
'EqC
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 13
[here a= distance OO and b= distance OE] The magnetic field acts
in a direction of perpendicular to the electric field and produces
the same dispersion in a distance b but in the opposite direction.
Dispersion produced by magnetic field )6(bd As all the ions are
focused to the same position.
)7()ba(
b
d
dbdd)ba(
From eq. (4) and (7),we have
)8(2
2
b
a1
2b
a
21
b
a
2b
)ba(b2)ba(
)ba(
b2
)ba(
b2
This is the required condition for focusing. If b , then
)9(2202
20
a
This gives the position of photographic plate. The modified
apertures gives an accuracy of 1 part in 104 and resolvability of
600. 12. What is ultrasonic? Answer: Ultrasound is a cyclic sound
pressure wave with a frequency greater than the upper limit of the
human hearing range. Ultrasound is thus not separated from "normal"
(audible) sound based on differences in physical properties, only
the fact that humans cannot hear it. Although this limit varies
from person to person, it is approximately 20 kilohertz (20,000
hertz) in healthy, young adults. Ultrasound devices operate with
frequencies from 20 kHz up to several gigahertzes. Ultrasonics is
the application of ultrasound. Ultrasound can be used for imaging,
detection, measurement, and cleaning. At higher power, levels
ultrasonics are useful for changing the chemical properties of
substances. 13. Explain Magnetostriction Method of producing
ultrasonic waves and hence describe its advantages. Answer:
Production of ultrasonic waves:- There are two major types of
method-
I. Magnetostriction oscillator and II. Piezo-electric
oscillator
Magnetostriction oscillator (Pierce Oscillator):-
Magnetostriction is a property of ferromagnetic materials that
causes them to change their shape or dimensions during the process
of magnetization. The variation of material's magnetization due to
the applied magnetic field changes the magnetostriction strain
until reaching its saturation value, . James Joule when observing a
sample of iron first identified the effect in 1842.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 14
This effect causes losses due to frictional heating in
susceptible ferromagnetic cores. The effect is also responsible for
the high-pitched buzzing sound that can be heard near transformers
on alternating current carrying pylons. Principle: The general
principle involved in generating ultrasonic waves is to cause some
dense material to vibrate very rapidly. The vibrations produced by
this material than cause air surrounding the material to begin
vibrating with the same frequency. These vibrations then spread out
in the form of ultrasonic waves. When a magnetic field is applied
parallel to the length of a ferromagnetic rod made of material such
as iron or nickel, a small elongation or contraction occurs in its
length. This is known as magnetostriction. The change in length
depends on the intensity of the applied magnetic field and nature
of the ferromagnetic material. The change in length is independent
of the direction of the field. When the rod is placed inside a
magnetic coil carrying alternating current, the rod suffers a
change in length for each half cycle of alternating current. That
is, the rod vibrates with a frequency twice that of the frequency
of A.C. The amplitude of vibration is usually small, but if the
frequency of the A.C. coincides with the natural frequency of the
rod, the amplitude of vibration increases due to resonance.
Construction: the coils L1 and L. wind the ends of the
ferromagnetic rod A and B. The coil L is connected to the collector
of the transistor and the coil L1 is connected to the base of the
transistor as shown in the figure. The condenser C can adjust the
frequency of the oscillatory circuit (LC) and the milliammeter
connected across the coil L. can note the current. The battery
connected between emitter and collector provides necessary biasing
i.e., emitter is forward biased and collector is reverse biased for
the NPN transistor. Hence, current can be produced by applying
necessary biasing to the transistor with the help of the battery.
Working: The rod is permanently magnetized in the beginning by
passing direct current. The battery is switched on and hence
current is produced by the transistor. This current is passed
through the coil L, which causes a corresponding change in the
magnetization of the rod. Now, the rod starts vibrating due to
magnetostriction effect.
Fig.1.
When a coil is wounded over a vibrating rod, then e.m.f. will be
induced in the coil called as converse magnetostriction effect. Due
to this effect, an e.m.f. is induced in the coil L1. The induced
e.m.f. is fed to the base of the transistor, which act as a
feedback continuously. In this way the current in the transistor is
built up and the vibrations of the rod is maintained. The condenser
C adjusts the frequency of the oscillatory circuit and when this
frequency is equal to the frequency of the vibrating rod, resonance
occurs. At resonance, the rod vibrates longitudinally with larger
amplitude producing ultrasonic waves of high frequency along both
ends of the rod. Condition for resonance: Frequency of the
oscillatory circuit = Frequency of the vibrating rod
E
lLC 2
1
2
1
where, l=the length of the rod. E=the youngs modulus of the
material of the rod. = is the density of material of the rod.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 15
Merits: 1. Magnetostriction materials are easily available and
inexpensive. 2. Oscillatory circuit is simple to construct. 3.
Large output power can be generated. Limitations 1. It can produce
frequencies upto 3 MHz only. 2. It is not possible to get a
constant single frequency, because rod depends on temperature and
the degree of magnetization. 3. As the frequency is inversely
proportional to the length of the vibrating rod, to increase the
frequency, the length of the rod should be decreased which is
practically impossible. 14.Explain Piezoelectric Method of
producing ultrasonic waves and hence describe its advantages.
Answer: Jacques and Pierre Curie discovered piezoelectricity in
1880. Paul Langevin first investigated quartz resonators for use in
sonar during World War I. The first crystal-controlled oscillator,
using a crystal of Rochelle salt, was built in 1917 and patented in
1918 by Alexander M. Nicholson at Bell Telephone Laboratories,
although Walter Guyton Cady disputed his priority. Cady built the
first quartz crystal oscillator in 1921. Other early innovators in
quartz crystal oscillators include G. W. Pierce and Louis Essen.
Quartz crystal oscillators were developed for high-stability
frequency references during the 1920s and 1930s. By 1926 quartz
crystals were used to control the frequency of radio broadcasting
stations and were popular with amateur radio operators.
Piezoelectric effect: When crystals like quartz or tourmaline are
stressed along any pair of opposite faces, electric charges of
opposite polarity are induced in the opposite faces perpendicular
to the stress. This is known as piezoelectric effect.
(A) ( B) Piezoelectric effect- Mechanism: Piezoelectric and
inverse piezoelectric effects are only exhibited by certain
crystals, which lack centre of symmetry. In a piezoelectric
crystal, the positive and negative electrical charges are
separated, but symmetrically distributed, so that the crystal
overall is electrically neutral. Each of these sides forms an
electric dipole and dipoles near each other tend to be aligned in
regions called Weiss domains. The domains are usually randomly
oriented, but can be aligned during
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 16
poling, a process by which a strong electric field is applied
across the material, usually at elevated temperatures. When a
mechanical stress is applied, this symmetry is disturbed, and the
charge asymmetry generates a voltage across the material. For
example, a 1 cm cube of quartz with 2 kN (500 lbf) of correctly
applied force can produce a voltage of 12,500 V. Piezoelectric
materials also show the opposite effect, called converse (inverse)
piezoelectric effect, where the application of an electrical field
creates mechanical deformation in the crystal. Inverse
piezoelectric effect: When an alternating e.m.f is applied to the
opposite faces of a quartz or tourmaline crystal, it undergoes
contraction and expansion alternatively in the perpendicular
direction. This is known as inverse piezoelectric effect. This is
made use of in the piezoelectric generator. Piezoelectric
generator: A slab of piezoelectric crystal is taken and using this
parallel plate capacitor is made. Then with other electronic
components an electronic oscillator is designed to produce
electrical oscillations >20 kHz. Generally one can generate
ultrasonic waves of the order of MHz using piezoelectric
generators. Quartz slabs are preferred because it possesses rare
physical and chemical properties. A typical circuit diagram is
given below. The tank circuit has a variable capacitor 'C' and an
inductor 'L' which decides the frequency of the electrical
oscillations. When the circuit is closed current rushes through the
tank circuit and the capacitor is charged, after fully charged no
current passes through the same. Then the capacitor starts
discharging through the inductor and hence the electric energy is
in the form of electric and magnetic fields associated with the
capacitor and the inductor respectively. Thus, we get electrical
oscillations in the tank circuit and with the help of the other
electronic components including a transistor, electrical
oscillations are produced continuously. This is fed to the
secondary circuit and the piezoelectric crystal (in our case a slab
of suitably cut quartz crystal) vibrates, as it is continuously
subjected to varying (alternating) electric field, and produces
sound waves. When the frequency of electrical oscillations is in
the ultrasonic range then ultrasonic waves are generated. When the
frequency of oscillation is matched with the natural frequency of
the piezoelectric slab then it will vibrate with maximum amplitude.
The frequency generated is given as follows:
E
l
P
LCf
22
1
where E= The Youngs modulus of the piezoelectric material and =
The density of the piezoelectric material. 15. What is
reverberation? Answer: Reverberation means the prolonged refraction
of sound from the walls, floor and ceiling of room. It is also
defined as the persistence of audible sound after the source has
stopped to emit sound. The time of reverberation is also defined as
the time taken for the sound of fall below the minimum audibility
measured from the instant when the source stopped sounding. The
Sabine defined the standard reverberation time as the time taken by
sound to fall to one millionth of the intensity just before the
source is cut off. 16. Derive the expression for Sabine formula for
reverberation time.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 17
Answer: The derivation for the formula is based on assumption,
that there is a uniform distribution of sound energy inside the
room. Now, first, we shall calculate the rate at which the energy
is incident upon the walls and other surfaces and hence the rate at
which it is being absorbed.
Fig.1.
Consider the reception of sound energy by small element ds of a
plane wall AB as shown in fig.1. from the centre of ds and with
redii r and r+dr , draw two circles such that they lie in the plane
containing the normal to the element ds .Now consider the area
between the circles and also lying between the angles +d which the
two redii make with the direction of normal at the surface ds .
This area is shown shaded in fig.1. the arc of this area is rd and
the radial length is dr . Hence its area is rd.dr. If the whole
fig. is rotated about the normal through an angle d, the distance
travelled by this area will be the circumferential distance rsind.
Volume traced out by this area element=rd.dr(rsind) Suppose E is
the value of sound energy density, the energy contained in the
above volume is given by=EdV This sound energy is travelling
through the element equally in all directions. Energy travelling
per unit
solid angle any direction
4
EdV
But the solid angle subtended by ds at the element of volume
considered 2r
cosds
the energy in the element of volume that is travelling towards
ds is given by
)1(d.d.cossindr.4
Eds
r
cosds.
4
d.dr.d.sinEr
r
cosds.
4
EdV
2
2
2
In order to find out the value of the energy reaching ds in unit
time, the above expression should be integrated from r=0 to r=C is
velocity of sound, and it is the distance within the volume is
lying. Integrating eqn. (1), first with respect to , we get
)2(dcossin.dr.ds.2
E2.dcossin.dr.ds.
4
E
2dddcossin.dr.ds.4
Edcossin.dr.ds.
4
E2
0
2
0
Now, integrating with respect to , we have energy received at
d
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 18
)3(dr.ds.2
E1.dr.ds.
2
E
1d2sind2sindr.ds.2
E
dcos.sin2dr.ds.2
Ed
2
cos.sin2dr.ds.
2
Edcos.sindr.ds.
2
E
2
0
2
0
2
0
2
0
2
0
Now integrating again with respect to r whose values varies
between 0 to C, we get
)4(C.ds2
Edrds
2
EC
0
If a be the absorption coefficient of the wall AB whose value ds
is a part, then Energy absorbed by ds per unit time
ds.a.EC4
1
Hence total absorption all the surfaces of the wall where the
sound is falling
)5(4
ECAds.aC.E
4
1
where a.ds=A, the total absorption on all the surfaces on which
sound falls. (i). Rate of Growth of Sound Energy in a Room:- Let P
be the power output i.e. rate of emission of energy from the source
and V, the total volume of the room. Then the total energy in the
room at the instant when energy density is E will be EV. Rate of
growth of energy
dt
dEVEV
dt
d
But at any instant, rate of growth of energy in space= Rate of
supply of energy from the surface- Rate of absorption by the entire
surface
)6(4
ECAP
dt
dEV
when steady state is attained dE/dt=0, and if the steady state
energy density is denoted by Em, then its is given by
4
CAEP
4
CAEP0 mm
)7(CA
P4Em
From equ.(6) A.V4
EC
V
P
dt
dE
CA
P4E
dt
dEE
CA
P4
dt
dE ; where
CA
4
V
1and
V4
CA
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 19
Multiplying both sides by te , we have tttt eCA
P4Ee
dt
de
CA
P4eE
dt
dE
Integrating the above equation, we get )8...(..........KeCA
P4Ee tt
where K is a constant of integration. Initial condition is that
at t=0, E=0 .Applying this condition to above equation, we get
CA
P4K
Putting this value in equation (8),we get
)9(e1Eme1CA
P4Ee
CA
P4
CA
P4E
CA
P4e
CA
P4Ee ttttt
The equation shows the growth of energy with time t. (ii). Decay
of energy Density in a room:- Let the source is cutoff when E has
reached the maximum value of Em, so that P=0 and t=0 when
CA
P4EE max
From equation (9) mmax EEK
0P,EEe mt
Or tmeEE .(10)
Equation (10) shows the decay of the energy density with time
after the source is cutoff. (iii). Standard Reverberation Time
(T):- Reverberation time T is defined as the time taken by sound to
fall to one millionth of its value before the cutoff. Hence, to
calculate T.
we put Ttand10EE 6m in eq.(10)
6TT6T
max
10ee10eE
E
Taking log, we have 63026.210log6T e
A
V.
340
63026.24T63026.2T.
V4
CA
(Taking the velocity of sound in air C=340 m/sec at room
temperature)
Or VaS
165.0V
A
165.0T
..(11)
This equation is in good agreement with the experimental values
obtained by Sabine. 17. Define Absorption Coefficient. Derive the
expression for Absorption Coefficient of a Sound. Answer: The
coefficient of absorption of a material is defined as the ratio of
the sound energy absorbed by the surface to that of the surface to
that of the total incident sound energy on the surface i.e.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 20
Hence, the
absorption coefficient of a material is defined as the rate of
the sound energy absorbed by a certain area of the surface to that
of an open window of same area. The absorption coefficient of a
surface is also defined as the reciprocal of its area which absorbs
the same sound energy as absorbed at a unit area of an open
windows. Measurement of Absorption Coefficient: The determination
of standard times of reverberation in the room without and with a
standard large sample of the material sample of the material inside
the chamber. If the reverberation times are T1 and T2 respectively,
then by applying Sabines formula, we have
V165.0
aS
V165.0
A
T
1
1
and V165.0
SaaS
T
1 11
2
where a1is the absorption coefficient of the area S1. From the
above equations, we have
12
11
T
1
T
1
V165.0
Sa
Hence, knowing the terms on the right hand side of this equation
a can be calculated. 18. Obtain an expression for fringe width in
wedge shaped thin film. Answer: A wedge shaped film is one whose
surfaces are inclined at certain small angle. Where- reflective
index of wedge shaped film= . The path difference between the rays
reflected at the upper
and lower surfaces= t2 . The additional path difference= 2
Fig.1.
The effective path difference between the two rays= 2
t2
The condition for the maximum intensity (bright fringes) is
n2
t2
Or 2
1n2t2
(1)
surfacetheonincidentenergysoundTotal
surfacethebyabsorbedySoundenergacoefficentAbsorption
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 21
The condition for minimum intensity (dark fringes) is
2
1n22
t2
Or nt2 . (2)
For the nth dark fringes, we have nt2 .(3)
Let this fringes be obtained at a distance nx from the edge
tanxt n .(4)
As is small
tan (5) nxt (6)
From eqn. (2) and (6),
nx2 n (7)
Similarly, if (n+1)th dark fringes is obtained at a distance,
then
)1n(x2 1n (8)
From eqn. (7) and (8), we get
n)1n(x2x2 n1n
n1n xx2
Fringe width
2
xxX n1n ..(9)
Where is measured in radian. Similarly, we can calculate fringe
width for bright fringes. 19. Give the theory of formation of
Newtons Ring by reflected light. Explain why central ring is dark
and higher order rings come closer to each other. Answer: When a
plano- convex lens of large radians of curvature is placed with its
convex surface in constant with a plane glass plate, an air film of
gradually increasing thickness from the point of contact is formed
between the upper surface of the plate and lower surface of the
lens. If monochromatic light is allowed to fall normally on this
film, then alternate bright and dark concentric rings with their
centre dark are formed. These rings are known as Newtons rings.
Experimental Arrangement: G= glass plate inclined at angle of 45 to
the incident beam. L= plano-convex lens P= plane glass plate, S=
monochromatic light source.
Fig.1. Experimental Arrangement
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 22
Theory:
Fig.2.
The condition of maximum intensity of bright rings 1airfor is
given by
n2
t2
Or
2122
nt (1)
Where ...........3,2,1n
The condition of minimum intensity of dark rings 1airfor is
given by
212
22
nt
Or nt2 (2) Where .....,.........3,2,1n
Determination of Diameters of Bright & Dark Rings: Let r be
the radius of Newtons ring. R be the radius of curvature of the
circle. t be the thickness of wedge shaped film. From the
geometrical property of a circle, DEODDBAD ODR2ODrr tR2tr2 22 tRt2r
.(3)
Since t is very small compared to R , hence 2t can be neglected.
Hence
Rt2r2
Or R
rt
2
2 .. (4)
From eqn.(1) and (4), we have
2
1n2R
r2
Radius of thn bright ring
2
R1n2rn
(5)
The diameters of thn bright ring nn r2D ,
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 23
2
R1n22Dn
R21n2R21n22
R1n24D n
Or 1n2Dn .(6) Thus, the diameter of bright rings are
proportional to the square roots of the odd natural numbers.
From eqn. (2) and (4), radius of thn dark ring, we have
nR
rt2
2
n
Rnrn .(7)
The diameters of thn dark ring is given by
RnRnRnrD nn 4422
Or nDn (8)
Thus, the diameter of dark rings are proportional to the square
roots of natural numbers. Determination of wavelength of sodium
light using Newtons rings: From eqn.(8), we have
Rn4D2n (9)
Similarly, the diameter of thpn ring is given by R)pn(4D2 pn
(10)
From eqn. (9) and (10), we have
Rp4Rn4R)pn(4DD 2n2
pn
pR4
DD 2n2
pn
(11)
20. Explain the formation of interference fringes by means of
Fresnel biprism using monochromatic source of light. Answer: The
biprism is a device to obtain two coherent sources to produce
sustained interference. It is a combination of two prisms of very
small refracting angles, placed base to base. Actually, it is
constructed as a single prism with one of its angle about 179 and
the other two about 03 each. Production of Fringes: Let, S= A
narrow adjustable slit; P= biprism; S1 and S2= virtual image &
coherence source
(a) (b)
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 24
Theory: Let, S= A narrow adjustable slit; P= biprism; S1 and S2=
virtual image & coherence source
Hence 2
dOROQ;xPO;DRSQS;dSS 2121
From PQS1 and PRS2 , According to Pythagoras theorem
4
dxdxD
2
dxDOQPODQPQSPS
2
22
2
22222
1
2
1
(1)
And
4
dxdxD
2
dxDORPODPRRSPS
222
2
22222
2
2
2
..(2)
From eqn. (1) and (2), we have
xd24
dxdxD
4
dxdxDPSPS
2
2
2
22
1
2
2
xd2PSPSPSPS 1212
PSPS
xd2PSPS
12
12
..(3)
But DPSPS 12 (approximately)
Path difference = D
xd
D2
xd2PSPS 12 .(4)
For P be the centre of bright fringes; nD
xdPSPS 12 where n= 0,1,2,3,.
Or nd
Dx .(5)
For the P to be the centre of a dark fringes; 2
1n2D
xdPSPS 12
where n= 0,1,2,3,..
Or 2
1n2d
Dx
(6)
Let nx and 1nx denote thn and th1n bright fringes. Then the
distance between th1n and thn bright
fringes is given by
d
Dn1n
d
Dn
d
D1n
d
Dxx n1n
Or d
DxxX n1n
.(7)
And let nx and 1nx denote thn and th1n dark fringes. Then the
distance between th1n and thn
dark fringes is given by
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 25
2
1n23n2d
D
21n211n2
d
D
21n2
d
D
211n2
d
Dxx n1n
Or d
DxxX n1n
. (8)
Where X is known as width of fringes. Experimental Arrangement
and Adjustment: The apparatus used for this purpose are optical
bench with four uprights, biprism sodium lamp, a convex lens, plumb
line, bench error rod and reading lamp. Following procedure is
adopted for obtaining sharp fringes:-
1. The optical bench is made horizontal with the help of sprit
level. 2. Widen the slit and focus the eyepiece on the cross wires.
One of the cross wires is set vertical by
means of a plumb-line. 3. Rotate the slit with the help of
tangent screw provided in the plane of the slit so that its
image
coincided with the vertical cross-wires. Hence the slit becomes
vertical. 4. Using the tangent screw provided with the biprism
mount and the slit will be nearly parallel. 5. The line joining
slit and the edge of the biprism is made parallel to the length of
the optical bench
by removing the lateral shift. The following measurements are
now made:-
1. Measurements of Fringe Width X :-Adjust the vertical cross
wire of the eyepiece on a bright fringe. Take the reading. Then the
eyepiece is moved laterally so that the vertical cross-wire
coincides with successive bright fringes and the corresponding
readings are noted. From these
readings the fringe-width X is determined. 2. Measurement of D
:- Takes the reading of position of the slit and the eyepiece on
the optical
bench. The difference these readings gives D . 3. Measurement of
d the distance between Virtual Sources in Biprism:- the distance
between
the two virtual sources in biprism is determined by any of the
following two methods:- I. Displacement Method and
II. Deviation Method 15. Discuss the Fraunhofer diffraction at a
single slit. Derive the necessary conditions for minima and maxima
produced. Answer: This phenomenon of bending of light corners and
spreading the region of geometrical shadow of an object is called
Diffraction. There are two types:-
I. Fresnel diffraction and II. Fraunhofer diffraction
Fraunhofer Diffraction at a Single Slit: Let wavelength of
parallel beam of monochromatic light, e width of narrow slit AB, L
Convex lens, angle of diffraction
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 26
Fig.1.
Let the disturbance caused at P by the wavelet from width of the
slit at C be
tcosAy0 . (1)
Then the wavelet from with dx at C, when it reaches P has the
amplitude Adx and phase
sin2
t .
Let this small disturbance be dy, we have
sinx2tcosAdxdy .(2)
For the total disturbance at the point of observation at an
angle , we get
2
e
2
e
dxsinx2
tcosAy
But we know that Bsin.AsinBcos.AcosBAcos
2
e
2
e
2
e
2
e
2
e
2
e
dxsinx2
sintsinAdxsinx2
costcosA
dxsinx2
sin.tsinsinx2
cos.tcosAy
2
e
2
e
2
e
2
e
sin2
sinx2cos
tsinAsin2
sinx2sin
tcosAy
sine2
sine2cos
sine2cos
tsinAsine2
sine2sin
sine2sin
tcosAy
We know that coscosandsinsin
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 27
sine2
sine2cos
sine2cos
tsinAsine2
sine2sin
sine2sin
tcosAy
sine2
sine2sin2
tcos.Ae0sine2
sine2sin2
tcosAy
Let
sine2and 0AAe , therefore we get tcos
sinAy 0
.(3)
Resultant amplitude
sinAR 0
The resultant intensity at P is given by 2
2
02
22
0
2 sinIsin
ARI
(4)
Where 200 AI represent the intensity at 0 .
21. Discuss the Fraunhofer diffraction at a N- slit. Derive the
necessary conditions for minima and maxima produced. Answer: A
diffraction grating is an arrangement equivalent to N number of
parallel slits of equal widths and separated from one another by
equal opaque space. Diffraction grating is made by ruling large
number of fine, equidistant and parallel lines on an optically
plane glass plate with a diamond point makes diffraction grating.
The ruled portion scatters the light while unruled portion
transmits light.
Fig.1.
Let amplitude of each N-waves
sinAA 0 ; Phase difference
sinde2; Grating element
de ; Width of each silt d The resultant disturbance y is given
by
termsN..............2tcostcostcosAy
We know that 2
eecos
ii ; therefore
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 28
termsN............2
ee
2
ee
2
eeAy
2ti2tititititi
termsN.............2
e
2
e
2
e
2
e
2
e
2
eAy
2iti2itiitiitititi
termsN.......2
e
2
e
2
e
2
e
2
e
2
e
2
e
2
e
2
e
2
eAy
i2tii2tiitiitititi
termsN.......ee12
eAtermsN.......ee1
2
eAy i2i
ti
i2i
ti
Using only real part of above equation, therefore
termsN.......ee1eAy i2iti The above equation is GP (Geometric
Progression) series, so we get
i
iNti
e1
e1eAy (1)
Intensity yyI (2)
where y is the complex conjugate of y.
i
iN
ti
i
iN
ti
e1
e1eA
e1
e1eAI
iN
i
iN
i
iNtiti2
i
iN
i
iNtiti e
e1
e11
e1
e1eA
e1
e1
e1
e1eeAAI
1ee1
1ee1A
eee1
eee1eA
eeee1
eeee1eAI
ii
iNiN2
0ii
0iNiN02
iiii
iNiNiNiN02
cos1
Ncos1A
2
ee1
2
ee1
Aee2
ee2A
ee2
ee2
4
1AI 2
ii
iNiN
2
ii
iNiN2
ii
iNiN2
....(3)
We know that 2sin212cos , therefore we get
sindesin
sindeNsin
Asinde2
2
1sin
sinde2N
2
1sin
A
2sin
2
Nsin
A
2sin2
2
Nsin2
AI2
2
2
2
2
2
2
2
2
2
2
2
(4)
Since
sinAA 0 ;
Therefore
2
2
2
22
0sin
Nsin.
sinAI .(5)
where
sin)de(
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 29
Hence the intensity distribution is product of two terms. The
first term 2
22
0
sinA
represent the
diffraction pattern due to a single slit and second term
2
2
sin
Nsinrepresent the interference pattern due
to N-slits.
For N=1 2
22
02
2
2
22
0
sinA
sin
sinsinAI
(single slit diffraction)
For N=2
2
22
2
2
2
02
2
2
2
2
02
2
2
2
2
0
sin
cossin4sinA
sin
cossin2sinA
sin
2sinsinAI
Or
2
2
22
0 cossin
A4I (double slit diffraction)
Hence for N-slit
2
2
2
22
0cos
NcossinAI (generalized expression)
Principal Maxima Intensity would be maximum when
0sin Or when n ; where n=0,1,2,3,
Also 0Nsin
Thus 0
0
sin
Nsin
i.e. indeterminate.
To solve this, we will use L Hospitals rule i.e.
N1.Nsin
N
Nsin
Lim.Nsin
N
Nsin
NLimsin
Nsin
N
NLim
sin
NsinLim
nnnn
22
22
0
p NsinA
I
..(6)
Theses maxima are most intense and are called Principal maxima.
Now as n
n
sinde
Or nsinde (7) where n=0,1,2,3.. This is known as grating
equation. 22. What is ruby laser? Describe the construction and
action of the ruby laser. Answer: A ruby laser is a solid-state
laser that uses a synthetic ruby crystal as its gain medium. The
first working laser was a ruby laser made by Theodore H. "Ted"
Maiman at Hughes Research Laboratories on May 16, 1960.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 30
Fig.1. Ruby Laser
Ruby lasers produce pulses of visible light at a wavelength of
694.3 nm, which is a deep red color. Typical ruby laser pulse
lengths are on the order of a millisecond. Working and theory: A
ruby laser most often consists of a ruby rod that must be pumped
with very high energy, usually from a flashtube, to achieve a
population inversion. The rod is often placed between two mirrors,
forming an optical cavity, which oscillate the light produced by
the ruby's fluorescence, causing stimulated emission. Ruby is one
of the few solid-state lasers that produce light in the visible
range of the spectrum, lasing at 694.3 nanometers, in a deep red
color, with a very narrow line width of 0.53 nm. The ruby laser is
a three level solid-state laser. The active laser medium is a
synthetic ruby rod that is energized through optical pumping,
typically by a xenon flashtube. Ruby has very broad and powerful
absorption bands in the visual spectrum, at 400 and 550 nm, and a
very long fluorescence lifetime of 3 milliseconds. This allows for
very high energy pumping, since the pulse duration can be much
longer than with other materials. While ruby has a very wide
absorption profile, its conversion efficiency is much lower than
other mediums.
Fig.2. Energy Level Diagram
Modern lasers often use rods with antireflection coatings, or
with the ends cut and polished at Brewster's angle instead. This
eliminates the reflections from the ends of the rod. External
dielectric mirrors then are used to form the optical cavity. Curved
mirrors are typically used to relax the alignment tolerances and to
form a stable resonator, often compensating for thermal lensing of
the rod. Ruby also absorbs some of the light at its lasing
wavelength. To overcome this absorption, the entire length of the
rod needs to be pumped, leaving no shaded areas near the mountings.
The active part of the ruby is the dopant, which consists of
chromium ions suspended in a sapphire crystal. The dopant often
comprises around 0.05% of the crystal, and is responsible for all
of the absorption and emission of radiation. Depending on the
concentration of the dopant, synthetic ruby usually comes in either
pink or red. 23. What is He-Ne laser? Describe the construction and
action of the He-Ne laser.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 31
Answer: A heliumneon laser or He-Ne laser is a type of gas laser
whose gain medium consists of a mixture of helium and neon inside
of a small-bore capillary tube, usually excited by a DC electrical
discharge. Working and theory: The gain medium of the laser, as
suggested by its name, is a mixture of helium and neon gases, in
approximately a 10:1 ratio, contained at low pressure in a glass
envelope. The gas mixture is mostly helium, so that helium atoms
can be excited. The excited helium atoms collide with neon atoms,
exciting some of them to the state that radiates 632.8 nm. Without
helium, the neon atoms would be excited mostly to lower excited
states responsible for non-laser lines. A neon laser with no helium
can be constructed but it is much more difficult without this means
of energy coupling. Therefore, a He-Ne laser that has lost enough
of its helium will most likely not lase at all since the pumping
efficiency will be too low. The energy or pump source of the laser
is provided by a high voltage electrical discharge passed through
the gas between electrodes within the tube. A DC current of 3 to 20
mA is typically required for CW operation. The optical cavity of
the laser usually consists of two concave mirrors or one plane and
one concave mirror, one having very high (typically 99.9%)
reflectance and the output coupler mirror allowing approximately 1%
transmission.
Fig.1. Schematic diagram of a heliumneon laser
Commercial He-Ne lasers are relatively small devices, among gas
lasers, having cavity lengths usually ranging from 15cm to 50cm,
and optical output power levels ranging from 0.5 to 50 mW. The red
He-Ne laser wavelength of 633 nm has an actual vacuum wavelength of
632.991 nm, or about 632.816 nm in air. The wavelength of the
lasing modes lie within about 0.001 nm above or below this value,
and the wavelengths of those modes shift within this range due to
thermal expansion and contraction of the cavity.
Frequency-stabilized versions enable the wavelength of a single
mode to be specified to within 1 part in 108 by the technique of
comparing the powers of two longitudinal modes in opposite
polarizations. Absolute stabilization of the laser's frequency (or
wavelength) as fine as 2.5 parts in 1011 can be obtained through
use of an iodine absorption cell.
Fig.2. Energy level diagram of a He-Ne laser
The mechanism producing population inversion and light
amplification in a He-Ne laser plasma originates with inelastic
collision of energetic electrons with ground state helium atoms in
the gas mixture. As shown in the accompanying energy level diagram,
these collisions excite helium atoms from the ground state to
higher energy excited states. This excitation energy transfer
process is given by the reaction equations:
He*(23S1) + Ne 1S0 He (1S0) + Ne* 2S2 + E and
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 32
He*(21S) + Ne 1S0 + E He (1S0) + Ne*3S2 where (*) represents an
excited state, and E is the small energy difference between the
energy states of the two atoms, of the order of 0.05 eV or 387 cm1,
which is supplied by kinetic energy. The medium becomes capable of
amplifying light in a narrow band at 1.15 m and in a narrow band at
632.8 nm. The remaining step in utilizing optical amplification to
create an optical oscillator is to place highly reflecting mirrors
at each end of the amplifying medium so that a wave in a particular
spatial mode will reflect back upon itself, gaining more power in
each pass than is lost due to transmission through the mirrors and
diffraction. When these conditions are met for one or more
longitudinal modes then radiation in those modes will rapidly build
up until gain saturation occurs, resulting in a stable continuous
laser beam output through the front (typically 99% reflecting)
mirror. Spectrum of a helium neon laser illustrating its very high
spectral purity. The 0.002 nm bandwidth of the lasing medium is
well over 10,000 times narrower than the spectral width of a
light-emitting diode, with the bandwidth of a single longitudinal
mode being much narrower still. The visible output of the red He-Ne
laser, long coherence length, and its excellent spatial quality,
makes this laser a useful source for holography and as a wavelength
reference for spectroscopy. Prior to the invention of cheap,
abundant diode lasers, red He-Ne lasers were widely used in barcode
scanners at supermarket checkout counters. Laser gyroscopes have
employed He-Ne lasers operating at 0.633 m in a ring laser
configuration. He-Ne lasers are generally present in educational
and research optical laboratories. 24. Describe the construction
and working of Nd:YAG laser. Answer: Nd:YAG (neodymium-doped
yttrium aluminum garnet; Nd:Y3Al5O12) is a crystal that is used as
a lasing medium for solid-state lasers. The dopant, triply ionized
neodymium, Nd(III), typically replaces a small fraction of the
yttrium ions in the host crystal structure of the yttrium aluminium
garnet (YAG), since the two ions are of similar size. It is the
neodymium ion which proves the leasing activity in the crystal, in
the same fashion as red chromium ion in ruby lasers. Generally the
crystalline YAG host is doped with around 1% neodymium by atomic
percent. Laser operation of Nd:YAG was first demonstrated by J. E.
Geusic et al. at Bell Laboratories in 1964.
Fig.1. Nd:YAG Laser
Working & Theory: Nd:YAG lasers are optically pumped using a
flashtube or laser diodes. These are one of the most common types
of laser, and are used for many different applications. Nd:YAG
lasers typically emit light with a wavelength of 1064 nm, in the
infrared. However, there are also transitions near 940, 1120, 1320,
and 1440 nm. Nd:YAG lasers operate in both pulsed and continuous
mode. The high-intensity pulses may be efficiently frequency
doubled to generate laser light at 532 nm, or higher harmonics at
355 and 266 nm. Nd:YAG absorbs mostly in the bands between 730760
nm and 790820 nm. At low current densities krypton flashlamps have
higher output in those bands than do the more common xenon lamps,
which
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 33
produce more light at around 900 nm. The former are therefore
more efficient for pumping Nd:YAG lasers.
Fig.2. Energy level diagram of Nd:YAG Laser
The amount of the neodymium dopant in the material varies
according to its use. For continuous wave output, the doping is
significantly lower than for pulsed lasers. Nd:YAG lasers and
variants are pumped either by flashtubes, continuous gas discharge
lamps, or near-infrared laser diodes (DPSS lasers). Prestabilized
laser (PSL) types of Nd:YAG lasers have proved to be particularly
useful in providing the main beams for gravitational wave
interferometers such as LIGO, VIRGO, GEO600 and TAMA. 25. Give
principle of propagation of light through optical fiber. Derive an
expression for acceptance angle. Answer: Let us consider light
propagation in an optical fibre. The end at which the light enters
the fibre
is called launching end. Let the refracting index of the core be
1n and the refractive index of cladding be
)nn(n 122 . Let the outside medium from which the light is
launched into the fibre have a refractive
index 0n . Let the refracted angle r , with the axis and strikes
the core-cladding interface at an angle .
Fig.1. Light propagation in fibers
If c (critical angle), the ray undergoes total internal
reflection at the interface. As long as the angle
c , the light remains within the fibre.
Applying Snells law to the fibre, we get
0
1i
n
n
sin
sin
r
..(1)
Now largest value of i occurs when c
From ABC we have
cos)90sin(sin r ..(2)
From eqn.(1).
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 34
cosn
nsin
n
nsin
0
1r
0
1i ..(3)
When maxic,
c0
1
max cosn
nsin ..(4)
But 2
1c
n
nsin
Therefore
0
2
2
2
1
cn
nncos
.(5)
Substituting the expression (5) into (4), we get
0
2
2
2
1
maxn
nnsin
(6)
If 202221 nnn , then for all values of i , total internal
reflection will occur. Assuming 1n0 , the maximum value of isin for
a ray to be guided is given by
222
1max nnsin .........(7)
22211max sin nn (8)
The angle m is called the acceptance angle of the fibre.
Acceptance angle may be defined as the
maximum angle that a light ray can have relative to the axis of
the fibers and propagates down the fibre. 26. Give the construction
and theory of Huygens eyepiece. Why cannot across wire be used with
it? Answer: An eye-piece is a combination of lenses designed to
magnify the image already formed by the objective of a telescope
and microscope. An eyepiece consists of two planoconvex lenses. F
is called the field lens and E the eye lens (Fig. 1.). The field
lens has large aperture to increase the field of view. The eye lens
mainly magnifies the image. To reduce the spherical aberration, the
lenses taken are plano-convex lenses. Further the focal lengths of
the two lenses and their separation are selected in such a way as
to minimize the chromatic and spherical aberrations. A combination
of lenses is used in an eyepiece of a simple lens magnifier for the
following reasons: (i) The field of view is enlarged by using two
or more lenses. (ii) The aberrations can be minimized.
Fig.1. Eyepiece
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 35
Construction: It consists of two plano- con vex lenses of focal
lengths 3f (field lens) and f (eye lens), placed a distance 2f
apart [Fig.2.]. They are arranged with their convex faces towards
the incident rays. The eye-piece satisfies the following conditions
of minimum spherical and chromatic aberrations. (i). The distance
between the two lenses for minimum spherical aberration is given by
a=f1-f2. In Huygen's eyepiece, a=f1-f2=3f-f=2f. Hence this
eye-piece satisfies the condition of minimum spherical
aberration.
(ii). For chromatic aberration to be minimum 2
ffa 21
. In Huygens' eyepiece, f2
2
ff3a
. Hence
this eyepiece satisfies the condition of minimum chromatic
aberration.
Fig.2.
Working: An eye-piece forms the final image at infinity. Thus
the field lens forms the image I2 the first Field Lens focal plane
of eye lens, i.e., at a distance f to the left of eye-lens. Now the
distance between the field lens and eye-lens is 2f. Therefore, the
image I2 lies at a distance f to the right of field lens. The image
I1 formed by the objective of microscope or telescope acts as the
virtual object for the field lens. Thus we treat I1 as the virtual
object for the field lens, and I2 as the image of I1 due to it
(Fig.2.) or
?uandf3F,fv . We have
2
f3u
f3
f2u
u
1
f3
ff3
u
1
f3
1
f
1
f3
1
u
1
f
1
F
1
u
1
v
122
------- (1)
i.e. I1 should be formed at a distance 2/f3 from the field lens.
Therefore, the rays coming from the objective which converge
towards I1 are focused by the field lens at I2. The rays starting
from I2 emerge from the eye-lens as a parallel beam.
Fig.3.
Cardinal Points of Huygens Eyepiece: The equivalent focal length
F of this eyepiece is
2
f3F
f3
2
ff3
f2
f
1
f3
1
ff
a
f
1
f
1
F
1
2121
The second principal point is at a distance from the eye
lens.
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 36
ff2
f2
f2ff3
f2f
aff
af 2
21
2
The first principal point is a distance from the field lens.
Fig.4.
The position of the principal points P1 and P2 and the principal
focal point F1 and F2 are shown in Fig.4. Since the system is in
air, the nodal points coincide with the principal points. 27.
Obtain the expression for the focal length of combination of two
thin lenses. Also find the positions of cardinal points for a lens
combination. Answer: Consider two thin convergent lenses L1 and L2
of focal length f1 and f2 placed coaxially at a distance d apart.
Let a ray AB parallel to the axis be incident at B on the lens L1
at a height C1B=h. It surfers deviation through an angle 1 at the
first lens and proceeds towards the principal focus F1. It means
the second lens at a point D at a height h2 from the axis on
refraction it is further deviated through an angle 2 and meets
finally the axis at F2. As the incident parallel ray PA after
refraction through both the lenses meets the axis at F2, the point
F2 must be the second principal focus of the combination. The lens
of focal F placed at H2 is called the equivalent lens, which can be
replace the combination.
Fig.1
Deviation produced by the equivalent lens F
h1
Deviation produced by the first lens 1
11
f
h
Deviation produced by the second lens 2
22
f
h
As both deviations are in the same direction, the total
deviation is
f3f2
f6
f2ff3
f2f3
aff
af 2
21
1
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Subject: Applied Physics [New] Code 300218 (15)
Aloke Verma, Department of Applied Physics, SRIT, New Raipur
Page 37
2
2
1
1
21f
h
f
h
From fig.1 111112 dhtanBChCDhh
11tan
1
112
f
hdhh
2121212121211
11
21
11
ff
d
f
1
f
1
F