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Page # 1
WAVE OPTICS
Coherent and Incoherent SourcesWhy do we not commonly see
interference effects with visible light? With light from a source
such as theSun, an incandescent bulb, or a fluorescent bulb, we do
not see regions of constructive and destructiveinterference;
rather, the intensity at any point is the sum of the intensities
due to the individual waves.Light from anyone of these sources is,
at the atomic level, emitted by a vast number of
independentsources. Waves from independent sources are incoherent;
they do not maintain a fixed phase relation-ship with each other.
We cannot accurately predict the phase (for instance, whether the
wave is at amaximum or at a zero) at one point given the phase at
another point. Incoherent waves have rapidlyfluctuating phase
relationships. The result is an averaging out of interference
effects, so that the totalintensity (or power per unit area) is
just the sum of the intensities of the individual waves. Only
thesuperposition of coherent waves produces interference. Coherent
waves must be locked in with a fixedphase relationship. Coherent
and incoherent waves are idealized extremes; all real waves fall
some-where between the extremes. The light emitted by a laser can
be highly coherent-two points in the beamcan be coherent even if
sep arated by as much as several kilometers. Light from a distant
point source(such as a star other than the Sun) has some degree of
coherence.
Thomas Young (1773-1829) performed the first visible-light
interference experiments using a clever technique toobtain two
coherent light sources from a single source. When a single narrow
slit is illuminated, the lightwave that passes through the slit
diffracts or spreads out. The single slit acts as a single coherent
sourceto illuminate two other slits. These two other slits then act
as sources of coherent light for interference.
INTENSITY OF TWO SOURCE INTERFERENCEWe now obtain an expression
for the distribution of intensity of two coherent sources that are
in phase.The wave function in this case is the electric field. We
assume that the slits are narrow enough fordiffraction to spread
light from each slit uniformly over the screen. Thus, the amplitude
of the fields anypoint on the screen will be equal. At a given
point of the screen the fields due to S1 and S2 are
E1 = E0 win (t) ; E2 = E0 sin (t + )where the phase difference
depends on the path difference = r2 r1. Since one wavelength
corresponding to a phase change of 2, a distance corresponds to a
phase change given by / 2 = / . If the screen is far from the
slits, d sin (see the ref. figure for YDSE), therefore
= 2
=
sind2
The resultant field is found from the principle of superposition
:E = E1 + E2 = E0 sin (t) + E0 sin (t + )
TEACHING NOTESTEACHING NOTES
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Page # 2
By using the trigonometric identity sin A + sin B = 2 sin [(A +
B)/2] cos [(A B)/2], we obtain
E = 2E0 cos
2 sin
2t
The amplitude of the resultant wave is 2E0 cos (/ 2). The
intensity of a wave is proportional to thesquare of the amplitude,
so from equation wave have
I = 4I0 cos2 cos2
2where I0 E0
2 is the intensity due to a single source. This function is
plotted in fig. of pt (iv). The maximaoccur when = 0, 2, 4, ... =
2m. At these points I = 4I0 ; that is the intensity is four times
that of asingle source. The minima (I = 0) occur when = , 3, 5,
....... = (2m + 1).
(i) If amplitudes of waves arriving at point P on the screen are
different then resultant intensity is given by
I = I1 + I2 + 21II2 cos
Also, Imax = 221 II , when cos = 1Imin = 221 II , when cos =
1
(ii)min
max
II
=
2
21
21
IIII
=
2
21
21
AAAA
= 2
1r1r
where r = 2
1
2
1
II
AA
(iii) The phenomenon of interference is based on conservation of
energy. There is no destruction of energy inthe interference
phenomenon. The energy which apparently disappears at the minima,
has actually beentransferred to the maxima where the intensity is
greater than that produced by the two beams actingseparately.
Iav = 21
2
021212
0
2
0 IIcosII2II(21
d
Id
2
0
0dcos
as the average value of intensity is equal to the sum of
individual intensities, therefore the energy is notdestroyed but
merely redistributed in the interference pattern.
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Page # 3
(iv) All maxima are equally spaced and equally bright. This is
true for minima as well. Also interferencemaxima and minima are
alternate. The intensity distribution in interference pattern is
shown in figure.
O /2 In
tens
ity (I
) 221max III
21av III 2
21min III
Path difference ( x)
(iv) Path difference (x) and phase difference () are related as
given below : path difference = 2 phase difference
or x = 2
YOUNGS DOUBLE SLIT EXPERIMENT (YDSE)The experiment set up for
Youngs double slit experiment is shown in figure. Light after
passing througha pin hole S is allowed to fall on thin slits S1 and
S2 placed symmetrically w.r.t. S. A screen isplaced at a distance D
from S1 and S2.
y
Q O
D
S1
S2
P
d
S1
S2x
MonochromaticSource Screen
Geometric construction for describing Youngs double-slit
experiment
S
(D >> d)
Let P be the point, at which we want to investigate the
intensity. Two rays S1P and S2P starting from S1and S2 reach P and
interfere with each other.If x is the path difference between two
rays,
x = S2P S1P d sin Ddy
DytansinanglesmallFor
This equation assumes that S1P and S2P are parallel, which is
approximately true because D is muchgreater than d.
(a) Maxima : Point P will be a bright spot if the path
difference x is integral multiple of .
yn = dnD
where, n = 0, 1, 2, 3, .......
Thus, bright spots are obtained at distances, 0, dD
, dD2
, dD3
....... from O.
(b) Minima : Point P will be a dark spot if the path difference
x is an odd multiple of 2
.
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Page # 4
i.e., if Dyd
= 2)1n2(
yn = d2D)1n2(
where, n = 0, 1, 2, 3, ........
Thus, dark spots are obtained at distances, d2D5,
d2D3,
d2D
....... from O.
(c) Fringe width ()It is the distance between two consecutive
bright or dark fringes.
Let yn and yn 1 respectively, be the distances of nth and (n
1)th bright fringe from O,
= [yn yn1] = n dD
(n 1) dD
= dD
(n n + 1)
or = dD
Similarly, it can be proved that distance between two
consecutive dark fringes, is given by
= dD
= = D / dHence, the bright and dark fringes are equally
spaced.
[Home Work : HCV : 1 to 11Sheet Ex.1 : 1 Ex. 3 : 1, 6Q. Bank :
1, 3 AR : 2 ]
Remarks :(i) If whole apparatus is immersed in liquid of
refractive index then,
= dD
i.e., fringes width decreases
(ii) Some times in numerical problems, angular fringe width ()
is given which is defined as angular separationbetween two
consecutive maxima or minima
= dD
In medium, other than air or vacuum,
= d
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Page # 5
(iii) x = Dyd
is valid when angular position of maxima or minima is less than
6
. However x = d sin is
valid for larger values of provided d
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Page # 6
SHAPE OF INTERFERENCE FRINGES IN YDSEWe discuss the shape of
fringes when two pinholes are used instead of the two slits in
YDSE.Fringes are locus of points which move in such a way that its
path difference from the two slits remainsconstant.
y
S1
S2
O x
Px
S2P S1P = D = constant ........(1)
If = 2
, the fringe represents 1st minima.
S2
S1
Y
X
= 3 = 2 =
= = =
If = 23
it represents 2nd minima
If = 0 it represents central maxima,If = , it represents 1st
maxima etc.Equation (1) represents a hyperbola with its two foci at
S1 and S2The interference pattern which we get on screen is the
section of hyperboloid of revolution when werevolve the hyperbola
about the axis S1S2.
(A) If the screen is to the X-axis, i.e. in the YZ plane, as is
generally the case, fringes are hyperbolic witha straight central
section.
(B) If the screen is in the XY plane, again fringes are
hyperbolic.(C) If screen is to Y-axis (along S1S2) i.e. in the XZ
plane, fringes are concentric circles with center on the
axis S1S2 ; the central fringe is bright if S1S2 = n and dark if
S1S2 = (2n 1) 2
.
Z
A
Y
Y
B
X
X
Y
C[Home Work :HCV : 29 to 32Sheet Ex.1 : 2 to 6 Ex.2 : 1 Ex.3 :
7, 12, 13Q Bank MCQ : 6, 7, 8, 29 ]
{tell students to attempt from the book or can be given as
HW}HCV-Ex. 33
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Page # 7
Ex. Consider the situation shown in figure. The two slits S1 and
S2 placed symmetrically around the centralline are illuminated by a
monochromatic light of wavelength . The separation between the
slits is d. Thelight transmitted by the slits falls on a screen 1
placed at a distance D from the slits. The slit S3 is at thecentral
line and the slits S4 is at a distance z from S3. Another screen 2
is placed a further distance Daway from 1. Find the ratio of the
maximum to minimum intensity observed on 2 if z is equal to
S1
S2
S4
S3
1
z
2D D
d
(a) d2D
(b) d4D
[Sol.(a) Let I is intensity due to slits S1 and S2 on screen S1.
Further, intensity at any point on screen 1 is givenby
IP = 4I cos2
2At slit S3, = 0
3SI = 4I
At slit S4, x = Ddz
= 2
= 4SI = 0
Now on screen 2 Imax = 2SS 43 II = 4IImin = 2SS 43 II = 4I
min
max
II
= 1 Ans
(b) z = d4D
3SI = 4I
At slit S4, x = Ddz
= 4
= 2
4
= 2
4SI = 4I cos2
4 = 2I
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Page # 8
Imax = 2SS 43 II = 2I2I4 = I (2 + 2 )2
Similarly, Imin = 2SS 43 II = 2I2I4 = I 222
min
max
II
= 2
2222
Ans.
HCV-Ex. 28Ex. Figure shows three equidistant slits being
illuminated by a monochromatic parallel beam of light. Let
BP0 AP0 = /3 and D >> . (a) Show that in this case d =
3/D2 . (b) Show that the intensity at P0is three times the
intensity due to any of the three slits individually.
d
d
DA
B
C
P0
Sol.
d
d
DA
B
C
P0
BP0 AP0 = 3
or d sin = 3
or d tan = 3
(For small angle tan sin )
d
D2/d
= 3
or d = 3F2
(b) xA/B = path difference between waves coming from A and B =
3
A/B = phase difference
= 2xA/B = 3
2
Similarly, xB/C = d sin
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Page # 9
= d
D2/d3
= D2d3 2
= 1
B/C = 2Now, phase diagram of the waves arriving at P0 is as
shown below :
AA
A
120
2A
A
120
Amplitude of resultant wave is given by
A = 120cos)A2)(A(2)A2(A 22 = 3 AAs intensity (I) A2 Intensity at
P0 will be three times the intensity due to any of the three slits
individually.
[Home Work : HCV : 20, 25, 26, 27, 28, 33, 34Sheet Ex-1 : 11,
12, 13 Ex.3 : 4, 14, 18Q. Bank : Single correct 4, 5, 9, 10, 11,
13, 14, 15, 16, 24
MCQ 2, 3, 4, 10, 11, 12, 13 AR 3 ]
GEOMETRICAL PATH AND OPTICAL PATHActual distance travelled by
light in a medium is called geometrical path (x). Consider a light
wavegiven by the equation.
E = E0 sin (t kx + )
If the light travels by x, its phase changes by kx = vx, where ,
the frequency of light does not
depend on the medium, but v, the speed of light depends on the
medium as v = c
.
Consequently, change in phase,
= kx = C
(x)
It is clear that a wave travelling a distance x in a medium of
refractive index suffers the same phasechange as when it travels a
distance x in vacuum. i.e. a path length of x in medium of
refractive index is equivalent to a path length of x in vacuum.
The quantity x is called the optical path length of light, xopt.
And in terms of optical path length, phasedifference would be given
by,
= Cxopt = 0
2
xopt
where 0 = wavelength of light in vacuum.However in terms of the
geometrical path length x,
= C
(x) = 2x
where = wavelength of light in the medium
0 .
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Page # 10
1. Displacement of fringe on introduction of a glass slab in the
path of the light coming out of the slits.
D
S2
S1
P
O
O
d
t
On intrudction of the thin glass-slab of thickness t and
refractive index , the optical path of the ray S1Pincreases by t(
1). Now the path difference between waves coming from S1 and S2 at
any point P is
p = S2P (S1P + t( 1))= (S2P S1P) t( 1)
p = d sin t ( 1) if d
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Page # 11
The phase change on reflectionA ray of light is incident on
air-water interface ; let the amplitude reflection and transmission
coefficientsbe r1 and t1 respectively. The amplitudes of reflected
and transmitted waves are ar1 and at1 respectively.From the
principle of reversibility of light, the system retraces its whole
previous motion. The wave ofamplitude ar1 gives a reflected wave of
amplitude ar1t1. The wave of amplitude at1
r ,t1 2
r ,t2 1
at t1 2
gives a reflected wave of amplitude ar2t1 and transmitted wave
amplitude at1t2.So ar1
2 + at1t2 = at1t2 = 1 t1
2 ... (1)Further, the waves of amplitudes at1r2 and ar1t1 must
cancel each other.
at1r2 + ar1t1 = 0r2 = r1 ... (2)
equation shows a difference of phase of between the two cases ;
a reversal of sign means a displacementin the opposite sense. If
there is no change of phase on reflection from above, there must be
a phasechange of from below and vice-versa.When light gets
reflected from a denser medium there is an abrupt phase change of ;
no phase changeoccurs when reflection takes place from rarer
medium.
Thin film interferenceWhen light passes the boundary between two
transparent media, some light is reflected at the boundary.As shown
in the figure some light is reflected from first interface and some
from second interface. If weconsider a monochromatic incident light
the two reflected waves are also monochromatic and coherentbecause
they arise from the same monochromatic incident light wave via
amplitude division. Thesewaves interfere, since they are superposed
along the same normal line.The phase difference between two
interfering waves is due to :(1) Optical path difference (due to
distances travelled),(2) Reflection from a denser medium.
The second factor is irrelevant for reflection at rarer
medium.Three situations may arise :(1) Neither wave experiences a
phase change upon reflection.(2) Both the waves suffer a phase
change upon reflection.In either of these two cases the phase
change due to reflection is irrelevant ; no difference in phase
resultsdue to reflection.In either of these cases phase change is
determined solely from optical path difference.
Condition for constructive interference :
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Page # 12
2t = mCondition for destructive interference :
2t =
21m
where m = 0, 1, 2, .......(3) One of the reflected waves
experiences a phase change of radian upon reflection and the
otherwaves does not.It is material which wave suffers a phase
change ; the conclusions in the previous case are first
reversed.
Condition for destructive interference :2t = m
Condition for constructive interference :
2t =
21m
where m = 0, 1, 2, .......
Ex. A light ray is incident normal to a thin layer of glass.
Given the figure,what is the minimum thickness of the glass that
gives the reflected lightan orangish color ( air = 600 nm)?(A) 50
nm (B*) 100 nm (C) 150 nm (D) 200 nm (E) 500 nm
[Sol. For reflected light to have orangish color, rays from A,
C, E must be out of phase for l = 600 nmor = (2n + 1)
or 2gt (2n + 1) 2
i.e. t = g4
)1n2(
A C Et
DB
Transmitted light
(dense)(rare) water
air(rare)glass
Reflected light
or tmin = g4
= 100 nm ]
(OPTIONAL) INTERFERENCE DUE TO REFLECTED LIGHT :Consider a
transparent film of thickness t and refractive index . A ray SA
incident on the upper surfaceof the film is partly reflected along
AR1 and partly refracted along AB. At B part of it is reflected
alongBC and finally emerges out along CR2. The difference in path
between the two rays. AR1 and CR2 iscalculated as given below :Let
CN and BM be perpendicular to AR1 and AC. As the paths of the rays
AR1 and CR2 beyond CNare equal. The path difference between them
is
x = Path ABC in film Path AN in air= (AB + BC) AN = 2 AB AN
r
i C Air
R2R1N
i
S
A
B
rM
t
Air
T1
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Page # 13
Now, AB = BC = BMAB
BM = BM sec r = t sec r
and AN = ACAN
AC = AC sin i = 2 AM sin i
= 2 BMAM
BM sin i = 2 (tan r) t
rsin
rsinisin
= 2t rcosrsin 2
= 2t sec r sin2 r
Then, = 2AB AN = 2t sec r 2 t sec r sin2 r= 2 t sec r (1 sin2 r)
= 2 t cos r
The ray AR1 having suffered a reflection at the surface of
denser medium undergoes a phase change or
path diff. of 2
.
At B the reflection takes place when the ray is going from a
denser to rarer medium and there is no phasechange.Hence, the
effective path difference between AR1 and CR2 is given by
Path Diff. (x) = 2 t cos r 2
(i) If the path difference x = n where n = 0, 1, 2, 3, 4 etc.,
constructive interference takes placeand the film appears
bright.
2t cos r 2
= n
or 2t cos r = (2n + 1) 2
(ii) If the path difference x = (2n + 1) 2
where n = 0, 1, 2, ........... etc., destructive
interference
takes place and the film appears dark.
2t cos r 2
= (2n + 1) 2
or 2t cos r = n
Remarks :(i) If the thickness of the film is very small as
compared to the wavelength of light used, so that 2t cos r can
be neglected, then the total path difference between AR1 and CR2
will reduce to 2
. Thus two rays will
interfere destructively and darkness will result.
(ii) It should be remembered that the interference pattern will
not be perfect because the intensities of the rayAR1 and CR2 will
not be the same and their amplitudes are
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Page # 14
Fringes of equal thicknessSoap bubbles and oil films on a road
do not have uniform thickness of the film at any given
pointdetermines whether the reflected light has a maximum or
minimum intensity. When white light is used,each wavelength has its
own fringe pattern. At a given point of the film, one wavelength
may be enhancedand / or another wavelength suppressed. This is the
source of the colors in soap bubbles an oil films onthe road.A
wedge-shaped film of air may be produced by placing a sheet ofpaper
or a hair between the ends of two glass plates, as in fig. With
flatplates, one sees a series of bright and dark bands, each
characteristic ofa particular thickness. If the plates are not
flat, the fringes are not straight; each is locus of points with
the same thickness. If one plate is knownto be flat, the fringes
display the irregularities of the other, as shown infigure. The
pattern shows where the plate needs to be polished for it tobe made
optically flat.
Ex. A wedge-shaped film of air is produced by placing a fine
wire of diameter D between the ends of two flatglass plates of
length L = 20 cm, as in fig. When the air film is illuminated with
light of wavelength = 550nm, there are 12 dark fringes per
centimeter. Find D.
Sol. A indicated in fig. only one of the reflected ray suffers a
phase inversion. At the thin end of the wedge,where the thickness
is less than /4, the two rays interfere destructively. This region
is dark in thereflected light. The condition for destructive
interference in the reflected light is
2t = m m = 0, 1, 2, .....the change in thickness between
adjacent dark fringes is t = / 2. The horizontal spacing
betweenfringes d = 1/12 cm = 8.3 104 m. From figure we see that D /
L = t / , so
= d2L
= m106.16)m2.0()m105.5(
4
7
Thus D = 6.6 105 m
Newtons RingsWhen a lens with a large radius of curvature is
placed on a flat plate, as to fig. a thin film of air is
formed.When the film is illuminated with monochromatic light,
circular fringes, called newtons rings, can be withthe unaided eye
or with a low power microscope (figure). An important feature of
Newtons rings is thedark central spot. Newton tried polishing the
surfaces to get rid of it, The dark spot was also initiallypuzzling
to Young. It implied that the light wave suffers a phase inversion
on reflection at a medium witha higher refractive index. Young
tested this idea by placing oil of sassafras between a lens of
crown glassand a plate of flint glass. The refractive index of the
oil is between the values for these two glasses. Sinceboth
reflections occur at a medium with a higher refractive index, they
should both suffer a phase inversionand therefore be in phase be in
phase. This is precisely what happened : The central spot became
brightand undoubtedly gave Young much satisfaction.
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Page # 15
Ex. In an experiment on Newtons rings the light has a wavelength
of 600 nm. The lens has a refractive indexof 1.5 and a radius of
curvature of 2.5 m. Find the radius of the 5th bright fringe.
Sol. If R is the radius of curvature of the lens, then from fig.
we see that r2 = R2 (R t)2, where r is the radiusof a fringe and t
is the thickness of the film. Since t is very small, we may drop
terms in t2 to obtain.
r2 2Rt ... (i) In order to find r, we must first find t. The
condition for a bright fringe is
2t = F21m
... (ii)
We note that n = 1 for the air film (the index for the glass is
irrelevant) and that m = 4 for the fifth brightfringe. Thus, from
(ii)
t = 2
)106()5.4( 7 = 1.35 106 m
Substituting this into (i), we find r = Rt2 = 2.6 103 m.
[ Home Work : HCV : 35, 36, 37, 38Sheet Ex.1 : 14 to 17Ex.2 : 8,
9Q.Bank : Single correct - 25, 26, 27
MCQ - 7 ]
HUYGENS CONSTRUCTION(This matter is from NCERT)Huygens, the
Dutch physicist and astronomer of the seventeenth century, gave a
beautiful geometricaldescription of wave propagation. We can guess
that he must have seen water waves many times in thecanals of his
native place Holland. A stick placed in water and oscillated up and
down becomes a sourceof waves. Since the surface of water is two
dimensional, the resulting wavefronts would be circlesinstead of
spheres. At each point on such a circle, the water level moves up
and down. Huygens idea isthat we can think of every such
oscillating point on a wavefront as a new source of waves.
According toHuygens principle, what we observe is the result of
adding up the waves from all these different sources.These are
called secondary wave or wavelets. Huygens principle is illustrated
in (figure 10.1) in thesimple case of a plane wave.
(i) Consider a plane wave passing through a thin prism. Clearly,
the portion of the incoming wavefrontwhich travels through the
greatest thickness of glass has been delayed the most. Since light
travels moreslowly in glass. This explains the tilt in the emerging
wavefront.
(ii) Similarly, the central part of an incident plane wave
traverses the thickest portion of a convex lens and isdelayed the
most. The emerging wavefront has a depression at the centre. It is
spherical and convergesto a focus.
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Page # 16
(iii) A concave mirror produces a similar effect. The centre of
the wavefront has to travel a greater distancebefore and after
getting reflected, when compared to the edge. This again produces a
converging sphericalwavefront.
(iv) Cocave lenses and convex mirror can be understood from time
delay arguments in a similar manner. Oneinteresting property which
is obvious from the pictures of wavefronts is that the total time
taken from apoint on the object to the corresponding point on the
image is the same measured along any rays goingthrough the centre
are shorter. But because of the slower speed in glass, the time
taken is the same as forrays travelling near the edge of the
lens.
[Home Work :Sheet : Ex.3 : 3, 15, 16, 17Q. bank : single correct
- 2, 12, 30 ]