1
M. Bhojaraj, Veerashiava College, Bellary
INTERFERENCE OF LIGHT
When two or more wave trains of light pass through the same medium and cross one another, then
the net effect [displacement] produced in the medium is the sum of the effects due to all the waves. At
any instant, the resultant displacement of the particles of the medium and hence the intensity of light
depends upon (1) the phase difference between the waves and (2) the algebraic sum of the displacement
due to the individual waves. Hence there is a modification in the intensity of light in the region of
superposition of the two waves and is known as interference of light.
Thus, modification in the distribution of light energy due to the superposition of two or more
waves is called interference of light. With a single source, the distribution of energy in the medium is uniform. But, when there are two
sources nearby, giving out continuous waves of the same wavelength and amplitude and having the same
phase or a constant phase difference, the distribution of the energy in the medium is no longer uniform.
At some points where the crest of one wave falls on the crest of the other wave (or trough of one wave
falls on the trough of the other wave), the resultant displacement doubles and the intensity at that point
become four times [Intensity is directly proportional to the square of the amplitude] and it is called
constructive interference. On the other hand, if the crest of one wave falls on the trough of the other wave
and vice versa, the resultant displacement is zero and hence the intensity becomes zero. This is known as
destructive interference.
Due to interference of light, there is only a redistribution of the energy in the medium and there is
no creation or destruction of energy in the medium. The energy is only transferred from some points to
neighboring points in the medium.
Interference of light obeys the law of conservation of energy:
We can prove that interference obeys law of conservation of energy. If A is the amplitude of each
of individual interfering waves, then the sum of intensity in the absence of interference is
I = A2 + A2 = 2 A2 - - - (1)
On the other hand, in interfering waves, at points of constructive interference, I1 = (2A)2 = 4 A2
and at adjacent points of destructive interference, I2 = (A- A)2 = 0. Hence the average intensity in the
region of superposition is 22
211 22
04
2A
AIII
- - - (2)
From equations (1) and (2) it follows that the law of conservation of energy is obeyed.
Let the two sources A and B emit continuous waves of monochromatic light of wavelength and
of the same amplitude and in the same phase. At some point “O” equidistant from A and B, the two wave
trains always arrive in the same phase and reinforce each other producing maximum intensity at that
point. At some other point P, where AP and BP are not equal, there will be a difference in phase, which
depends upon the path difference [BP – AP]. If this path difference is an even multiple of /2, [BP – AP
= 2n(/2) = n] the waves arrive at the point in the same phase and reinforce each other producing
maximum intensity.
When the path difference [BP – AP] = [2n + 1](/2) [odd multiple of /2] the waves arrive at P in
opposite phase and one wave cancel the effect of the other wave resulting in zero intensity at such points.
CONDITIONS FOR PERMANENT INTERFERENCE OF LIGHT - COHERENT SOURCES:
To obtain sustained interference of light waves at a point in the medium the following conditions
should be satisfied.
1. Two sources should continuously emit waves of the same wavelength and frequency.
Interference is a result of millions of waves crossing a point in the medium and hence the resultant
displacement is a constant only if the periodic time is same for both the interfering waves.
To find the position of maximum and minimum energy, consider
two sources of light A and B, as in the following figure (1), lying close to
one another.
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M. Bhojaraj, Veerashiava College, Bellary
2. For observing interference fringes, the amplitude of the two interfering wave trains should be
equal or very nearly equal. This will result in distinct visibility of bright and dark fringes.
3. Two sets of interfering wave trains should either have the same phase or a constant phase
difference. This means that the two waves must originate from a single parent source. The phase
difference between waves arriving at any point in the medium depends upon (i) path difference of
the point from the two sources and (ii) the phase relation between the waves at the time of
emission from the source.
The path difference is constant for a given point in the medium as it depends only on the
distance, while the latter rapidly changes with time not only in different sources, but even in
different parts of the same source. Due to this reason it is not possible to observe interference
phenomenon using two independent sources.
In addition to the above three conditions, the following conditions should be maintained to observe
clear, well defined fringe pattern in the region of superposition.
4. Two sources should be very narrow. A broad source is equivalent to a large number of narrow
sources lying side by side. Each set of these sources produces their own interference pattern,
which overlap on one another to such an extent that the interference pattern is replaced by general
illumination.
5. Two sources emitting a set of interfering beams should be very close to each other. That
means the two interfering wavefronts must intersect at a very small angle. Otherwise, the path
difference between the interfering waves is large and owing to small wavelength of the waves the
interference bands are formed so close to one another that the fringes may not be distinctly
visible.
Coherent sources:
Two sources of light are said to be coherent if they emit (1)waves continuously of same
frequency and wavelength (2) amplitude of the two waves must be equal or nearly equal and (3) they
must emit waves of same phase or waves having a constant phase difference.
Methods for producing coherent sources:
There are two methods for producing coherent sources. They are (1) by the method of Division of
wavefront and ( 2) by method of division of amplitude.
Division of wave front:
In division of wavefront, the wavefront is divided into two parts, such that each part act as
independent source emitting light having the same phase or constant phase difference.
Examples:
(1) In Young’s double slit method, a single wavefront is divided in to two parts by allowing light
from a single slit to pass through two slits
(2) In Lloyd’s method, the light is reflected from a plane mirror at grazing incidence. The source and
its reflected image act as two coherent sources. In this case the waves emitted from the source and
its virtual image emit waves of constant phase difference of Π.
(3) Fresnel’s Bi-prism method, light from a narrow source is passed through the bi-prism. The two
refracted images act as two virtual coherent sources
Division of Amplitude:
In division of amplitude, the wavefront is divided in amplitude and the divided amplitude is made
to interfere. The divided amplitudes act as two coherent sources.
Examples:
(1) In thin films, the light falling on the film is partly reflected form the upper and lower surfaces.
These reflected waves from the two surfaces act as two coherent sources.
(2) In Newton’s rings, the light reflected from the upper and lower surfaces of the air film formed
between the lens and the glass plate act as two coherent sources.
(3) In Michelson interferometer, the beam is divided in amplitude by using a glass plate and again
they are re-united. The re-united light act as coherent sources.
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M. Bhojaraj, Veerashiava College, Bellary
Young’s double slit experiment: -
Thomas Young demonstrated the interference
phenomenon in 1801. The arrangement is as shown in
figure (1). Light from a monochromatic source is incident
on a screen having a narrow slit S in it. The cylindrical
waves emerging from it are allowed to fall on another
screen containing two close, narrow and parallel slits S1
and S2. The slits S1 and S2 are parallel to the slit S. The
cylindrical waves emerging from the two slits S1 and S2
super-impose and the resultant intensity of light is studied
by arranging another screen at a suitable distance as shown
in figure (1). On the screen bright and dark bands known as
interference bands are seen. The waves arriving in phase at
a point on the screen form a bright band and the waves
arriving out of phase at a point on the screen form a dark
band at that point.
Theory of interference: - Let us consider two light waves of same angular frequency traveling in a medium along the
same direction. Let ‘a’ and ‘b’ represents the amplitudes of the individual waves. The displacement of
any particle of the medium due to the two waves at any instant t are given by,
y1 = a sin t and ---------------------------------------- (1)
y2 = b sin (t +) -------------------------------------- (2)
where is the phase difference between the waves at the point where they meet each other.
According to the principle of superposition, the net displacement is given by,
y = y1 + y2 -------------------------------------------- (3)
= a sin t + b sin (t +)
= a sin t +b (sin t cos + cos t sin)
= (a + b cos) sin t + (b sin) cos t
= R sin t cos + R cost sin ---------------- (4)
= R {sin (t +)} ----------------------------- (5)
Where, a + b cos = R cos , ----------------------------------------- (6)
b sin = R sin -----------------------------------------(7)
Squaring and adding equations (6) and (7), we get
R2 = a2 + b2 + 2ab cos -------------------------------- (8)
and, tan = b sin /(a + b cos ) -------------------------------------(9)
Where R is the resultant amplitude and θ is the phase difference between the resultant and first wave.
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M. Bhojaraj, Veerashiava College, Bellary
Conditions for constructive interference: - The resultant amplitude R has maximum value when cos = +1 or when =2n where n is 0, 1,
2, 3, 4……….
The value of R2max = a2 + b2 + 2ab = (a+b)2 Rmax = (a + b) ---------(10)
The corresponding path difference for maximum value of R is given by,
= (/2) = (/2) 2n = n ----------------------- (11)
If a = b, then Rmax = 2a. The intensity of light Imax = R2max.= 4 a2. The two waves interfere
constructively and hence the resultant intensity is 4 times the intensity of the individual waves.
Condition for Destructive interference:- The resultant amplitude R has minimum value when cos = zero or when = (2n+1) where n is
0, 1, 2, 3, 4 ……….
The value of R2min = a2 + b2 - 2ab = (a-b)2 Rmin = (a - b) --------- (12)
The corresponding path difference for maximum value of R is given by,
= (/2) = (/2) (2n + 1) = (2n +1)/2 -------- (13)
If a = b, then Rmin = zero. The intensity of light Imin = zero. The two waves interfere destructively
and hence the resultant intensity is zero.
NOTE: - During interference the two waves does not destroy each other, but, their energy is only
redistributed in the region of superposition. The average of the intensity of maximum and the minimum is
equal to Iav = (Imax+ Imin)/2 = {(2a)2+02}/2 = 2 a2, which is equal to the sum of the intensity of the two
individual waves (a2+a2). In other words, this phenomenon is supporting the law of conservation of
energy.
Expression for fringe width: -
Figure (2)
Let ‘d’ be the distance between the two slits A & B and D is
the distance between the screen and the plane of the slits. Let O
be a point on the screen such that AO=BO. The waves from A
and B arrive in phase at O since they travel the same path and
hence a bright band is formed at O. It is called central bright
band (centre of the fringe pattern).
Let P be a point on the screen at a small distance x from O
as shown in figure 2. The path difference between the waves
from A and B reaching the point P is BP – AP =
From BFP, BP2 = BF2 +FP2 = D2 + (x + d/2)2---- (1)
From AEP, AP2 = AE2 +EP2 = D2 + (x - d/2)2---- (2)
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M. Bhojaraj, Veerashiava College, Bellary
Now, BP2 – AP2 = (BP+AP) (BP – AP)
= [D2 + (x + d/2)2] – [D2 + (x - d/2)2]
= x2 + d2/4 + xd - x2 - d2/4 + xd = 2xd
Therefore, the path difference is given by,
= BP – AP = 2xd/ (BP+AP) = 2xd/ (D+D) [BP AP D]
= xd/D --------------------------------- (3)
For bright fringe = xd/D = n -------------------------------------------- (4)
The distance of the nth bright fringe is given by, xn = nD/d
The width of a dark fringe = xn – xn-1 = [n – (n -1)] D/d = D/d----- (5)
For dark fringe = xd/D = (2n+1)/2 --------- (6)
The distance of the nth dark fringe is given by,
xn = (2n+1) D/2d
The width of a bright fringe = xn – xn-1
= [(2n+1)-(2{n-1}+1)] D/2d
= [2n+1–2n+1] D/2d
= D/d ------------ (7)
Equations (4) and (5) show that, the width of a bright or a dark fringe (band) is a constant and is
independent of the fringe number.
The width of a bright or a dark fringe is known as fringe-width and is denoted by .
Thus, = D/d. --- (8)
Note : -
1. Interference pattern consists of bright and dark fringes of equal width.
2. If white light sources are used, the fringe pattern is coloured with the central bright fringe white
and the other fringes are coloured in the order of VIBGYOR.
BIPRISM
A biprism consists of two thin prisms with their bases joined and their two faces making an obtuse
angle of about 1790 so that the other angles are each of about 301. In practice the biprism is constructed
from a single glass plate so that it is ground to have the required angles.
When biprism is held symmetrically at a short distance from an illuminated slit, the wavefront is
divided into two parts. Each half of the Bi-prism intercepted by one half of the wavefront produces a
virtual image of the source S by refraction. The distance between the source S and the Biprism P is so
adjusted that the two virtual images S1 & S2 are quite close to one another. The waves from the two
virtual sources superimpose and produce interference fringes in the region of superposition AB on the
screen as shown in the following figure (3)
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M. Bhojaraj, Veerashiava College, Bellary
A closely spaced fringe system is produced only in the region AB while the wide set of fringes are
formed at the edges of the pattern on account of diffraction [in regions AE & BF].
Since interference fringes are narrow they are generally observed using a low power microscope
so that the fringes formed in the focal plane of the microscope are observed.
Determination of wavelength of monochromatic light:
Theory of Bi-prism is same as the theory of interference discussed in Young’s method. The fringe
width is given by the formula,
= D/d ------- (1)
Where, “” is the wavelength of the radiation, “D” is the distance between the slit and the screen & “d” is
the distance between the two virtual sources.
Procedure:
The slit S, the Bi-prism P and a micrometer eyepiece E are mounted at the same height along a
straight line using three uprights capable of vertical and transverse adjustments, fitted on an optical
bench. The slit is illuminated by the monochromatic light of wavelength (), which is to be determined.
The bi-prism is moved at right angels to the optical bench, so that on looking through the eyepiece
holder, two bright virtual slits S1 & S2 are observed. The eyepiece is moved perpendicular to the bench so
that overlapping region AB is brought in the field of view and the fringes are obtained in the focal plane.
Well-defined fringes are obtained by arranging the refracting edge of the bi-prism parallel to the slit by
rotating the bi-prism suitably.
(1) Measurement of fringe width : The centre of the particular bright fringe is brought on the cross-wire of the micrometer
eyepiece and the reading for this fringe is noted using micrometer eye piece. The position of
the cross wire is moved to one side until 5 fringes cross the field of view using slow motion
screw and the reading for the 5th bring fringe is noted. The same procedure is repeated to note
the readings for 10th, 15th, 20th, 25th and 30th fringes. The readings are tabulated and the width
of 20 fringes and hence with its mean value is calculated. From the mean width (β1) of 20
fringes, the fringe width () is calculated using the relation, 20
1
Trial No Fringe Number Reading for Fringes Width of 20
fringes
Mean width of 20
fringes
1 0
2 5
3 10
4 15
5 20
6 25
7 30
\
Mean fringe width, 20
1 = _______ mm
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M. Bhojaraj, Veerashiava College, Bellary
(2) Measurement of distance D: The distance D between the slit and the focal plane of the eyepiece is measured with the help
of the scale on optical bench.
(3) Measurement of distance between the virtual sources:
The position of the lens is again adjusted to get diminished image of the slits. The distance
between the two diminished images ‘y’ is determined. The magnification of the diminished
image is given by, m2 = y/d = u/v --------- (3)
Therefore, m1 x m2 = [x/d] [y/d] =[v/u] [u/v] = 1
And hence, xy = d2 or d = [xy]1/2 ------(4)
Knowing ‘x & y’ the distance between the slits can be calculated using the equation (4)
Thus by knowing fringe width (), d and D the wavelength of the monochromatic source
is calculated using equation,
= d/D ---------- (5)
LLOYD’S SINGLE MIRROR
Lloyd’s mirror is an optical device for producing
interference fringes devised by Dr. H. Lloyd in 1834. It
is a single mirror of length about 30 cm and breadth
about 5 cm made of either polished metal or highly
polished black glass so that reflection occurs only from
the front surface.
When light from a slit ‘S’ [with the slit parallel to
the breadth of the mirror] is made to fall at grazing
incidence, the light reflected from the mirror appears to
come from the virtual image S1. Thus the direct light
from S and the light from the virtual slit S1 superimpose
in the region AB on the screen to produce interference
pattern with straight fringes parallel to the slit as shown
in figure (4).
Note 1: Fringes are formed only on one side of the
central fringe [not visible] corresponding to point C of
equal path difference from the two sources.
Central fringe can be made visible by
introducing a thin mica sheet of a certain thickness‘t’ so
that the extra path difference of ( - 1) t is introduced in
the path one beam.
Note: The central fringe in this case is dark,
which according to theoretical consideration
must be bright. The reason for central dark
fringe is due to a phase change of on
reflection from the denser glass plate. Thus
the waves from the two sources have an
initial phase difference of at the time of
emission itself.
Note: Due to reason mentioned in note (2),
the condition for constructive interference to
get bright fringe is = xd/D = (2n+1)/2 and
the condition for destructive interference is
= xd/D = n
A convex lens of suitable focal length
is placed between the eyepiece and the bi-
prism without disturbing their positions.
The position of lens is adjusted to get
magnified image of the slits. The
separation of the two magnified images
‘x’ is noted. If, ‘d’ is the distance between
the two slits, then the magnification is
given by, m1 = x/d = v/u ------(2)
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M. Bhojaraj, Veerashiava College, Bellary
Thickness of a transparent plate using Biprism:
Due to the introduction of the glass plate, the ray from A travels a path = AP + (μ – 1) t.
In the absence of the plate, the central maximum is formed at O. Due to the introduction of the
plate, the central maximum shifts to a point P at which the optical paths travelled by the rays from A and
B are equal.
In the absence of the plate, the path difference = D
dxAPBP .
In presence of the glass plate, the path difference = tD
dxtAPBP 11
If P corresponds to the position of nth maximum, then,
d
tD
d
nD
d
Dtnxnt
D
dx 111
In the absence of the plate, the nth maximum is at a distance =d
nD
The shift in the position of nth maximum is
d
tD
d
nD
d
tD
d
nDx
11
It is evident that the shift in the nth fringe is independent of the fringe number and is same for all
fringes. Thus the introduction of glass plate does not change the fringewidth.
The thickness of glass plate is related to the shift in fringe pattern by the relation,
1
D
dxt
Thus, by determining ‘x’,‘d’ & ‘D’ from experiment and by assuming the value of μ, it is possible
to determine the thickness of a thin glass plate by using interference phenomenon. Alternately, if
thickness is known, the refractive index of the material of the medium can be calculated.
INTERFERENCE BY DIVISION OF AMPLITUDE
Beautiful colour observed in thin films of oil spread on water, soap bubbles and heated metal
surface is a result of interference phenomenon. In all these cases the division of amplitude produces
interference phenomenon. The divided amplitude reunites in the region of superposition to produce
interference.
When light from the source passes through the
glass plate, an extra path difference is introduced by the
glass plate.
The distance travelled in the glass plate = t.
The distance travelled in air = AP – t
If ‘c’ and ‘v’ are the velocities of light in air and
glass respectively, then the time taken to travel the
distance from A to P
c
tAP
c
t
c
tAP
v
t
c
tAP )1(
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M. Bhojaraj, Veerashiava College, Bellary
Interference in thin film of uniform thickness [plane parallel thin films]
Let monochromatic light of wavelength from an
extended source S fall on the surface XY of the
thin film of uniform thickness “t” and refractive
index “” as shown in figure (1). The ray SA of
the incident wavefront is partly reflected along AP
and partly refracted along AE. At E, the ray is
partly reflected along EB and partly refracted
along the emergent ray EP1. Similar reflection and
refraction occurs at B, F, C, G, D and so on as
shown in figure. This successive reflection and
refraction of the incident light results in two
system of parallel beam. One of the beams is due
to reflection of light & the other is due to
transmission of light. In both the system of parallel
beam intensity of successive rays falls off rapidly
from one ray to the next ray. These reflected and
transmitted parallel rays produce interference
phenomenon.
[A] CONDITION FOR CONSTRUCTIVE & DESTRUCTIVE INTERFERENCE FOR REFLECTED
SYSTEM OF RAYS:
Figure (2)
ABN = i
BAM = r
Also,
AGM = r
AG = 2t
Let us consider two reflected rays AP and BQ
in reflected system. AP is reflected from the first
surface so that the reflection occurs from denser
medium of refractive index “” and BQ, after
reflection from the lower surface from a rarer
medium is transmitted from the first surface.
Draw BN perpendicular to AP and AM
perpendicular to EB. Draw AL normal to the first
surface at the point A & produce it to meet BE
produced at G. From figure, it follows that,
ABN = i, BAM =AGM = r.
Also, the triangles ALE & GLE are congruent.
Therefore, EA = EG & AL = LG
If “” is the refractive index of the material of the film, then the path difference measured in air between
the two reflected rays AP & BQ is given by,
Path difference, = [AE +EB] – AN = [GE +EB] – AN = GB – AN
Therefore, path difference, = GB - MB = [GB – BM] = GM
In triangle AGM, Cos r = GM/AG GM = AG Cos r = 2t Cos r
Thus, the path difference between the reflected rays, = 2t Cos r -------- (1)
Since the ray AP suffers reflection from denser medium, it undergoes a phase change of “” or a path
increase of [/2], Where is the wavelength of light in air. The ray AE, EB and BQ do not undergo any
phase change as it is reflected from a rarer medium.
Thus, the effective path difference between AP and BQ = 2t Cos r - /2
Consider the reflected rays BQ and CR. These have a path difference of 2t Cos r as they have
only internal reflection. The same path difference exists between successive pairs of reflected rays.
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M. Bhojaraj, Veerashiava College, Bellary
CASE (1): When the thickness of the film is negligibly small compared to .
If 2t Cos r = n, except the first two reflected rays all other successive pair of rays will be in
phase, but their amplitudes rapidly decreases. If the thickness‘t’ of the film is negligibly small, then the
total path difference between AP & BQ equals /2 and hence they interfere destructively and darkness
will result. [due to the effect of AP cancelled by all other rays taken together].
CASE (2): When the thickness of the film is not negligible compared to .
(a) The film appears bright if 2t Cos r - /2 = n, where, n = 0, 1, 2, 3, -------
Or, 2t Cos r = [2n +1] /2 -------- (2)
When the above condition is satisfied, the first and the second rays interfere constructively. The third,
fifth, seventh etc., rays are out of phase with the first two rays, where as, the fourth, sixth, eighth etc.,
rays are in phase with the second ray. The fourth has larger intensity than the fifth, sixth has larger
intensity than the seventh and so on. Due to this the rays of stronger intensity in the series combine with
the first ray to give maximum intensity in the field of view of reflected beam.
(b) The film appears dark if 2t Cos r - /2 = [2n – 1] /2
Or, 2t Cos r = [2n –1 +1]/2 = n. ----------- (3)
In this case the first ray is out ;of phase with the second, while the other rays are in phase with the
second ray. The second and the other reflected rays cancel the effect of the more intense first ray to
produce darkness in the field of view of the reflected beam.
As the source is extended, the rays of light are incident at various angles and hence the path
difference has variable values for different set of reflected rays. Hence the field of view in the
reflected beam consists of alternate bright and dark band. Also, the fringe pattern is not very well
defined as the intensity of minima is not equal to zero.
[B] CONDITION FOR INTERFERENCE IN THE CASE OF TRANSMITTED SYSTEM OF
FRINGES:
figure (3)
The rays emerging from the lower surface of the film may be
brought together by a lens to produce interference.
The path difference between the rays EP1 & FQ1 is given by,
= [EB + BF] – EN
= [EB + BG] – EN
= EG - EM = [EG – EM]
= GM = GF Cos r = 2t Cos r
= 2t Cos r ----(4)
EN = EM
Since all reflections occurs from the rarer medium, no extra path difference is introduced.
Case (i) If ‘t’ is very small compared to , 2t Cos r is negligible and the two waves interfere
constructively so that the film appears bright in transmitted light.
Case (ii) If t is not negligible compared to , [2t Cos r] increases with the increase in the thickness of
the film.
(a) When 2t Cos r = (2n+1)/2, the film appears dark
(b) When 2t Cos r = n, the film appears bright.
As the source is extended, the rays of light are incident at various angles and hence the path
difference has variable values for different set of transmitted rays. Hence the field of view in the
transmitted beam consists of alternate bright and dark band. Also, the fringe pattern is not very well
defined as the intensity of minima is not equal to zero.
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M. Bhojaraj, Veerashiava College, Bellary
FRINGES PRODUCED BY WEDGE SHAPED FILM:
Consider two optically flat similar glass plates in contact at one end using rubber band and
inclined to one another at an angle using a blade or wire kept in between the glass plates at the other
end. If a liquid drop of refractive index is kept in between the plates, then it encloses liquid film of
variable thickness with zero thickness at one end and a thickness equal to the thickness of the object kept
at the other end as shown in the following figure (1).
Fig. (1) Interference using wedge shaped film
Let the film is illuminated by monochromatic
light from a slit arranged parallel to the edge of the
wedge. Interference occurs between the rays
reflected at the upper and the lower surfaces of the
film respectively and parallel interference bands
result. These bands are alternately dark and bright.
Let‘t’ is the thickness of the film at a distance ‘x’
from the edge.
The geometric path difference between the two
rays at the this position is given by
= 2 t Cos r = 2 t
--[because r 0 & Cos r =1]
Since the angle of the wedge ‘’ is very small, we can write, t = x
Therefore, the path difference = 2 x.
Condition for bright fringe:
At the position ‘x’ nth bright fringe if formed when
the following condition is satisfied.
2 x = [n + ½ ]
The distance of the nth bright maximum is,
Condition for dark fringe:
At the position ‘x’ nth dark fringe if formed when
the following condition is satisfied.
2 x = n
The distance of the nth bright maximum is,
Fringe width: If x1 is the distance of the nth dark band and If x2 is the distance of the [n+m]th dark band, then
Therefore, the width of m fringes is given by,
If ‘d’ is the thickness of the object used to form the wedge shaped film and ‘l’ is the length of the air
wedge, then = [d/l] or, [1/] = [l/d]
Thus by finding experimentally and by knowing , & l, the thickness of the object to form the
wedge shaped film can be calculated.
By finding‘d’ by using screw gauge and by knowing , & l, the refractive index of the liquid used
as wedge shaped film can be calculated.
Also, by finding‘d’ and knowing and l, the wavelength of monochromatic source can be
calculated.
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M. Bhojaraj, Veerashiava College, Bellary
Determination of thickness of a blade/diameter of a wire using interference of light
formed in air wedge:
Fringe no.
n
Microscope
reading
R in cm
Fringe no.
n
Microscope
reading
R in cm
Width of
30 fringes
β1
Mean width
of 30 fringes
0 30 Mean [β1]
= _____ cm 5 35
10 40
15 45
20 50
25 55
The fringe width is calculated using the relation,
30
1
Mean
The length of the air wedge is measured by taking the reading for the edge of the
glass plate [R0] and the reading for the object [R] and taking the difference l = [R – R0].
Knowing β, l and the wavelength of the source of light, the thickness of the object is
calculated using the relation,
2
ld
NEWTON’S RINGS Newton has observed circular interference fringes in the light reflected from the air film enclosed
between the slightly curved surface of a convex lens and a plane glass plate in contact with it and made a
careful study of the diameter of the rings thus formed and hence they are known as NEWTON’S RINGS. An arrangement to study
Newton’s rings is as shown in the
following figure (1). Light from a
monochromatic source S rendered
parallel by a convex lens L1 and
reflected by a glass plate G held at 450
to the rays. These reflected rays fall on
a thin air film enclosed by a long focus
plano-convex lens L2 and a plane glass
plate. The light reflected from the upper
and lower surface of the thin air film
superimpose on one another to produce
interference rings as shown in figure (2)
Figure (1)
Theory of Newton’s rings:
The experimental arrangement to get interference fringes
using air wedge formed by a wire is shown in figure (2). The
position of crosswire in microscope is adjusted to one of the
dark fringe and treating this fringe number as zero, the
microscope reading is noted. The position of the cross wire is
shifted to the 5th, 10th. . . . . . , 55th fringe and the readings are
tabulated.
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M. Bhojaraj, Veerashiava College, Bellary
The interference is produced due to the rays reflected from the upper and lower surface of the film. If
“t” is the thickness of the air film at AA1 or BB1, the path difference between the two rays, one reflected
from A and the other reflected from A1 will be
x =2t cos --------- (1)
Where “” is the angle of refraction. For air, = 1 & since is nearly zero, Cos = 1.
Hence x =2t.
Since the ray reflected from the upper surface of the air film is reflected from air [a rarer medium]
and the other ray reflected from the lower surface of the air film [a denser medium], the second ray
experiences a further path change of [/2]. Due to this, the effective path difference between the
interfering rays will be x =2t + /2.
Therefore, the points A and B lie on a bright ring of diameter AB = d, if the path difference
satisfies the following condition.
2μt + /2. = n
OR, 2t = [2n – 1] /2 ----------- (2)
Let us consider the vertical section ACBE of the
plano-convex lens through the centre of curvature
of the lens surface as shown in figure (3). Let R be
the radius of curvature of the convex surface of the
lens and C be the point of contact of the lens with
the plane glass plate such that the points A and B
are equidistant from C.
Draw AA1 and BB1 perpendicular to the glass
plate P.
Let AA1 = BB1 = t be the thickness of the film at
distance r = CA1 = CB1 from the point of contact C
of the lens with the plate.
Figure (3)
From the geometry of the circle, we have
DE x CD = AD x DB
OR (CE – CD) CD = AD2, [because AD = DB]
OR (2R – t) t = r2 ----------- (3)
Where, r is the radius of the ring.
Since the radius of curvature R of the lens is very large compared to t, we can neglect “t”
compared to 2R.
2Rt = r2 = [AB/2]2 = [d/2]2 = d2/4
2t = d2/4R ---------- (4)
Where d is the diameter of the ring.
Comparing equations (2) & (4), the diameter of the nth bright ring is given by the following equation.
2
124
2
nR
dn or
Rndn
2122 or
Rnd n
212 - - - (5)
Where n is an integer taking values 1,2,3,…..for first, second, third …..rings respectively.
Since, R and are constants dn [2n – 1]1/2. Thus, the diameter of the bright rings is proportional
to the square root of odd natural numbers.
Also for the nth dark ring,
μ nR
dn 4
2
or
nRd n
42 or
nRd n
4
ndn , where, n = 1, 2, 3 etc.
Thus the diameter of the dark rings is proportional to the square root of the natural numbers.
Experimental determination of refractive index of liquids:
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M. Bhojaraj, Veerashiava College, Bellary
Figure (4) Graph 1 Graph 2
A few drops of liquid whose RI is to be found is placed between lens and glass plate to form a thin
film of liquid and arranged as in figure (4) to get Newton’s rings. The vertical cross wire of travelling
microscope is arranged to the position of 18th dark ring, on left side, so that it is tangent to the ring. The
reading of the microscope is noted. The readings are also noted successively for 16th, 14th - - - 8th dark
ring on left side and on the right side, the readings for 8th, 10th, - - - -18th dark ring are noted.
The diameter [D] of the ring is calculated for each ring by finding difference of reading for each ring.
The square of the diameter [d2] is calculated and tabulated as in the table. A graph of d2 versus ring
number is plotted. The graph is a straight line. The slope 1 of the graph is calculated. Ring
number
n
Microscope reading
[x 10-2 m]
Diameter [Dn,l] of the ring
[x 10-2 m]
2
,lnD [x 10-4 m2] Left edge Right edge
18
16
14
12
10
8
The experiment is repeated in a similar way with air film formed in between the lens and the glass
plate. The slope 2 of the second graph is calculated. Ring
number
n
Microscope reading
[x 10-2 m]
Diameter [Dn,a] of the ring
[x 10-2 m]
2
,anD [x 10-4 m2] Left edge Right edge
18
16
14
12
10
8
Since,
RmnSlope
)(41
and
1
)(42
RmnSlope
as μ = 1 for air. In these equations, (n-m) is
difference in ring number. Therefore, the 1
2
Slope
Slopegives the refractive index of the liquid. Thus, the
refractive index of the liquid is calculated using the relation, 1
2
Slope
Slope
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M. Bhojaraj, Veerashiava College, Bellary
MICHELSON’S INTERFEROMETER
Michelson’s interferometer consists of two highly polished mirrors M1 and M2 and two plane
glass plates A and C arranged as shown in the following figure (1). The rear side of the mirror A is
half silvered so that light from the source “S” is equally reflected and transmitted by it.
Michelson’s Interferometer
Light from a monochromatic source rendered in to a
parallel beam by passing it through a lens L is made to fall on
the plate “A” inclined at an angle of 450 to the incident light.
The light is partially reflected along AM1 and partly
transmitted along AM2. These two beams moving mutually
perpendicular to each other are reflected back by the mirrors
M1 and M2 and retrace their path and reach the plate A. The
ray reflected from M2 is again reflected from A and the ray
from M1 is transmitted through A and these two rays travel
along the same direction and are received by the eye.
The ray reflected from M1 passes through A thrice, while
the reflected ray from M2 passes only once through A. To
compensate this optical path, a glass plate C of same material
and same thickness is arranged parallel to A in the path of the
second beam. The mirror M1 is fixed on a carriage and can be
moved with the help of a handle H. The distance through
which the mirror M1 is moved can be measured with the help
of micrometer screw attached to the Handle.
The plane of the two mirrors can be made perfectly perpendicular to one another with the help of
the screws attached to them. The plate C is necessary for white light fringes and can be dispensed
with while using monochromatic light.
If the mirrors M1 and M2 are perfectly perpendicular, the observer’s eye will see the image of M1
and M2 through A. There will be an air film between the two images and the distance between them
can be varied with the help of the handle H. In this case, the fringes will be perfectly circular. If the
path traveled by the two rays is same, then the field of view will be completely dark. If the two
images of M1 and M2 are inclined, the air film between them has wedge shape and straight fringes are
observed. When the handle H is rotated, the fringes cross the centre of the field of view of the
observer’s eye. If M1is moved through /2, one fringe will cross the field of view.
TYPES OF FRINGES:
Fig (2) Formation of Circular fringes
Circular fringes:
When the mirror M1 and the virtual image |
2M of the
mirror M2 are exactly parallel, in general circular fringes are
formed. In this case the source is extended one and S1 and
S2 are the virtual images of the source due to the mirrors M1
and M|2.
If the distance between |
21MM = d, then the distance
between S1 and S2 = 2d. The path difference between the
beam will be 2d Cos .
When 2d Cos = n, the beam will reinforce to produce maximum. These fringes are circular
and are formed due to phase difference determined by the inclination “”, they are known as the
fringes of equal inclination or Haidinger’s fringes. When M1 and |
2M coincide, the path difference
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M. Bhojaraj, Veerashiava College, Bellary
become zero and the field of view is perfectly dark [due to path change occurs for one of the beam
due to reflection at denser surface].
Localized fringes:
When the mirror M1 and the virtual image |
2M of
M2 are inclined, the air film enclosed is wedge shaped.
The fringes formed depends on the nature of
inclination of the two mirror images. If the two mirrors
intersect at the centre, the fringes are straight and
parallel as shown in figure [3(ii)] If the mirrors are
inclined to one another and does not intersect, then the
fringes are curved with the convex towards the thin
edge of the wedge.
White light fringes:
With white light, the fringes are observed only when the path difference is small. The
different fringes overlap on one another and only a few coloured fringes are visible. The central fringe is
dark and a few fringes near the central dark fringe are coloured. After about 8 to 10 fringes a number of
colours overlap at a point and as a result general illumination is observed. White light fringes are useful
for the determination of zero path difference between the interfering rays. It is useful in the
standardization of a metre scale.
Applications of Michelson’s Interferometer:
(1)Determination of wavelength of monochromatic light: When the two mirrors are equidistant from A, the field of view is perfectly dark. The mirror M2 is
fixed at this position and M1 is moved to get circular fringe pattern [or straight and parallel fringes].
The reading of the micrometer eyepiece is noted [R1]. The handle H is rotated until 50 fringes appear
or disappear at the centre [50 fringes cross the reference cross wire in the case of parallel fringes].
The reading of the micrometer eyepiece is noted again [R2]. If d = R2 – R1 is the displacement of the
mirror position for the crossing of 50 fringes, then, d =
250
.
Thus knowing d, the wavelength of the source is calculated by using the relation,
50
2d
(2)To determine the wavelength difference between two monochromatic, closely lying spectral lines :
[Resolution of spectral lines]
In the case of Sodium source, we get two spectral lines D1 and D2having wavelengths 1 and
2 such that the difference in wavelength is very small. When observed in a Michelson’s
interferometer, each wavelength forms its own fringes, which overlap on one another. By
adjusting the position of mirror m1, a single fringe pattern having maximum brightness can be
obtained. In this position, the bright fringe due to D1 coincides with the bright fringe due to D2
line. A reading [R1] of micrometer screw is noted. When the handle is turned to move the
mirror M1, the fringes go out of step and on further moving the mirror, the fringes with
maximum brightness is again formed. The reading [R2] for this position is noted. This is
possible only when (n + 1) th order of shorter wavelength coincides with the nth order of longer
wavelength.
If “d” is the distance between two positions of M1 for maximum brightness, then
2d = n1 1 = (n1+1) 2 21
21
n - - - - -(1)
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M. Bhojaraj, Veerashiava College, Bellary
Substituting for n1 in the above equation, we get, 21
212
d - - - - - - (2)
dd 22
2
2121
- - - - (3)
Where, λ is the mean wavelength of Sodium light. Thus, by knowing the distance [d] moved by
mirror for positions of distinct fringe pattern, wavelength difference can be calculated.
FABRI PEROT INTERFEROMETER
It consists of two glass plates E1 and E2 separated by a distance “d”. The inner surfaces of the two
plates are parallel and thinly silvered to reflect 70% of the incident light. The outer surfaces of the plates
are also parallel to each other, but, inclined to their respective inner faces. This is to avoid interference
effects due to multiple reflections and refractions that all faces are not made parallel.
Light from a broad monochromatic source S1 is made parallel using a collimating lens L1. Each
parallel beam suffers multiple reflections in the air film between the silvered surfaces of the plates E1 and
E2 and emerges from E2 as a parallel beam, so that for interference the rays must be brought to focus at P
by a convex lens L2. The fringes formed are seen in the focal plane of L2. The fringes are circular and due
to path difference of equal inclination known as Haidinger’s fringes.
If “” is the angle of incidence on the silvered surface of E1 and “d” is the distance between E1
and E2, then the path difference between successive rays = 2d Cos . The condition for the rays to
produce maximum is given by,
2d Cos = n - - - - - -(1)
In equation n is an integer giving the order of interference for the particular bright fringe.
All the points on a circle passing through the point “P” with the centre at O2 on the axis O1O2
satisfy this condition and hence the maxima consists of a series of concentric rings on the screen with O2
as centre.
In the interferometer one the plate is fixed, while, the other is capable of motion with the help of
slow motion screw, so that the thickness of the air film can be changed. When the thickness changes by
“/2”, one fringe disappear at the centre. Hence by knowing the number of fringes disappearing at the
centre when the mirror moves through a distance “d”, the wavelength of the monochromatic source can
be calculated using the equation (1).
In this case the radius of a ring is a function of “” and hence two sets of fringe pattern is obtained
if there are two different wavelengths in a source.
Uses:
(1) It can be used to determine the wavelength of a monochromatic source of light.
(2) It can be used to determine the wavelength difference between two monochromatic sources of
light.
(3) It can be used for the calibration of a metre scale.
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M. Bhojaraj, Veerashiava College, Bellary
Advantages over Michelson’s interferometer::
(1) Fringes obtained are very sharp as compared to the fringes formed in Michelson’s interferometer.
(2) The resolving power of the microscope is very large.
(3) Clear and distinct fringes are formed even when the separation between the plates is large.
INTERFERENCE FILTERS
Interference filters are the optical devices, which work on the principle of interference of
transmitted light in thin films, and transmit a narrow band of about 100 0
A around a chosen wavelength.
In simplest form the arrangement is very much similar to a Fabri-Perot etalon as shown in the following
figure.
In this arrangement, two thin transparent layers ‘a’ and ‘b’ of good
reflecting material like silver separated by an optical thickness ‘d’ equal
to one or more half wavelengths of a dielectric material like magnesium
fluoride. To construct such a filter one layer of silver is deposited on a
solid transparent substance like glass. Then a coating of the dielectric
(acts as a spacer layer) is deposited on the silver and finally the second
layer of silver is added. The thin ‘sandwich’ of dielectric metal
constitutes the filter. A second glass plate is added to the silver coating
to give mechanical protection.
When the path difference between the successive transmitted pairs of
rays is equal to 2d Cos or simply 2d [as is nearly equal to zero,
Cos = 1]. Therefore, the emergent rays are all in phase when 2d is
equal to one wavelength 0 or an integral multiple of 0.
Therefore, the condition for maximum is, 2d = m 0
With white light the intensity of the transmitted beam will be maximum for only those
wavelengths, which satisfy the above equation. Therefore, a spectrum of white light consists of a series of
sharp bright bands separated by wide dark regions. Bright bands are obtained for those wavelengths
satisfying the above equation and a dark region for all those, which do not satisfy the above equation.
The maxima will occur at wavelengths given by the equation, ,2
0m
d where m is any integer
value.
If the thickness of dielectric is such that 2d = 0 and 0 = 5000 0
A , then only a single narrow band in this
region is transmitted, because the other peaks in the spectrum would occur at 25000
A , 1670 0
A and these
are in Ultra Violet region and hence the device acts as optical filter.
In a similar way, if the filter is constructed to transmit a wavelength 0 = 2500
0
A , then it also
transmits 5000 0
A and 6250 0
A also and hence it allows a band of wavelength around these wavelengths.
Thus, the filtering action mainly depends on the thickness of dielectric film between the silvered surfaces.