interference of light - veerashaivacollege.org€¦ · interference of light when two or more wave...

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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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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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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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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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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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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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(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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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

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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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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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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.

11


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.

12


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.

13


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:

14


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

15


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

16


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)

17


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.

18


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.

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