1

OPTICS

Interference of Light:

Many theories were put forward to explain the nature of light:

Newton’s corpuscular theory

Huygens’ wave theory

Electromagnetic theory

Quantum theory

The above-mentioned theories explain that some of the

properties like refraction, reflection, interference etc. can be

explained by considering the light as a wave while the

phenomenon like photoelectric effect, Compton Effect etc. may

be described by considering the light as particle.

Interference:

It is the phenomenon of superposition of two coherent*

waves in the region of superposition.

At some points in the medium, the intensity of light is

maximum while at some other points the intensity is

minimum.

The positions of maximum intensity are called maxima,

while those of minimum intensity are called minima.

Principle of Superposition:

In a medium, the resultant displacement of a particle acted upon

by two or more waves simultaneously is equal to the algebraic

sum of the displacements of the particle due to individual

waves.

Thus if the displacement due to a single wave train at a point be

y1 and y2 be the displacement due to the another wave train in

the same direction, then the resultant displacement y may be

written as,

2

21 yyy

If, tay sin11 and tay sin22

Where δ is the phase difference between two waves.

tatay sinsin 21

tataa cossinsincos 221

Let coscos21 Aaa and sinsin2 aa ; we have,

tAtAy cossinsincos

tAsin

Where, 2/1

21

2

2

2

1 cos2 aaaaA

and

cos

sintan

21

2

aa

a

The intensity at any point is proportional to the square of the

amplitude, i.e. 2AI , then

cos2 21

2

1

2 aacaAI …………(1)

Since 2

1a and 2

2a are the intensities of the two interfering waves,

the resultant intensity at any point is not just the sum of the

intensities due to the separate waves 2

2

2

1 aa .

Condition for maximum and minimum intensities:

From (1) it is clear that I will be maximum at points where

.........2,1,0;21cos nnor

Therefore, the intensity will be maximum when phase

difference is even multiple of , but we know that,

Path difference

2phase difference

3

n2

2 or 2

2

n

Therefore, 2

21max aaI

Similarly I will be minimum at points where cos δ=-1and

therefore,

12 n

2

12

n n=0,1,2……….

And, 221min aaI

*Coherent Sources: The light waves of practically the same

intensity, of exactly the same wavelength and of exactly a

constant initial phase difference are known as Coherent

Sources. For Interference of light to take place the two light

waves must have the same plane of polarization. Two light

waves from independent sources can never be coherent. Thus a

pair of coherent waves may be obtained from a single source of

light.

Young’s Double Slit experiment: In 1801, Young first

demonstrated experimentally the phenomenon of interference.

4

In his experiment light from a monochromatic source is

allowed to fall on a slit S and then through a double slit S1

and S2.

The interference pattern was observed on a screen XY

where few dark and bright bands or fringes are observed.

These fringes are of equal distances.

The bold lines show the amplitude in positive direction

(crest) where the dotted lines show the amplitude in the

negative direction (troughs).

At points where crests fall on crests or troughs fall on

troughs the amplitudes add up and the resultant intensity

increases (I α A2) and thus we obtain constructive

interference.

At points where crests fall on the troughs or the troughs

fall on the crests, the amplitudes are reduced (the resultant

amplitude will be zero if the two lights from S1 and S2 have

equal amplitudes and so the intensity is also zero). This is

known as destructive interference.

S1

S2

S

I

X

Y

5

Thus on the screen alternate bright and dark regions of

equal width, called interference fringes are observed.

Analytical treatment for Young’s double Experiment:

Let S1and S2 be two coherent sources separated by a

distance d and made from a monochromatic source S.

Let a screen be placed at a distance D from the coherent

sources.

O is a point on the screen which is equidistant from S1and

S2.

Therefore, path difference at O will be zero and the

intensity at O will be maximum.

Let P be a point on the screen such that,

OP= y

Therefore,

2,

2

dyPR

dyPQ

S1

S2

d

P

R

C O

d/2

d/2

D

Q y

6

D

dy

Dd

yDPS2

2,

2

2

2/12

2

1

and D

dy

DPS2

2

2

2

The path difference of two light rays emerging from S1and S2

and reaching to a point P on the screen is,

D

ydPSPS 12

Therefore the phase difference,

2path difference=

D

yd

2

Position of bright fringes: As mentioned earlier that for

bright fringes path difference is even multiple of λ/2, i.e.

nnD

yd

22

Thus the distance of nth bright fringe from point O is given by:

d

Dnyn

.

Position of dark fringes: For dark fringe or the minimum

intensity at P, the path difference must be an odd multiple

of half wavelength, i.e.

2

12

nD

yd

7

Thus the distance of the nth dark fringe from point O is given

by,

d

Dnyn

2

12

.

Fringe Width: The distance between any two consecutive

dark fringes or bright fringes is same and known as fringe

width. It is given by:

d

Dyy nn

1

Conclusions: 1. There is a bright fringe at the centre of the screen.

2. Alternatively dark and bright fringes on both the sides

of the sides of the central maximum occur.

3. Fringe widths of dark and bright fringes are same.

4. Fringe width depends upon the wavelength of light,

separation of sources and source to screen separation.

Condition for Sustained Interference: To obtain a well

defined and observable interference the condition for

interference of light are given as under:

The beams from two sources must propagate along the same

line; otherwise the vibrations cannot be along a common line.

The two sources must be very narrow because a broad source

is equivalent to a no. of sources and so positions of darkness

and brightness cannot be well defined.

The two sources must have equal intensities.

The two sources should have equal amplitudes and

frequencies. In case of unequal frequencies the phase

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difference between the interfering waves will vary

continuously and as a result, intensity at any point will not be

constant but vary with time. If the amplitudes are different

then a clear cut distinction between maximum (a1+ a2) and

minimum (a1- a2) is not possible.

The common original source must be monochromatic, i.e.,

emitting light of single wavelength.

The light from the two sources must have either zero phases

or a constant difference of phase. This condition is very well

observed in the case of coherent sources.

The separation between source and screen should be large to

get wide fringes.

If the interference pattern has to be obtained by polarised

light then the polarised waves must be in the same state of

polarization.

Two Classes of Interference The interference is divided into

two classes on the basis of the way of obtaining the two

coherent sources:

1. Division of wavefront:

In this case the wavefront originating from one source is

divided into two parts. This division can be made with the

help of any optical system like Fresnel’s biprism, Fresnel’s

S1

S

S2

9

mirror etc. The two wavefronts after traveling certain

distances are brought together to give interference pattern.

2. Division of amplitude:

In this process the amplitude is divided into two or more parts

either by partial reflection or refraction and which after

traveling different paths produce interference at a point.

Newton’s rings, Michelson interferometer are the examples of

this class.

Fresnel’s Biprism: However Young demonstrated the

phenomenon of interference, but it was doubted that the fringes

are not due to interference of light waves but due to some

modification of light at the slits. These doubts are removed by

the Fresnel’s biprism experiment. By using biprism, Fresnel

obtained two coherent sources from a single source by

refraction.

Construction: It consists of two acute angle prisms with their

bases in contact forming a single obtuse angle prism ABC. In

actual practice the prism is grounded from a single optically

S Reflected

System

Transmitted

system

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true glass plate. The obtuse angle of the prism is nearly 1790, so

that each of acute angles is nearly 30’.

Working:

A narrow vertical slit S is illuminated by a source of

monochromatic light.

The prism is placed with its refracting edge parallel to the

line joining the sources.

The light from the slit S is allowed to fall symmetrically on

the biprism ABC with its refracting edge vertical (i.e.,

parallel to the line source S).

When the light from S falls on the upper half of the prism,

after refraction it appears to come from virtual source S1.

Similarly after refraction from the lower half of the prism

it appears to be coming from virtual source S2.

The distance between the biprism and the source S is so

adjusted that the sources S1 and S2 are very close to each

other.

The interference fringes are observed in the region EF of

the screen.

S

S2

S1

d

a b

D

B

O’

A

C

G

E

H

O

F

Screen

11

The analytical treatment is same as for Young’s Double

Slit experiment.

If d be the distance between the sources S1 and S2 and D

between the source and the screen, then the fringe width is

given by,

d

D

And thus

ba

d

D

d

Moreover the distance of n th bright fringe from O.

,)(d

Dny brightn

Similarly

d

Dny darkn

2

12)(

.

Where )1(2 ad

Applications: a) Determination of wavelength of light:

D

d

b) Determination of thickness of a thin sheet of transparent

material:

a) Biprism can be used to determine the thickness of a

given thin sheet of transparent material e.g., glass,

mica.

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b) Suppose A and B are two virtual coherent sources.

c) The point C is equidistant from A and B.

d) When a transparent plate G of thickness t and

refractive index µ is introduced in the path of one of

the beams, the fringe which was originally at C shifts

to P.

e) The time taken by the wave from B to P in air is the

same as the time taken by the wave from A to P partly

through air and partly through the plate.

f) Suppose c0 is the velocity of light in air and c its

velocity in medium then,

c

t

c

tAP

c

BP

00

t

c

ctAPBP 0

, But

c

c0

tttAPBP 1

If P is the point originally occupied by the n th fringe, then the

path difference

nAPBP

nt 1 ……..(i)

P t

d

x

D

C

A

B

G

13

Also the distance x through which the fringe shifted

d

Dn

Where

d

D, the fringe width.

d

Dnx

Also, D

xdn

Or,

D

xdt 1

g) Therefore, knowing x, the distance through which the

central fringe is shifted, D, d and µ, the thickness of

the transparent plate can be calculated.

h) If a monochromatic source is used, the fringes will be

similar and it is difficult to locate the position where

the central fringe shifts after the introduction of the

transparent plate. Hence white light is used. The

fringes will be coloured but the central fringe is white.

Interference in Thin films: It has been observed that

interference in the case of thin films takes place due to (1)

reflected light and (2) transmitted light. Newton was able to

show the interference rings when a convex lens was placed on a

plane glass-plate.

Interference Due To Reflected Light(Thin films):

Consider a transparent film of thickness t and refractive

index µ.

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A ray SA incident on the upper surface of the film is partly

reflected along AT and partly refracted along AB.

At B part of it is reflected along BC and finally emerges

out along CQ.

The difference in path between the two rays AT and CQ

can be calculated.

CN is normal to AT and AM is normal to BC.

The angle of incidence is i and the angle of refraction is r.

CB and AE are extended to meet at P.

rAPC The optical path difference

ANBCABx

Here, CM

AN

r

i

sin

sin

CMAN .

CMBCABx .

CMPCCMBCABx .

PM.

In the ΔAPM,AP

PMr cos

or rEPAErAPPM coscos.

rt cos2 tEPAE

rtPMx cos.2. ……………(1)

This eq. (1), in the case of reflected light does not represent the

correct path difference but only the apparent. It has been

established on the basis of electromagnetic theory that, when

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light is reflected from the surface of an optically denser

medium (air-medium interface) a phase change equivalent to

a path difference 2

occurs.

Therefore, the correct path difference in this case,

2cos2

rtx ………..(2)

1) If the path difference x = n where n = 0,1,2,3….etc.,

constructive interference takes place and the film appears

bright.

nrt 2

cos2 or, 2

12cos2

nrt …..(3)

2) If the path difference is 2

12

nx where n = 0,1, 2,

3….etc., destructive interference takes place and the film

appears dark. 2

122

cos2

nrt

or, 1cos2 nrt ………….(4)

Here n is an integer only, therefore (n+1) can also be taken

as n.

nrt cos2 ………….(5)

Where n = 0,1, 2, 3……………etc.

i

i

r

S T Q

C A

E

P F

M

N

B

r

AIR

AIR

µ t

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Interference Due To Transmitted light (Thin Film):

Let there be a thin transparent film of thickness r and

refractive index μ.

A ray SA after refraction goes along AB.

At B it is partly reflected along BC and partly refracted

along BR.

The ray BC after reflection at C, finally emerges along DQ.

Here at B and C reflection takes place at the rarer medium

(medium-air interface).

Therefore, no phase change occurs.

BM is normal to CD and DN is normal to BR.

The optical path difference between DQ and BR is given

by

BNCDBCx

also MD

BN

r

i

sin

sin or MDBN .

from fig. rBPC and CP = BC = CD

t

B

R Q

i N

D M

C

P

A

AIR

AIR

μ r

i

r

r r r

i

S

17

PDCDBC

PMMDPDMDPDx

In the BP

PMrBPM cos, or rBPPM cos.

But, tBP 2

rtPM cos2 rtPMx cos2.

1) For bright fringes, the pat difference nx

nrt cos2

where n = 0, 1, 2, 3...etc.

2) For dark fringes, the path difference 2

12

nx

2

12cos2

nrt

where n = 0, 1, 2, 3...etc.

In the case of transmitted light, the interference fringes

obtained are less distinct because the difference in amplitude

between BR and DQ is very large.

Important points: When the film is exposed to white light

then only those wavelengths will be present for which the

condition of maxima is satisfied or which have the path

difference satisfying the condition for bright fringe.

Newton’s Rings: When a Plano convex lens of long focal

length is placed on a plane glass plate, a thin film of air is

enclosed between the lower surface of the lens and the upper

surface of the plate. The thickness of the air film is very small

at the point of contact and gradually increases from the centre

outwards. The fringes produced with the monochromatic light

are circular. The fringes are concentric circles, uniform in

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thickness and with the point of contact as the centre. These

fringes are known as Newton’s Rings.

Construction:

S is a source of monochromatic light at the focus of the

lens L1.

A horizontal beam of light falls on the glass plate B at 450.

The glass plate B reflects a part of the incident light

towards the air film enclosed by the lens L and the plane

glass plate G.

The reflected beam from the air film is viewed with a

microscope.

Working:

Interference takes place and dark and bright circular

fringes are produced.

L

G

AIR FILM

M

S

L1

B 45

0

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This is due to the interference between the light reflected

from the lower surface of the lens and the upper surface of

the glass plate G.

Theory: 1. Newton’s ring by reflected light: Let the radius of curvature

of the lens is R and the air film is of thickness t at a distance

OQ= r, from the point of contact O.

Here interference is due to reflected light. Therefore for the

bright rings

2

12cos2

nrt ….(i) where n = 1,2,3……….etc.

For normal incidence r = 00 therefore, cos r = 1

For air, μ = 1

Hence 2

122

nt …..(ii)

For the dark rings, nrt cos2

Or, nt 2 ……(iii) where n = 0,1,2,3…….etc.

From fig. in ΔCEP 222 EPCECP

Or 222 EPOECOCP

L

G

REFLECTED

LIGHT

AIR FILM

C

R

O Q

H E P

rn

t

20

or 222

nrtRR

or Rtrn 22 (ignoring t2)

hence R

rt n

2

2

substituting the value of t in (ii) and (iii) we get,

for bright rings

2

122 Rnrn

……..(iv)

or

2

12 Rnrn

………..(v)

for dark rings Rnrn 2……..(vi)

or Rnrn ………..(vii)

Result:

The radius of nth dark ring is proportional to

i. n

ii.

iii. R

Similarly the radius of nth bright ring is proportional to

i. 2

)12( n

ii.

iii. R

If Dn is the diameter of the nth dark ring,

RnrD nn 42 ……..(viii)

If Dn is the diameter of the nth bright ring,

21

2

1222

RnrD nn

…..(ix)

2. Newton’s ring by transmitted light:

In case of transmitted light the interference fringes are

produced such that for bright rings, nrt cos2 ………(x)

For air, μ = 1and cos r = 1 then nt 2 ……….(xi)

For dark rings,

2

122

nt ………(xii)

Hence in this case the radius of nth bright is

Rnrn …….(xiii)

And the radius of the nth dark ring is

2

12 Rnrn

…….(xiv)

Applications of Newton’s Ring:

L

G

AIR FILM

TRANSMITTED

LIGHT

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Determination of wavelength of light: For nth dark ring we

have

RnDn 42 ……..(i)

or nR

Dn 4

2

…….(ii)

again for (n+p)th dark ring,

pnR

D pn

4

2

……(iii)

pR

DD npn

4

22

……(iv)

Determination of Refractive Index of a liquid: The

refractive index of a liquid, forming film between lens and

glass plate, can be obtained from the following eq.

pR

DDliquidnpn

4

22

………..(i)

pR

DDairnpn

4

22

Since for air μ= 1

Hence

pR

DDairnpn

4

22

……..(ii)

Dividing (ii) by (i) we get

liquidnpn

airnpn

DD

DD

22

22

23

Newton’s Rings with white light: In case of white light the

diameter of the rings of different colours will be different and

there will be an overlapping of the rings of different colours

over each other. The only first few rings will be clear while

other rings cannot be viewed.

Michelson’s Interferometer: Principle: The amplitude of light beam from a source is

divided into two parts of equal intensities by partial reflection

and transmission. These beams are sent in two directions at

right angles and are brought after they suffer reflection from

plane mirrors to produce interference fringes.

Construction:

A

G1 G2

M1

M2

T

S

C

B Transmitted

Reflected

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It consists of two highly polished plane mirrors M1 and M2

two plane parallel glass plates G1 and G2 of exactly the same

thickness.

The plate G1 is semi silvered at the back so that the

incident beam is divided into reflected and transmitted

beams of equal intensities.

M1 and M2 are mutually at right angles to each other.

The plates G1 and G2 are parallel to each other and are at an

angle of 450 to M1 M2.

The mirrors M1 and M2 can be adjusted exactly perpendicular

to each other with the help of the screws on their backs.

The mirror M1 can be moved exactly parallel to itself with a

carriage on which it is moved.

Working:

The light falling on the glass plate G1 is partly reflected and

partly transmitted.

The reflected ray AC travels normally towards plane mirror

M1 reflected back to the same path and comes out along AT.

The transmitted ray after reflection from mirror M2 follows

the same path and then moves along AT after reflection from

the back surface of G1.

So the rays received at T are produced from a single source

by the division of amplitude and we get an interference

pattern.

The reflected ray AC travels through glass plate G1 twice and

in the absence of plate G2 the transmitted ray does not cross

plate G1.

To compensate it a second plate G2 is introduced in the path

of transmitted ray.

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Adjustment:

Distance of the mirrors M1 and M2 are adjusted to be nearly

equal from glass plate G1.

In order to obtain a parallel ray of light a lens is adjusted

between the source S and plate G1.

Formation of fringes:

1. Circular fringes:

If eye is placed to see mirror M1 directly then in addition it

will see virtual image of M2 formed by reflection in glass

plate G1.

Thus one of the interfering beams cones from M1 and other

appears to come from the virtual image of M2 i.e. M2’.

If M2 is exactly at right angles to M1 and the plate G1 is at

450 to each of then, the air film formed between M1 and

M2’ will have uniform thickness.

Let the two interfering beams appear to come to the eye

from two virtual images S1 and S2 of the original source

and let 2d be the distance between them.

Now since reflected ray from M2 suffers reflection at the

silvered surface of plate G1, an additional path difference

λ/2 is introduced.

Therefore, the total path difference 2/cos2 d .

Therefore, for brightness,

.....2,1,0,2/cos2 etcnnd

If M2’coinsides with M1 then total path difference will be

λ/2 which is the condition for destructive interference.

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If the light is obtained from an extended monochromatic

light source then for a given value of n, the value of α is

constant and so the locus of the fringes is circle.

2. Localized fringes:

If M1 and M2’ are not exactly parallel the path difference is

very small; we observe fringes as in the case of a wedge

shaped film.

In this case the locus of points of equal thickness is a

straight line parallel to the edge of the wedge.

M1

M2’

2d α

S

S1

S2

2d cos α

27

For small path difference, the fringes are nearly straight but

for larger path difference the fringes are generally curved

with the convex surface of the film towards the edge of the

wedge.

Important points: 1. As the circular fringes are formed due to two parallel

interfering waves, hence they are formed at infinity.

2. Here the circular fringes are formed at same inclination

hence they are called equal inclination fringes or Haidinger

fringes.

Applications:

1. Determination of wavelength of Monochromatic light:

Mirror M1 and M2 are adjusted so that circular fringes are

visible in the field of view.

If M1 and M2 are equidistant from the glass plate G1, the

field of view will be perfectly dark.

The mirror M2 is kept fixed and the mirror M1 is moved

with the help of the handle of the micrometer screw and

the number of fringes that cross the field of view is

counted.

M1

M2’ M1

M2’

M1

M2’

28

If mirror is moved through a distance of d for a

monochromatic light of wavelength λ and no. of fringes

that cross the centre of the field of view is n then 2

nd ,

because for one fringe shift the mirror moves through a

distance equal the half the wavelength. Hence λ can be

determined.

2. Determination of the difference in wavelength between

two neighbouring spectral lines:

Set the apparatus for circular fringes.

Let the source has two wavelengths λ1 and λ2 (λ1> λ2)

which are very close to each other.

The two wavelengths form their separate fringe patterns,

but because of very small change in wavelength, the two

patterns overlap.

As the mirror M1 is moved slowly the two patterns separate

out slowly and when the path difference is such that the

dark fringe of λ1 falls on the bright fringe of λ2 maximum

indistinctness occurs.

Again when the path difference is such that the bright

fringe of λ1 falls on bright fringe of λ2 maximum two

successive positions of maximum distinctness occurs.

Let the mirror M1 is moved through a distance of d

between distinctness.

This will happen only when nth

fringe of λ1 coincides with

n+1 th

fringe of λ2 then

21 )1(2 nnd

1

2

dn and

2

21

dn

29

Subtracting we get,

12

221

dd =

21

212

d

Or d2

2121

21 hence 2

21

hence d2

2

21

Thus by measuring the distance d moved by M1 the

difference between two wavelengths can be determined.

3. Determination of thickness of a thin plate:

The apparatus is set for parallel fringes.

White light source is used instead of monochromatic

source.

The cross wire is set on the central fringe.

Now the thin plate whose thickness is to be determined is

introduced in the path of one of the interfering beams.

Due to the insertion of the plate path difference is changed

to 2(μ-1)t.

Thus a shift in the fringe system occurs.

Now the mirror M1 is moved till the central fringe

coincides with the cross wire.

The distance x moved by mirror M1 is measured using

micrometer screw. Hence we have,

tx )1(22

30

1

xt

Using above equation t can be determined.

Diffraction: The bending of light beam round the corners or

edge and spreading of light into the geometrical shadow of an

opaque obstacle placed in its path is known as diffraction. In

other words the diffraction may be defined as the encroachment

of light into the geometrical shadow region of small opaque

obstacle or aperture placed in the path of light.

Types of diffraction: There are two classes of diffraction:

1. Fresnel Diffraction

2. Fraunhofer Diffraction

1. Fresnel Diffraction:

In this case diffraction is considered to take place from

different parts of the aperture.

Here the source or screen or both are at finite distance from

obstacle.

In this class the incident wave front is divergent, either

spherical or cylindrical.

S

A

B

A’

B’

31

Here the observed pattern is a projection of diffracting

element modified by diffracting effects and the geometry

of the source.

Here the diffraction centre of the pattern may be either

bright or dark.

2. Fraunhofer Diffraction:

Here the source of light and screen are at infinite distance

from the obstacle.

Here the inclination is important.

Here the incident wave front is generally plane and the

centre of diffraction pattern is always bright.

Fraunhofer Diffraction At A Single Slit:

The light from a monochromatic point source S is

converted into parallel beams of light by a convex lens

L1L1’.

Now this beam is incident normally on a slit AB of width e.

θ

θ θ

θ

θ

e

A

θ

B

θ

B’

θ

A’

θ

P

θ

L2

θ

L2’

θ

W1

θ

W1’ θ

W2

θ

W2’

θ

L1’

L1

W

W’

O

Y

Y’

K

32

The incident light wave fronts are shown by W1W1’ and

W2W2’.

Every point on the wave front incident on the slit and lying

within the width of the slit emits secondary waves which

superimpose to give diffraction pattern on the screen YY’.

In this diffraction pattern a central bright band is obtained.

On either side of this band alternate dark and bright bands

of decreasing intensity are observed.

Analysis: AK is perpendicular from A on BB’. The path

difference of the rays meeting at P is clearly BK. Here,

sinsin eABBK Therefore, the corresponding phase difference

sin

22eencepathdiffer

If the slit is assumed to be made up of n equal parts, the

amplitude of the secondary waves originating from these parts

will be equal. Let the amplitude of each secondary wave be ‘a’

and then the phase difference between any two successive

waves will be

)(,sin21

sayden

Thus the resultant in a direction θ, due to the superposition of

these n secondary waves can be calculated in the following

way:

Let there be n-harmonic waves having equal amplitudes ‘a’ and

common phase difference ‘δ’ between successive vibrations. To

find the resultant we construct a polygon, the closing side of

which gives the resultant amplitude R and a resultant phase Ф.

33

Now resolving the amplitudes parallel and perpendicular, we

get,

)1....(..........)1cos(.........2coscoscos naaaaR

)2....(..........)1sin(.........2sinsinsin naaaR

Multiplying eq. (1) by 2sin

2

,we

)1cos(

2sin2.........2cos

2sin2cos

2sin2

2sin2

2sincos2 naR

From the trigonometrical relation )sin()sin(sincos2 BABABA we have,

]2

3sin

2

1sin...

2

3sin

2

5sin

2sin

2

3sin

2sin2[

2sincos2

nn

aR

2

1cos

2sin2

2

1sin

2sin

nna

na

R

a

a

a

1δ

2δ

3δ

a

Ф

O

34

or

2sin

2

1cos

2sin

cos

nna

R …………(3)

similarly, multiplying (2) by 2sin2

and applying

)cos()cos(sinsin2 BABABA

and 2

sin2

sin2coscosCDDC

DC

we get,

2sin

2

1sin

2sin

sin

nna

R……..(4)

from (3) and (4) we get

2sin

2sin

na

Rand

2

1

n

Hence for the present case

n

e

ea

en

en

n

a

d

nda

Rsin

sin

sinsin

2

sin21

sin

2

sin21

sin

2sin

2sin

Let

sinethen,

35

n

a

n

aR

/

sin

/sin

sin

As n/ is small,

nn

sin

Again since ‘n’ is infinitely large and ‘a’ is negligibly small,

Let na =A, a finite no.

Hence,

sinAR ...........(5)

The secondary waves diffracted by the slit in a direction θ are

focused by the lens L2L2’ at a point P on the screen. Therefore,

the resultant intensity at P, 2RI . i.e.

2

22 sin

AI ........(6)

Conditions for maxima and minima:

0d

dI

or 0sin

2

22

A

d

d

or 0sincossin2

2

2

A .....(7)

Case I: When 0sin

or sin α =0 or n ( o , if α

= 0 then (sin α)/α = indeterminate)

Therefore, .......3,2,1,0 nn

or

n

e

sin

or ne sin ..........(8)

36

which gives the condition for nth minima.

Case II: The second bracketed term in eq.(7) will give the

condition for maxima.

Thus 0sincos

2

or tan ..............(9)

The intersection of the curves tany and y will give

the solution of eq.(9). The points of intersection will give the

values of α as

,.......2

7,

2

5,

2

3,0

Here α = 0 corresponds to central maximum.

Hence for central maximum the intensity is given by,

0

2

2

22 sinIA

AI

y=

tan

α

y=

tan

α

y=

tan

α

y = α

450

O

y

α

π/2 3π/2 5π/2

37

e e e e e e

38

Important Results:

1. The width of central maximum is directly proportional to

the wavelength of light used and as the wavelength of red

light is more than the wavelength of violet light hence the

width of the central is more for red light than for violet

light.

2. The width of the central maximum is inversely

proportional to the width of slit e hence width of central

maximum is greater for narrow slits.

Diffraction At Double Slit:

Let there be two slits AB and CD of equal width ‘e’,

separated by a distance‘d’.

The incident light gets diffracted from these two slits and

focused on the screen XY.

The diffraction by two parallel slits is a case of diffraction as

well as interference.

A

B

C

D

S1

S2

L L

P

O

P’

M

S θ

θ

θ

e+d

Y

Y’

39

Thus the pattern obtained on the screen consists of equally

spaced interference fringes due to both the slits and their

intensity being modulated by the diffraction phenomenon

occurring due to individual slits.

The two slits AB and CD may be considered equivalent to

two coherent sources S1 and S2.

Now due to individual slit, the resultant amplitude due to all

secondary waves diffracted in the direction θ is given by:

sinAR

..................(1)

Where,

sine

For simplicity, the resultant of all the secondary waves may

be taken as a single wave of amplitude R.

The resultant at point P on the screen will be the result of

interference between these two waves, of same

amplitudes

sinA, starting from S1 and S2.

The path difference between the wavelets from S1 and S2 in a

direction θ is, sin)2 deMS and therefore the phase

difference

sin

222 deMS .

The resultant amplitude at P may be given as,

cos2 21

2

2

2

1

2 RRRRRr

cos

sin2

sinsin222

AAA

40

2cos

sin4 2

2

22

A

Hence resultant intensity,

2cos

sin4 2

2

222

ARI rr

or

2

2

222 cos

sin4

ARr ...........(2)

where

sin

2de

From the above expression, it is clear that the resultant

intensity depends on the two factors.

(a). Diffraction term

2

22 sin

A :This factor is same as in the

case of single slit. Thus this term corresponds to

diffraction pattern due to secondary waves from the two

slits individually. This term gives a central maximum

having alternately minima and subsidiary maxima of

decreasing intensity on either side. The direction of

minima are given by,

0sin or m or

m

e

sin

........3,2,1,sin mme .............(3)

and the position of secondary maxima approach to

..........2/7,2/5,2/3

(b). Interference term 2cos :This term corresponds to

the two diffraction patterns coming out from the two slits.

This term gives a set of equidistant dark and bright

fringes. The directions of the maxima are given by

41

1cos 2 or n or

nde sin

or nde sin where n=0,1,2,3,.....(4)

Thus in a direction θ = 0, the central maxima due to

interference and diffraction coincide. The minima due to

interference term is given by,

0cos 2 or 2/12 n

or 2/12sin nde

Missing Orders: In the double slit arrangement, we find that

some of the interference maxima are missing. Since for the

same value of θ, the following two relations hold true,

nde sin , Interference Maxima.....(5)

me sin , Diffraction Minima....(6)

When both the above conditions are satisfied simultaneously,

then the interference maxima will be absent in the direction for

which θ is common. From eq. (5) and (6) we have,

m

n

e

de

Now the following conditions may be considered:

1. If d = e then n = 2m = 2,4,6,..... Hence the 2nd

,4th

,6th

etc.

order interference maxima will be absent.

2. If d = 2e, then n = 3m = 3,6,9.... Thus 3rd

, 6th

, 9th

order of the

interference maxima will be absent.

3. If e +d =e, i.e. d = 0, then the diffraction pattern will similar

to as observed to a single slit of width equal to 2e.

42

Effect of Increasing the slit width e: On increasing the slit

width e, the central peak will become sharper, but the fringe

spacing remains unchanged. Hence less interference maxima

fall within the central diffraction maximum.

Effect of increasing the distance between slits d: On

increasing the separation between the slit d keeping the slit

width constant, the fringes become closer together but the

I

θ

4A2sin

2αcos

2β/α

Cos2β

β π 2π -π -2π 0

43

envelope of the pattern remains unchanged. Hence more

interference maxima fall within the central envelope.

Effect of increasing the wavelength : When the wavelength

of monochromatic light falls on the slit increases, the envelope

becomes broader. As a result, the fringes move farther apart.

Diffraction Due to N-Slits (Plane Transmission Grating): A

diffraction grating consists of a large no. of parallel slits of

equal width and equal separation. It generally consists of 10000

to 15000 lines per inch.

Theory: The diffraction grating is equivalent to N-slits

arrangements and the diffraction pattern we obtain will be the

combined effect of all such slits. Let ‘e’ be the width of each

slit and‘d’ that of opaque part between them, then (e + d) is

known as grating element.

θ P

(e+d)

S1

S2

SN

M1

M2

MN-1

L

X

Y

44

The diffracted rays from each of the slits are allowed to fall on

a convex lens which focuses all of them at a point P on the

screen. As in the single slit, the waves diffracted from each slit

are equivalent to a single wave of amplitude,

sinAR

…….(1)

where

sine ………(2)

The path difference between any two consecutive waves from

two slits sin)( de . Therefore, the corresponding phase

difference will be

sin)(

2de

. Since the phase difference

is constant between any two consecutive waves it can be taken

as

2sin)(

2 de

………(3)

Thus we have to find out the resultant of N vibrations in a

direction θ and each vibration is of amplitude

sinA

.

Now similar to the resultant of n-harmonic waves, the resultant

of N-slits may be given as,

sin

sin.

sin

sin

sin

2

2sin

2

2sin

'NANR

NR

R

Thus the resultant intensity at P may be given as,

45

2

2

2

222

sin

sin.

sin'

NARI

………..(4)

Here the factor 2

22 sin

Ais the intensity factor due to a single

slit while

2

2

sin

sin N is due to the interference from all the N-slits.

Principle Maxima: for maximum intensity, we have,

0sin or n , .....2,1,0n

But in this condition 0

0

sin

sin

N is in indeterminate form. Thus

to find its value we adopt the differential calculus method and

then we get,

NN

n

sin

sinlim

. Thus ,

2

2

22 sinN

AI

………(5)

The intensity at these maxima is maximum and that is why it is

principal maxima. Therefore, the condition for principal

maxima is

0sin or n or

nde sin)( or

nde sin ……….(6)

Now for n = 0, we get θ = 0, which gives the direction of the

central order maxima.

The values n = 1,2,3…….. correspond to the first, second, third

….order maxima.

46

Minima: For sin N β =0. But 0sin . We get the minimum

intensity. A series of minima, thus, obtained for

N

mormN

or

mdeN sin)(

mdeN sin)( ……….(7)

Thus for all integral values of m except 0,N,2N,3N…….we get

a minima, because for these values sin β = 0 and this will give

the position of principal maxima. Between m = 0 and N, i.e.

two maxima the no. of minima exist for m= 1,2,3,……,(N-1).

Since maxima and minima are obtained alternately there will be

(N-2) other maxima. These (N-2) ,maxima are known as

secondary maxima and the position of these maxima can be

obtained by differentiating eq. (4) w.r.t. β and equating it with

zero. Thus,

0sin

cossinsincos

sin

sin2.

sin2

2

NNNNA

d

dI

or 0cossinsincos NNN

or NN tantan ……..(8)

Therefore the roots of this equation except for which n

(central maxima) correspond to the positions of secondary

maxima.

47

If we construct a right angle triangle with its sides as 1, N tan β

and 2tan1 N then we have

22 tan1

tansin

N

NN

……….(9)

22 tan1 N

Hence from eq. (9) the intensity of secondary maxima may be

given as

222

22

2

22

sintan1

tan.

sin'

N

NAI

2222

2

2

22

sinsincos.

sin'

N

NAI

22

2

2

22

sin11.

sin'

N

NAI

………(10)

Hence from (5) and (10) we obtain

imafprincipalIntensityo

imaondaryfIntensityo

N

N

I

I

max

maxsec

sin11

'22

2

1

N tan β

Nβ

48

Condition For Absent Spectra With a diffraction grating: For a grating the direction of n

th order principal maxima is

given by

nde sin)( ............(1)

And the direction of minima in the diffraction pattern due to a

single slit is given by the eq.

me sin ...............(2)

The resultant intensity in the direction θ is given by

2

2

2

22

sin

sin.

sin NAI .........(3)

Intensity Distribution Curve

Central Maxima

Intensity

49

Where,

sine and

sin)( de

In eq. (3), 2

22 sin

Ais the diffraction term and represents the

intensity due to diffraction at a single slit while

2

2

sin

sin N is

interference term representing the intensity of waves interfering

from all the N slits. Therefore, if in any direction, diffraction

term is zero and interference term has maximum value, the

principal maxima will not be present in that direction. Form (1)

and (3) we get,

m

n

e

de

or

m

e

den

..............(4)

Eq (4) represents the condition for the nth

order to be absent

from the grating spectra. Thus,

1. If d = e then n = 2m = 2,4,6,..... Hence the 2nd

,4th

,6th

etc.

order interference maxima will be absent.

2. If d = 2e, then n = 3m = 3,6,9.... Thus 3rd

, 6th

, 9th

order of

the interference maxima will be absent.

Maximum No. Of Orders Of The Spectra With A Grating: In a diffraction grating the direction of the nth order principal

maxima for wavelength λ is given by

nde sin)(

or

sin)( den

Thus the maximum no. of possible orders nmax. is given by

)(max

den

1)(sin max

50

Dispersive Power Of A Diffraction Grating: The dispersive

power of diffraction grating is defined as the rate of change of

the angle of diffraction with the wavelength of light. It is

expressed as

d

d.

For a grating we have nde sin)(

Differentiating it w.r.t. λ we get,

nd

dde

cos)(

or

cos)( de

n

d

d

Linear Dispersive Power: Linear dispersive power is defined

asd

dx, where dx is the linear separation between two

wavelengths differing each other by dλ. If the focal length of

the lens used is f then.

Linear dispersive power

d

df

d

dx

f Angular dispersive power

cos)( de

nf

Characteristics Of Grating Spectra: 1. Spectra of different orders are situated symmetrically on

both sides of zero order image.

2. Spectral lines are almost straight and quite sharp.

3. Spectral colours are in order from violet to red.

51

4. Most of the intensity goes to zero order and rest is

distributed among the other orders.

5. The lines are more and more dispersed as we go to higher

orders.

Resolving Power Of An Optical Instrument: The ability of

an optical instrument to form separate images of two objects,

placed very close to each other, is known as its resolving power.

The minimum separation between two objects up to which they

can be seen as distinct objects by an optical instrument is called

the limit of resolution of that instrument.

Rayleigh’s Criterion Of Resolving: According to Rayleigh

“Two nearby point objects are just resolved by an optical

instrument when the principal maxima in the diffraction pattern

of one falls over the first minimum in the diffraction pattern of

the other and vice versa. The same thing happens in case of

spectral lines of two different wavelengths, the lines are

resolvable when the principal maximum due to one wavelength

falls over the first minimum due to the other or vice versa.

52

λ1 λ2

Not Resolvable

Just Resolvable

DIP Resultant

Intensity curve

λ1 λ2

Easily Resolvable

λ2 λ1

Principle Maxima Principle Maxima

53

Resolving Power Of A Grating: The resolving power of a

grating is defined as its ability to form separate diffraction

maxima of two wavelengths which are close to each other. If d

λ is the smallest difference in two wavelengths, which can be

just resolved by a grating and λ is the wavelength of either of

them or mean wavelength, then λ /d λ is known as the resolving

power of the grating.

Expression for resolving power:

Let (e + d) be the grating element and N the total number of

slits. Let P1 be position of the nth principal maximum of

spectral line of wavelength λ while nth principal maximum due

to wavelength (λ + d λ) be at P2.

θn

dθn P1

P2

Central Image

A

B

X

Y

54

Now according to Rayleigh’s criterion of resolution the two

wavelengths can be resolved if the position of P2 coincides with

the first minimum due to wavelength λ.

In fig. the dotted line represents the diffraction pattern due to

wavelength (λ + d λ) and the solid line represents the diffraction

pattern due to wavelength λ.

Now the principal maximum of wavelength λ in a direction θn

is given by

nde n sin)( ...........(1)

And the equation of minimum is given by

mdeN n sin)( .........(2)

Where m has all integral values except 0, N, 2N..........nN,

because for these values of m the condition for maxima is

satisfied.

Therefore, the first minimum adjacent to nth principal

maximum in the direction )( nn d may be obtained by putting

m=nN +1.

Thus the first minimum in the direction )( nn d is given by

)1()sin()( nNddeN nn ..........(3)

And the principal maximum of (λ + d λ) in the direction

)( nn d may be given as

)()sin()( dndde nn ............(4)

or )()sin()( dnNddeN nn ..........(5)

comparing (3) and (5), we get

)()1( dnNnN

or nNd / ......(6)

also from eq. (1) we have

55

nden

sin

Hence (6) can be modified as

ndeN

dsin

/

.......(7)

Here N(e + d) is the total width of the grating.

Resolving Power Of A Telescope: A telescope is used to see

distant objects. The ability of telescope to form separate images

of two distant point objects situated close to each other is

known as its resolving power. The resolving power of a

telescope is measured by means of the angle subtended by two

nearby point objects at its objective; it does not depend on the

linear separation between the objects. When the images of the

nearby distant objects are just resolved by the telescope then

the angle subtended by these two objects at the telescope

objective is called the limit of resolution of the telescope. Thus

the resolving power of telescope is defined as the reciprocal of

the smallest angle subtended at the objective by the two distant

objects, the images of which are just seen as separate ones

through a telescope.

Expression for resolving power:

dθ

O

dθ

A

C

P1

P2

56

Let a be the diameter of the objective of the telescope.

Consider the incident ray of light from two neighbouring

points of a distant object.

The image of each point is a Fraunhofer diffraction pattern.

According to Rayleigh, these two images are said to be

resolved if the position of the central maximum of the

second image coincides with the first minimum of the first

image and vice versa.

The path difference between the secondary waves traveling

in the directions AP1 and BP1 is zero and hence they

reinforce with one another at P1.

Similarly, all the secondary waves from the corresponding

points between A and B will have zero path difference.

Thus, P1 corresponds to the position of the central maxima

of the first image.

The secondary waves traveling in the directions AP1 and

BP2 will meet at P2 on the screen.

Let the angle P2AP1 be dθ. The path difference between the

secondary waves travelling in the directions BP2 and AP1

is BC.

From fig. ΔABC,

daABddABBC .sin

57

If this path difference da. , the position of P2

corresponds to the first minimum of the first image.

But P2 is also the position of the central maximum of the

second image.

Thus, Rayleigh’s condition of resolution is satisfied if

da.

or ad

............(1)

The whole aperture AB can be considered to be made up of

two halves AO and OB. The path difference between the

secondary waves from the corresponding points in the

halves will be2

.

All the secondary waves destructively interfere with one

another and hence P2 will be the first minimum of the first

image.

The eq. ad

holds good for rectangular apertures. For

circular aperture, this eq., according to Airy, can be written

as

ad

22.1 ............(2)

where λ is the wavelength of light and a is the aperture of

the telescope objective, which is equal to the diameter of

the metal ring in which the objective lens is mounted.

Here dθ refers to the limit of resolution of the telescope.

The reciprocal of dθ measures the resolving power of the

telescope.

58

22.1

1 a

d .........(3)

Resolving Power Of Microscope: The ability of a microscope

to form separate distinct images of two nearby small objects,

which cannot be seen by necked eye, is known as its resolving

power. The limit of resolution of a microscope is the smallest

separation between the two objects when their images are just

resolved.

Expression for resolving power:

In fig. MN is the aperture of the objective of a microscope

and A and B are two object points at a distance d apart.

A’ and B’ are the corresponding Fraunhofer diffraction

patterns of the two images.

A’ is the position of the central maximum of A and B’ is

the position of the central maximum of B.

A’ and B’ are surrounded by alternate dark and bright

diffraction rings.

The two images are said to be just resolved if the position

of the central maximum B’ also corresponds to the first

minimum of the image of A’.

The path difference between the extreme rays from the

point B and reaching A’ is given by

2α A

B

A’ A

B’

O

M

N

59

)'()'( MABMNABN

But NA’ = MA’

path MNBNdifference

In fig. AD is perpendicular to DM and AC is perpendicular

to BN

)()( DBDMCNBCBMBN

But DMAMANCN

path DBBCdifference

From Δs ACB and ADB

sinsin dABBC

and sinsin dABDB

Path difference sin2d

If the path difference 22.1sin2 d

Then limit of resolution

sin2

22.1d

Thus resolving power

22.1

.).(2

22.1

sin21..

AN

dPR , for self-

luminous object.

α

α

α

α

A

B

M

O C

D

60

If a medium of refractive index μ is introduced between the

object and objective then,

22.1

.).(2

22.1

sin21..

AN

dPR .

Resolving Power Of A Prism: Resolving power of prism is

defined as its ability to form separate images of two

neighbouring wavelengths.

Expression for resolving power:

Let S is a source of light; L1 is a collimating lens and L2 in

the telescope objective.

As the two wavelengths, λ and λ + d λ are very close, if the

prism is set in the minimum deviation position it would

hold good for both the wavelengths.

a A α

α α

δ dδ

S α

L1 α

L2 α

I1 α

I2 α t

61

The final image I1 corresponds to the principal maximum

for wavelength λ and image I2 corresponds to the principal

maximum for wavelength λ +d λ.

I1 and I2 are formed at the focal plane of the telescope

objective L2.

The face of the prism limits the incident beam to a

rectangular section of width a.

Hence, the Rayleigh criterion can be applied in the case of

a rectangular aperture.

In the case of diffraction at a rectangular aperture, the

position of I2 will corresponds to the first minimum of the

image I1 for wavelength λ provided

ad

or a

d

......................(1)

Here δ is the angle of minimum deviation for wavelength λ.

From fig. A

22

A

22sinsin

A

or

2cossin

A

But

l

asin

l

aA

2cos

............(2)

62

Also l

tA

22sin ............(3)

In the case of a prism

2sin

2sin

A

A

2sin

2sin

AA

...........(4)

Here μ and δ are dependent on wavelength of light λ.

Differentiating eq. (4) w.r.t. λ

2sin

2cos

2

1 A

d

d

d

dA

Substituting the values from eq. (2) and (4)

l

t

d

d

d

d

l

a

22

1

or

d

dt

d

da .. ...................(6)

substituting the values of dδ from eq. (1)

d

dt

d.

The expression

d measures the resolving power of the

prism.

Polarisation Of Light: In transverse waves there can be a no.

of directions perpendicular to the direction of propagation in

which the particles of the medium can execute periodic

vibrations. Transverse waves in which the vibrations are

restricted to one particular direction are known as polarised

waves and the phenomenon is as polarisation.

63

Unpolarised Light: The light having vibrations along all

possible straight lines perpendicular to the direction of

propagation of light is known as Unpolarised Light.

Polarised Light: The light, which has acquired the property of

one-sidedness, is called polarised light. Therefore, the polarised

light is not symmetrical about the direction of propagation but

its vibrations are confined only to a single direction

perpendicular to the direction of propagation. The crystal which

makes the light polarised is known as polariser and the crystal,

which analyses the incoming polarised light, is called analyser.

1. Plane Polarised Light: Light is said to be plane polarised

when vibrations take place only in one direction parallel to

the plane through the axis of a beam.

2. Circularly Polarised Light: Light is said to be circularly

polarised when the vibrations in transverse plane are circular

having constant period. Here the amplitude of vibrations

remains constant but the direction changes only.

64

3. Elliptically Polarised Light: Light is said to be elliptically

polarised when the vibrations are elliptical and having a

constant period and takes place in a plane perpendicular to the

direction of propagation. Here the amplitude of vibrations

changes in magnitude as well as in direction.

Plane of Polarisation and Plane of Vibration: The plane in

which the vibrations occurs known as plane of vibration and a

plane perpendicular to plane of vibration in which no vibration

occurs is known as plane of polarisation.

B C

G H

A

B C

D

E F

G H

65

Plane of Vibration

Plane of Polarisation

Brewster’s law: In 1811, Brewster discovered a simple relation

between angle of polarisation and refractive index μ of the

medium. He found that the refractive index of the material

medium is equal to tangent of the angle of polarisation. i.e.

pitan ........(1)

Thus when angle of incidence becomes equal to the angle of

polarisation, then from Snell’s law,

r

i p

sin

sin ........(2)

Thus from (1) and (2), rip sincos

or rip sin)90sin(

ABCD EFGH

ip ip

r

θ

X Y O

A

C

B

66

or 090 pir

Also from fig. if θ be the angle between reflected and refracted

rays then, 09090180)(180 pir

Therefore, the reflected and refracted rays are at right angle to

each to other.

Double Refraction: When an ordinary light beam after passing

through a crystal splits up into ordinary and extraordinary light

the crystal is known as double refracting crystal and the

phenomenon is known as Double Refraction.

Ordinary and Extraordinary Rays: When ordinary light

beam when passed through a uniaxial crystal splits up into two

refracted rays. One of them obeys laws of refraction known as

(o-Ray) ordinary ray while the other does not obey laws of

refraction known as (E-ray) extraordinary ray.

These two rays have the following characteristics:

1. The plane of polarisation for O-ray is same as the principle

plane. Therefore, it has vibrations perpendicular to the

principal plane. For E-ray the case is opposite.

2. The two rays emerge from the crystal along the parallel

directions.

re

ro

E-Ray

O-Ray

Incident

Ray

67

3. The refractive index of the material for the two rays is given

as:

o

or

i

sin

sin and

e

er

i

sin

sin

The value of μo is constant while μe varies with the angle of

incidence.

4. In a crystal if μo >μe the crystal is known as negative crystal

in which the velocity of O-ray is less than E- ray.

5. In a crystal if μo <μe the crystal is known as positive crystal in

which the velocity of E- ray is less than O-ray.

6. Since the refractive index of O- ray is constant, hence it

travels with same speed in all directions.

7. Unpolarised light incident along optic axis does not split into

O- ray and E- ray.

8. The difference between the refractive indices of O- ray and

E- ray is known as birefringence i.e. birefringence

= eo .

NICOL PRISM: It was invented by William Nicol in 1828. It

is an excellent optical device used for producing and analyzing

the plane polarised light.

Principle: 1. It works on the phenomenon of double refraction.

2. When Unpolarised light is passed through uniaxial crystal, it

splits up into ordinary and extraordinary light.

3. If by some optical method one of the two rays is eliminated

the ray, emerging through the crystal will be plane polarised.

4. In Nicol prism O- ray is eliminated by total internal reflection

and E- ray is transmitted through the crystal.

68

Construction: It is most commonly used device to get plane

polarised light. It is made from a calcite crystal in the following

manner:

1. The length of the calcite crystal is taken three times its width.

2. The end faces in the crystal are cut in such a manner that the

acute angle (710) reduces to 68

0.

3. The crystal is then cut into two parts along a plane which

passes through the blunt corners A’ and D’ and is

perpendicular to the principal section.

4. Now the two cut surfaces are polished and cemented back

with a transparent substance known as Canada balsam, the

refractive index of which is midway for O and E ray.

5. The two end faces of the crystal are left open while the other

faces are coated with lampblack.

6. Finally, the crystal is enclosed in a brass tube.

Action: When Nicol crystal is used for, producing polarised

light is known as polariser and when it is used for the analysis

of polarised light is known as analyser.

(a). Nicol as a Polariser: 1. A beam of Unpolarised light when falls on the face of the

crystal it splits up into O- and E- rays.

2. Now after traversing some distances into Nicol prism

both the rays reach to the Canada balsam layer.

680 71

0

140

140

O O’ A’ A

D’ P

S0

SE

S

69

3. Since this layer offers a rarer medium to O-ray coming

from a denser medium of calcite, the O- ray suffers total

internal reflection.

4. This totally reflected ray is absorbed by the lampblack

coating.

5. Here the E-ray emerges out of the crystal and in this way

the crystal produces plane polarised E-ray and thus known

as polariser.

(b). Nicol as a Analyser:

POLARISAER ANALYSER

E- Ray

O- Ray

E- Ray

E- Ray

O- Ray

E- Ray

POLARISAER ANALYSER

70

1. When the two Nicol prisms are arranged coaxially then

the first Nicol act as a polariser and produces plane

polarised E-ray.

2. The ray emerging from polariser falls on the second nicol

prism known as analyser.

3. When the principal section of the two nicols are parallel

to each other, the plane polarised ray coming out from the

first nicol is easily transmitted through the second nicol.

4. Now if we rotate the second nicol gradually, the intensity

of transmitted ray from the second nicol decreases.

5. The intensity becomes zero when principal axis of both

the nicol become perpendicular to each other. Hence the

second nicol works as an analyser.

Quarter wave plate: A plate of doubly refracting crystal viz,

calcite or quartz whose refracting faces are cut parallel to the

direction of optic axis and its thickness is such that it produces

a phase difference of π/2 and path difference of λ/4 between

emerging ordinary and extraordinary rays is known as Quarter

Wave Plate.

If μ0 and μe be the refractive indices for ordinary and

extraordinary rays then, for normal incidence, the path

difference introduced between emerging O- and E- rays for

negative crystal (calcite) 4)(

teo

or )(4 eo

t

(for negative crystal eo )

and for positive crystal (quartz) 4)(

toe

71

or )(4 oe

t

(for positive crystal oe ).

This plate is used to produce circularly and elliptically

polarised light. When a plane polarised light is allowed to fall

on a quarter wave plate such that its vibrations make an angle

450 with the optic axis then the emergent beam will be

circularly polarised. When the vibrations of incident light make

an angle other than 450 with the optic axis the emergent light

will be elliptically polarised.

Half Wave Plate: A plate of doubly refracting crystal viz,

calcite or quartz whose refracting faces are cut parallel to the

direction of optic axis and its thickness ‘t’ is such that it

produces a phase difference of π and path difference of λ/2

between emerging ordinary and extraordinary rays is known as

Half Wave Plate.

Now for negative crystals like calcite ( eo ), path difference

is

2)(

teo

or )(2 eo

t

and for negative crystals like quartz ( eo )

)(2 oe

t

When a plane polarised is allowed to pass through half wave

plate then the emergent light is also plane polarised.

These two plates are also known as retardation plates as they

retard the motion of one of the beam.

72

END