Unit2 Wave Optics

8/6/2019 Unit2 Wave Optics

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OPTICS

Interference of Light:

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

yNewtons corpuscular theoryyHuygens wave theoryyElectromagnetic theoryyQuantum 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:

y It is the phenomenon of superposition of two coherent*waves in the region of superposition.

yAt some points in the medium, the intensity of light ismaximum while at some other points the intensity is

minimum.y 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,


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21 yyy ! If, tay [sin11 ! and H! tay sin22 Where is the phase difference between two waves.

H[[ ! tatay sinsin21 tataa [H[H cossinsincos 221 !

Let JH coscos21 Aaa ! and JH sinsin2 aa ! ; we have,tAtAy [J[J cossinsincos !

J[ ! tAsin Where,

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21 cos2 HaaaaA !

and HH

Jcos

sintan

21

2

aa

a

!

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

amplitude, i.e.2

AI w , thenHcos2 21

2

1

2aacaAI !! (1)

Since2

1a and2

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 2221 aa .Condition for maximum and minimum intensities:

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

.........2,1,0;21cos !!! nnor THH Therefore, the intensity will be maximum when phasedifference is even multiple ofT , but we know that,

Path difference v!(T

P

2phase difference


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TT

Pn2

2v! or 2

2P

n!(

Therefore, 2

21max aaI ! Similarly I will be minimum at points where cos =-1andtherefore,

TH 12 ! n

212P

!( nn=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 apair of coherent waves may be obtained from a single source of

light.

Youngs Double Slit experiment: In 1801, Young first

demonstrated experimentally the phenomenon of interference.


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y In his experiment light from a monochromatic source isallowed to fall on a slit S and then through a double slit S1

and S2.

y The interference pattern was observed on a screen XYwhere few dark and bright bands or fringes are observed.

y These fringes are of equal distances.y The bold lines show the amplitude in positive direction(crest) where the dotted lines show the amplitude in thenegative direction (troughs).

yAt points where crests fall on crests or troughs fall ontroughs the amplitudes add up and the resultant intensity

increases (I A2) and thus we obtain constructive

interference.

yAt points where crests fall on the troughs or the troughsfall on the crests, the amplitudes are reduced (the resultantamplitude 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


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y Thus on the screen alternate bright and dark regions ofequal width, called interference fringes are observed.

Analytical treatment for Youngs double Experiment:

y Let S1and S2 be two coherent sources separated by adistance d and made from a monochromatic source S.

y Let a screen be placed at a distance D from the coherentsources.yO is a point on the screen which is equidistant from S1and

S2.

y Therefore, path difference at O will be zero and theintensity at O will be maximum.

y Let P be a point on the screen such that,OP= y

Therefore,

!

!2

,2

dyPR

dyPQ

S1

S2

d

P

R

C Od/2

d/2

D

Qy


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D

dy

Dd

yDPS2

2,

2

2

2/12

2

1

!

!

andD

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,

DydPSPS !!( 12

Therefore the phase difference,

v!P

TJ

2path difference=

D

ydv

P

T2

y Position of bright fringes:As mentioned earlier that forbright fringes path difference is even multiple of /2, i.e.

PP

nnD

yd!!!(

22

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

d

Dnyn

P!

.

y Position of dark fringes: For dark fringe or the minimumintensity at P, the path difference must be an odd multipleof half wavelength, i.e.

2

12P

!!( nD

yd


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

by,

d

Dnyn

2

12 P!

.

yFringe Width: The distance between any two consecutivedark fringes or bright fringes is same and known as fringewidth. It is given by:

d

Dyy nn

PF !! 1

y 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:

y The beams from two sources must propagate along the sameline; otherwise the vibrations cannot be along a common line.

y The two sources must be very narrow because a broad sourceis equivalent to a no. of sources and so positions of darknessand brightness cannot be well defined.

y The two sources must have equal intensities.y 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.y The common original source must be monochromatic, i.e.,

emitting light of single wavelength.

y The light from the two sources must have either zero phasesor a constant difference of phase. This condition is very well

observed in the case of coherent sources.

y The separation between source and screen should be large toget wide fringes.

y If the interference pattern has to be obtained by polarisedlight 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 Fresnels biprism, Fresnels

S1

SS2


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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.

Newtons rings, Michelson interferometer are the examples ofthis class.

Fresnels 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 Fresnels biprism experiment. By using biprism, Fresnel

obtained two coherent sources from a single source byrefraction.

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

SReflectedSystem

Transmittesystem


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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:

yA narrow vertical slit S is illuminated by a source ofmonochromatic light.

y The prism is placed with its refracting edge parallel to theline joining the sources.

y The light from the slit S is allowed to fall symmetrically onthe biprism ABC with its refracting edge vertical (i.e.,

parallel to the line source S).

y When the light from S falls on the upper half of the prism,after refraction it appears to come from virtual source S1.

y Similarly after refraction from the lower half of the prismit appears to be coming from virtual source S2.

y The distance between the biprism and the source S is soadjusted that the sources S1 and S2 are very close to each

other.

y The interference fringes are observed in the region EF ofthe screen.

S

S2

S1

d

ab

D

B

O

A

C

E

O

F

Screen


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y The analytical treatment is same as for Youngs DoubleSlit experiment.

y If d be the distance between the sources S1 and S2 and Dbetween the source and the screen, then the fringe width is

given by,

d

DPF !

And thus FF

Pba

d

D

d

!!

Moreover the distance ofn th bright fringe from O.

,)(d

Dny brightn

P!

Similarly

d

Dny darkn

2

12)(

P!

.

Where EQ )1(2 ! ad

Applications:a)Determination of wavelength of light:

D

dFP !

b)Determination of thickness of a thin sheet of transparentmaterial:

a)Biprism can be used to determine the thickness of agiven 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 is introduced in the path of one of

the beams, the fringe which was originally at C shiftsto P.

e)The time taken by the wave from B to P in air is thesame 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 itsvelocity in medium then,

c

t

c

tAP

c

BP

!

00

tc

ctAPBP 0!

, ButQ!

c

c0

tttAPBP 1!!@ QQ If P is the point originally occupied by the n th fringe, then the

path difference

PnAPBP !

PQ nt !@ 1 ..(i)

Pt

d

x

D

C

A

B

G


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Also the distance x through which the fringe shifted

d

DnP!

Where F

P

!d

D

, the fringe width.

d

Dnx

P!@

Also, D

xdn !P

Or, Dxdt!1Q

g)Therefore, knowing x, the distance through which thecentral fringe is shifted, D, d and , the thickness of

the transparent plate can be calculated.

h)If a monochromatic source is used, the fringes will besimilar and it is difficult to locate the position wherethe 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 aplane glass-plate.

Interference Due To Reflected Light(Thin films):

y Consider a transparent film of thickness t and refractiveindex .


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yA ray SA incident on the upper surface of the film is partlyreflected along AT and partly refracted along AB.

yAt B part of it is reflected along BC and finally emergesout along CQ.

y The difference in path between the two rays AT and CQcan be calculated.

y CN is normal to AT and AM is normal to BC.y The angle of incidence is i and the angle of refraction is r.y CB and AE are extended to meet at P.

rAPC! The optical path difference ANBCABx ! Q

Here, CM

AN

r

i!!

sin

sinQ

CMAN .Q!@

CMBCABx .QQ ! CMPCCMBCABx !! .QQ

PM.Q!

In the APM,AP

PMr!cos

or rEPAErAPPM coscos. !! rtcos2! tEPAE !!3

rtPMx cos.2. QQ !!@ (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 T equivalent to

a path difference2

P occurs.

Therefore, the correct path difference in this case,

2cos2

PQ ! rtx ..(2)

1)If the path difference x = nP where n = 0,1,2,3.etc.,constructive interference takes place and the film appears

bright. PP

Q nrt !@2

cos2 or, 2

12cos2P

Q !@ nrt ..(3)

2)If the path difference is

2

12P

!! nxwhere n = 0,1, 2,

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

appears dark. 2

122

cos2PP

Q !@ nrt

or, PQ 1cos2 !@ nrt .(4)Here n is an integer only, therefore (n+1) can also be taken

as n.

PQ nrt !@ cos2 .(5)Where n = 0,1, 2, 3etc.

i

i

r

S T Q

CA

E

PF

M

N

B

r

AIR

AIR

t


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Interference Due To Transmitted light (Thin Film):

y Let there be a thin transparent film of thickness r andrefractive index .

yA ray SA after refraction goes along AB.yAt B it is partly reflected along BC and partly refracted

along BR.

y The ray BC after reflection at C, finally emerges along DQ.yHere at B and C reflection takes place at the rarer medium(medium-air interface).

y Therefore, no phase change occurs.y BM is normal to CD and DN is normal to BR.y The optical path difference between DQ and BRis given

by

BNCDBCx ! Q

alsoMD

BN

r

i!!

sin

sinQ or MDBN .Q!

from fig.rBPC! and CP = BC = CD

t

B

RQ

i N

DM

C

P

A

AIR

AIR

r

i

r

r rr

i

S


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PDCDBC !@ PMMDPDMDPDx QQQQ !!!@

In theBP

PMrBPM !( cos, or rBPPM cos.!

But, tBP 2! rtPM cos2!@

rtPMx cos2. QQ !!@ 1)For bright fringes, the pat difference Pnx !

PQ nrt !@ cos2 where n = 0, 1, 2, 3...etc.

2)For dark fringes, the path difference 212 P! nx

2

12cos2

PQ

!@

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 BRand DQ is very large.Newtons 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

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

fringes are known as Newtons Rings.


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Construction:

y S is a source of monochromatic light at the focus of thelens L1.

yA horizontal beam of light falls on the glass plate B at 450.y The glass plate B reflects a part of the incident light

towards the air film enclosed by the lens L and the planeglass plate G.

y The reflected beam from the air film is viewed with amicroscope.

Working:

y Interference takes place and dark and bright circularfringes are produced.

y This is due to the interference between the light reflectedfrom the lower surface of the lens and the upper surface of

the glass plate G.

L

G

AIRFILM

M

S

L1B

450


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Theory:1.Newtons ring by reflected light: Let the radius of curvature

of the lens is Rand 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

12cos2P

Q ! nrt .(i) where n = 1,2,3.etc.

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

For air, = 1

Hence 2

122P

Q ! nt ..(ii)

For the dark rings, PQ nrt !cos2 Or, PQ nt !2 (iii) where n = 0,1,2,3.etc.

From fig. in CEP222 EPCECP !

Or 222

EPOECOCP !

or 222 nrtRR ! or Rtrn 2

2 ! (ignoring t2)

hence R

rt n

2

2 !

L

REFLECTED

LIGHT

AIRFILM

C

R

O Q

H E P

rn

t


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

for bright rings

2

122 RnrnP

! ..(iv)

or

2

12 Rnrn

P! ..(v)

for dark rings Rnrn P!2

..(vi)

or Rnrn P! ..(vii)

Result:

y The radius ofnth dark ring is proportional toi. n

ii. P iii. R

y Similarly the radius ofnth bright ring is proportional toi.

2

)12( n

ii. P iii. R

y If Dn is the diameter of the nth dark ring,RnrD nn P42 !! ..(viii)

y If Dn is the diameter of the nth bright ring,

21222 RnrD nnP!! ..(ix)


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2.Newtons ring by transmitted light:

In case of transmitted light the interference fringes areproduced such that for bright rings,

PQ nrt !cos2 (x)For air, = 1and cos r = 1 then

PQ nt !2 .(xi)For dark rings,

2122P

Q ! nt (xii)

Hence in this case the radius ofnth bright is

Rnrn P! .(xiii)

And the radius of the nth dark ring is

2

12 Rnrn

P! .(xiv)

Applications of Newtons Ring:

y Determination of wavelength of light: Fornth dark ring wehave

RnDn P42 ! ..(i)

L AIRFILM

TRANSMITTED

LIGHT


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or PnR

Dn !4

2

.(ii)

again for(n+p)th dark ring,

PpnRD pn

!

4

2

(iii)

pR

DD npn

4

22 !@ P (iv)

y Determination of Refractive Index of a liquid: Therefractive index of a liquid, forming film between lens and

glass plate, can be obtained from the following eq.

pR

DDliquidnpn

4

22 !

Q

P..(i)

pR

DDairnpn

4

22 !

Q

P

Since for air = 1

Hence

pR

DDairnpn

4

22 !

P ..(ii)

Dividing (ii) by (i) we get

Q!

liquidnpn

air

npn

DD

DD

22

22

Michelsons Interferometer:Principle: The amplitude of light beam from a source is

divided into two parts of equal intensities by partial reflection


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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:

y It consists of two highly polished plane mirrors M1 and M2two plane parallel glass plates G1 and G2 of exactly the same

thickness.

y The plate G1 is semi silvered at the back so that theincident beam is divided into reflected and transmitted

beams of equal intensities.y M1 and M2 are mutually at right angles to each other.y The plates G1 and G2 are parallel to each other and are at an

angle of450

to M1 M2.

A

G1 G2

M1

M2

T

S

C

BTransmitted

Reflected


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y The mirrors M1 and M2 can be adjusted exactly perpendicularto each other with the help of the screws on their backs.

y The mirror M1can be moved exactly parallel to itself with acarriage on which it is moved.

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

partly transmitted.

y The reflected ray AC travels normally towards plane mirrorM1 reflected back to the same path and comes out along AT.

y The transmitted ray after reflection from mirror M2 followsthe same path and then moves along AT after reflection from

the back surface of G1.

y So the rays received at T are produced from a single source by the division of amplitude and we get an interference

pattern.

y The reflected ray AC travels through glass plate G1 twice andin the absence of plate G2 the transmitted ray does not cross

plate G1.y To compensate it a second plate G2 is introduced in the path

of transmitted ray.

Adjustment:

y Distance of the mirrors M1 and M2 are adjusted to be nearlyequal from glass plate G1.

y In order to obtain a parallel ray of light a lens is adjustedbetween the source S and plate G1.

Formation of fringes:

1.Circular fringes:


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y If eye is placed to see mirror M1 directly then in addition itwill see virtual image of M2 formed by reflection in glass

plate G1.

y Thus one of the interfering beams cones from M1 and otherappears to come from the virtual image of M2 i.e. M2.

y If M2 is exactly at right angles to M1 and the plate G1 is at45

0to each of then, the air film formed between M1 and

M2 will have uniform thickness.

y Let the two interfering beams appear to come to the eyefrom two virtual images S1 and S2 of the original source

and let 2d be the distance between them.

yNow since reflected ray from M2 suffers reflection at thesilvered surface of plate G1, an additional path difference

/2 is introduced.

y Therefore, the total path difference 2/cos2 PE ! d .y Therefore, for brightness,

.....2,1,0,2/cos2 etcnnd !! PPE y If M2coinsides with M1 then total path difference will be

/2 which is the condition for destructive interference.

y If the light is obtained from an extended monochromaticlight source then for a given value of n, the value of is

constant and so the locus of the fringes is circle.


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2.Localized fringes:y 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.y In this case the locus of points of equal thickness is astraight line parallel to the edge of the wedge.

y For small path difference, the fringes are nearly straight butfor larger path difference the fringes are generally curved

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

wedge.

M1

M2

S

S1

S2

2d cos


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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 inclinationhence they are called equal inclination fringes orHaidingerfringes.

Applications:

1.Determination of wavelength of Monochromatic light:y Mirror M1 and M2 are adjusted so that circular fringes are

visible in the field of view.

yIf M1 and M2 are equidistant from the glass plate G1, thefield of view will be perfectly dark.

y The mirror M2 is kept fixed and the mirror M1 is movedwith the help of the handle of the micrometer screw and

the number of fringes that cross the field of view is

counted.

y If mirror is moved through a distance of d for amonochromatic light of wavelength and no. of fringes

that cross the centre of the field of view is n then2

Pnd ! ,

because for one fringe shift the mirror moves through a

distance equal the half the wavelength. Hence can be

determined.

M1

M2M1

M2

M1

M2


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2.Determination of the difference in wavelength betweentwo neighbouring spectral lines:

y Set the apparatus for circular fringes.y Let the source has two wavelengths 1 and 2 (1> 2)

which are very close to each other.y The two wavelengths form their separate fringe patterns,

but because of very small change in wavelength, the two

patterns overlap.

yAs the mirror M1 is moved slowly the two patterns separateout slowly and when the path difference is such that the

dark fringe of 1 falls on the bright fringe of 2 maximum

indistinctness occurs.

yAgain when the path difference is such that the brightfringe of 1 falls on bright fringe of 2 maximum two

successive positions of maximum distinctness occurs.

y Let the mirror M1 is moved through a distance of dbetween distinctness.

y This will happen only when nth fringe of 1 coincides withn+1

th fringe of 2 then

21 )1(2 PP !! nnd

1

2

P

dn ! and

2

21

P

dn !

Subtracting we get,

12

221PP

dd ! =

21

212PP

PPd

Or d2

21

21

PPPP !

21 PP }3 hence2

21 PPP !3


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hence d2

2

21

PPP !

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:y The apparatus is set for parallel fringes.y White light source is used instead of monochromatic

source.

y The cross wire is set on the central fringe.yNow the thin plate whose thickness is to be determined is

introduced in the path of one of the interfering beams.y Due to the insertion of the plate path difference is changed

to 2(-1)t.

y Thus a shift in the fringe system occurs.yNow the mirror M1 is moved till the central fringe

coincides with the cross wire.

y The distance x moved by mirror M1 is measured usingmicrometer screw. Hence we have,

tx )1(22 ! Q

1!

Q

xt

Using above equation t can be determined.

Diffraction: The bending of light beam round the corners oredge 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.


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Types of diffraction: There are two classes of diffraction:

1.Fresnel Diffraction2.Fraunhofer Diffraction

1.Fresnel Diffraction:y In this case diffraction is considered to take place from

different parts of the aperture.

yHere the source or screen or both are at finite distance fromobstacle.

y In this class the incident wave front is divergent, eitherspherical or cylindrical.

yHere the observed pattern is a projection of diffractingelement modified by diffracting effects and the geometryof the source.

yHere the diffraction centre of the pattern may be eitherbright or dark.

2.Fraunhofer Diffraction:yHere the source of light and screen are at infinite distance

from the obstacle.

yHere the inclination is important.yHere the incident wave front is generally plane and the

centre of diffraction pattern is always bright.

S

A

B

A

B


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Fraunhofer Diffraction At A Single Slit:

y The light from a monochromatic point source S isconverted into parallel beams of light by a convex lens

L1L1.

yNow this beam is incident normally on a slit AB of width e.y The incident light wave fronts are shown by W1W1 andW2W2.yEvery 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.

y In this diffraction pattern a central bright band is obtained.yOn either side of this band alternate dark and bright bands

of decreasing intensity are observed.

e

A

B

B

A PL2

L2

W1

W1

W2

W2L1

L1W

W

O

Y

Y

K


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

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

UU sinsin eABBK !! Therefore, the corresponding phase difference

UP

T

P

T sin22 eencepathdiffer !!

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)(,sin

21sayde

n!

! UP

T

Thus the resultant in a direction , 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 between successive vibrations. To

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

which gives the resultant amplitude Rand a resultant phase .

R

a

a

a

1

2

3

a

O


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33

Now resolving the amplitudes parallel and perpendicular, we

get,

....(..........)1cos(.........2coscoscos HHHJ ! naaaaR2....(..........)1sin(.........2sinsinsin HHHJ ! naaaR

Multiplying eq. (1) by 2sin

2

H,we

!

HH

HH

HHHJ )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

!

HH

HHHHHHJ

nn

aR

2

1

cos2sin22

1

sin2sin

HH

H

H

!

!

nna

na

or

2sin

2

1cos

2sin

cosH

HH

J

!

nna

R(3)

similarly, multiplying (2) by 2sin2

Hand applying

)cos()cos(sinsin2 BABABA !

and2

sin2

sin2coscosCDDC

DC

!


34/73

34

we get,

2sin

2

1sin

2sin

sinH

HH

J

!

nna

R..(4)

from (3

) and (4

) we get

2sin

2sin

H

H

!

na

Rand

2

1HJ

!

n

Hence for the present case

!

!!

P

UT

P

UT

UP

T

U

P

T

n

e

ea

en

e

n

n

a

d

nda

Rsin

sin

sinsin

2

sin21

sin

2

sin21

sin

2sin

2sin

Let EPUT !sine then,

na

n

aR

/

sin

/sin

sin

E

E

E

E!!

As n/E is small,

}

@

nn

EEsin

Again since n is infinitely large and a is negligibly small,Let na =A, a finite no.

Hence, EEsinA

R ! ...........(5)


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35

The secondary waves diffracted by the slit in a direction are

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

the resultant intensity at P,2

RIw . i.e.

2

22 sin

E

EAI! ........(6)

Conditions for maxima and minima:

0!Ed

dI

or 0sin

2

22

!

E

E

E

A

d

d

or 0sincossin2

2

2 !

EEEE

EE

A .....(7)

Case I: When 0sin

!E

Eor sin =0 or TE ns! ( o{E , if

= 0 then (sin )/ = indeterminate)

Therefore, .......3

,2,1,0 !{ nn or TP

UTn

es!

sin

or PU ne s!sin ..........(8)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!

E

EEE

or EE !tan ..............(9)


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36

The intersection of the curves Etan!y and E!y will givethe solution of eq.(9). The points of intersection will give the

values of as

,.......2

7,

2

5,

2

3,0

TTTE sss!

Here = 0 corresponds to central maximum.

Hence for central maximum the intensity is given by,

0

2

2

22 sinIA

AI !!!

E

E

y=tan

y=ta

n

y=tan

y =

450

O

y

/2 3/2 5/2


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37

e e e e e e


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38

Diffraction At Double Slit:

y Let there be two slits AB and CD of equal width e,separated by a distance d.

y The incident light gets diffracted from these two slits andfocused on the screen XY.

y The diffraction by two parallel slits is a case of diffraction aswell as interference.

y Thus the pattern obtained on the screen consists of equallyspaced interference fringes due to both the slits and their

A

B

C

D

S1

S2

L LP

O

P

M

S

e+d

Y

Y


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39

intensity being modulated by the diffraction phenomenon

occurring due to individual slits.

y The two slits AB and CD may be considered equivalent totwo coherent sources S1 and S2.

yNow due to individual slit, the resultant amplitude due to allsecondary waves diffracted in the direction is given by:E

EsinAR !

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

Where, P

UTE

sine!

y For simplicity, the resultant of all the secondary waves maybe taken as a single wave of amplitude R.

y The resultant at point P on the screen will be the result ofinterference between these two waves, of same

amplitudes E

EsinA, starting from S1 and S2.

y The path difference between the wavelets from S1 and S2 in adirection is, Usin)2 deMS ! and therefore the phasedifference U

P

T

P

TH sin

222 deMS !v! .

y The resultant amplitude at P may be given as,Hcos2 21

2

2

2

1

2 RRRRRr !

HE

E

E

E

E

Ecos

sin2

sinsin222

!

AAA

!

2cos

sin4

2

2

22 H

E

EA

yHence resultant intensity,


40/73

40

!!

2cos

sin4

2

2

222 H

EE

ARI rr

or FEE 2

2

222 cos

sin4

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

where UP

TF

Hsin

2de

!!

y From the above expression, it is clear that the resultantintensity depends on the two factors.

(a). Diffraction term

2

22 sin

E

EA :This factor is same as in the

case of single slit. Thus this term corresponds todiffraction 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 !E or TE ms! or TPUT

m

e

s!

sin

........3,2,1,sin !s! mme PU .............(3)and the position of secondary maxima approach to

..........2/7,2/5,2/3 TTTE sss!

(b).Interference term F2cos :This term corresponds tothe 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

1cos 2 !F or TF ns! or TUP

Tnde s! sin

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


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41

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

interference and diffraction coincide. The minima due to

interference term is given by,

0cos 2 !F or 2/12 TF s! n or 2/12sin PU s! nde

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

some of the interference maxima are missing. Since for the

same value of , the following two relations hold true,

PU nde s! sin , Interference Maxima.....(5)PU me s!sin , Diffraction Minima....(6)

When both the above conditions are satisfied simultaneously,then the interference maxima will be absent in the direction for

which 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 theinterference maxima will be absent.

3.If e +d =e, i.e. d = 0, then the diffraction pattern will similarto as observed to a single slit of width equal to 2e.


42/73

42


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43

Cos2

2--2 0

4A2sin

2cos

2/


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44

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 10000to 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.

P

(e+d

S1

S2

SN

M1

M2

MN-1

L

X

Y


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45

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

are equivalent to a single wave of amplitude,

EEsinA

R !.(1)

where P

UTE

sine!

(2)The path difference between any two consecutive waves from

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

difference will beU

P

Tsin)(

2de

. Since the phase difference

is constant between any two consecutive waves it can be taken

asFU

P

T2sin)(

2 ! de(3)

Thus we have to find out the resultant ofN vibrations in a

direction and each vibration is of amplitude

EEsinA

.

Now similar to the resultant of n-harmonic waves, the resultantofN-slits may be given as,

F

F

EE

F

F

F

F

sin

sin.

sin

sin

sin

2

2sin

2

2sin

'NANR

NR

R !!!


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46

Thus the resultant intensity at P may be given as,

F

F

EE

2

2

2

222

sin

sin.

sin'

NARI !!

..(4)

Here the factor 2

22

sinE

EA is the intensity factor due to a single

slit while F

F2

2

sin

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

Principle Maxima: for maximum intensity, we have,

0sin !F or TF ns! , .....2,1,0!n

But in this condition00

sinsin !

FFN is in indeterminate form. Thus

to find its value we adopt the differential calculus method and

then we get,N

N

n!

sp F

F

TF sin

sinlim

. Thus ,

22

22sinN

A

I EE

! (5)The intensity at these maxima is maximum and that is why it is

principal maxima. Therefore, the condition for principal

maxima is

0sin !F or TF ns! or TUPT

nd

e s! sin)( or PU nde s! sin .(6)Now for n = 0, we get = 0, which gives the direction of the

central order maxima.


47/73

47

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

.order maxima.

Minima: For sin N =0. But 0sin {F . We get the minimumintensity. A series of minima, thus, obtained for

s!s!

N

m

ormN TFTF

orTU

P

TmdeN s! sin)(

PU mdeN s! sin)( .(7)Thus for all integral values of m except 0,N,2N,3N.we geta minima, because for these values sin = 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 beobtained by differentiating eq. (4) w.r.t. and equating it with

zero. Thus,

0sin

cossinsincos

sin

sin2.

sin2

2

!

!

F

FFFF

F

F

E

E

F

NNNNA

d

dI

or 0cossinsincos ! FFFF NNN or FF NN tantan ! ..(8)


48/73

48

Therefore the roots of this equation except for which TF ns!

(central maxima) correspond to the positions of secondary

maxima.

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

and F2tan1 N then we have

F

FF

22 tan1

tansin

N

NN

! .(9)

F22 tan1 N

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

FFF

E

E222

22

2

22

sintan1

tan.

sin

N

NAI

!

FFFE

E2222

2

2

22

sinsincos.

sin'

N

NAI

!

FEE

22

2

2

22

sin11.

sin'

!

N

NAI

(10)

Hence from (5) and (10) we obtain

1

N tan

N


49/73

49

imafprincipalIntensityoimaondaryfIntensityo

N

N

I

I

max

maxsec

sin11

'22

2

!

!F

Intensity Distribution Curve

Central Maxima

Intensity


50/73

50

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

thorder principal maxima is

given by

PU nde ! sin)( ............(1)And the direction of minima in the diffraction pattern due to a

single slit is given by the eq.

PU me !sin ...............(2)The resultant intensity in the direction is given by

F

F

E

E2

2

2

22

sin

sin.

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

Where,P

UTE

sine! and

P

UTF

sin)( de !

In eq. (3), 2

22 sin

E

EAis the diffraction term and represents the

intensity due to diffraction at a single slit while F

F

2

2

sin

sin N

isinterference 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

me

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.


51/73

51

2.If d = 2e, then n = 3m = 3,6,9.... Thus 3rd, 6th, 9th order ofthe 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 is given byPU nde ! sin)(

or P

Usin)( den

!

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

P

)(

max

den

!

? A1)(sin

max !U3

Dispersive Power OfA 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 asP

U

d

d

.

For a grating we havePU nde ! sin)(

Differentiating it w.r.t. we get,

nd

dde !

P

UUcos)(

or UP

U

cos)( de

n

d

d

!

Linear Dispersive Power: Linear dispersive power is defined

asPd

dx, where dx is the linear separation between two

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

the lens used is fthen.


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52

@Linear dispersive powerP

UP d

df

d

dx!!

v! f Angular dispersive power

Ucos)( de

nf

v!

Characteristics Of Grating Spectra:

1.Spectra of different orders are situated symmetrically onboth sides of zero order image.

2.Spectral lines are almost straight and quite sharp.3.Spectral colours are in order from violet to red.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 higherorders.

Resolving Power OfAn 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 theycan be seen as distinct objects by an optical instrument is called

the limit of resolution of that instrument.

Rayleighs 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 ofspectral 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 orvice versa.


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53

1 2

Not Resolvable

Just Resolvable

DIP Resultant

Intensity curve

1 2

Easily Resolvable

21

Principle MaximaPrinciple Maxima


54/73

54

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. Ifd

is the smallest difference in two wavelengths, which can be

just resolved by a grating and is the wavelength of either ofthem or mean wavelength, then /d is known as the resolving

powerof 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 while nth principal maximum due

to wavelength ( + d) be at P2.

n

dn P1

P2

Central Image

A

B

X

Y


55/73

55

Now according to Rayleighs criterion of resolution the two

wavelengths can be resolved if the position of P2 coincides with

the first minimum due to wavelength.

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

wavelength ( + d) and the solid line represents the diffractionpattern due to wavelength .

Now the principal maximum of wavelength in a direction n

is given by

PU nde n ! sin)( ...........(1)

And the equation of minimum is given by

PU mdeN

n ! sin)( .........(2)Where m has all integral values except 0, N, 2N..........nN, because for these values ofm the condition for maxima is

satisfied.

Therefore, the first minimum adjacent to nth principal

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

m=nN+1.

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

PUU )1()sin()( ! nNddeN nn ..........(3)And the principal maximum of ( + d ) in the direction

)( nn dUU may be given as

)()sin()( PPUU dndde nn ! ............(4)

or )()sin()( PPUU dnNddeN nn ! ..........(5)comparing (3) and (5), we get

)()1( PPP dnNnN !

or nNd !PP / ......(6)also from eq. (1) we have


56/73

56

P

U ndensin

!

Hence (6) can be modified as

P

U

PPndeN

d

sin

/

! .......(7)

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

Resolving Power OfA 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 isknown 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. Thusthe resolving power of telescope is defined as the reciprocalof

the smallestangle subtendedatthe objective bythe two distant

objects, the images ofwhich are justseen asseparate onesthroughatelescope.


57/73

57


y 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.

y The image of each point is a Fraunhofer diffraction pattern.According to Rayleigh, these two images are said to beresolved if the position of the central maximum of the

second image coincides with the first minimum of the first

image and vice versa.

y The path difference between the secondary waves travelingin the directions AP1 and BP1 is zero and hence they

reinforce with one another at P1.

y Similarly, all the secondary waves from the correspondingpoints between A and B will have zero path difference.

y Thus, P1 corresponds to the position of the central maximaof the first image.

d

O

d

A

B

C

P1

P2

AB = a


58/73

58

y The secondary waves traveling in the directions AP1 andBP2 will meet at P2 on the screen.

y Let the angle P2AP1 be d. The path difference between thesecondary waves travelling in the directions BP2 and AP1

is BC.y From fig. ABC,

UUU daABddABBC .sin !!! y If this path difference PU !da. , the position of P2

corresponds to the first minimum of the first image.

y But P2 is also the position of the central maximum of thesecond image.

y Thus, Rayleighs condition of resolution is satisfied ifPU !da.

ora

dP

U ! ............(1)

y The whole aperture AB can be considered to be made up oftwo halves A

Oand

OB. The path difference between thesecondary waves from the corresponding points in the

halves will be2

P.

yAll the secondary waves destructively interfere with oneanother and hence P2 will be the first minimum of the first

image.

y The eq. ad PU ! holds good for rectangular apertures. Forcircular aperture, this eq., according to Airy, can be written

as

ad

PU

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


59/73

59

where 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.

yHere d refers to the limit of resolution of the telescope.The reciprocal ofd measures the resolving power of thetelescope.

PU 22.11 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 smallestseparation between the two objects when their images are just

resolved.


y In fig. MN is the aperture of the objective of a microscopeand A and B are two object points at a distance d apart.

yA and B are the corresponding Fraunhofer diffractionpatterns of the two images.

yA is the position of the central maximum ofA and B isthe position of the central maximum of B.

2A

B

A

B

O

M

N


60/73

60

yA and B are surrounded by alternate dark and brightdiffraction rings.

y The two images are said to be just resolved if the positionof the central maximum B also corresponds to the first

minimum of the image ofA.y The path difference between the extreme rays from the

point B and reaching A is given by)'()'( MABMNABN

But NA = MA

path@ MNBNdifference !

y In fig. AD is perpendicular to DM and AC is perpendicularto BN

)()( DBDMCNBCBMBN !@

ButDMAMANCN !!!

path@ DBBCdifference ! y From s ACB and ADB

EE sinsin dABBC !! and EE sinsin dABDB !!

A

B

M

OC

D


61/73

61

Path difference Esin2d!

y If the path difference PE 22.1sin2 !d yThen limit of resolution E

P

sin2

22.1!d

y Thus resolving powerPP

E22.1

.).(2

22.1

sin21..

AN

dPR !!! , for self-

luminous object.

y If a medium of refractive index is introduced between theobject and objective then,

PP

EQ22.1

.).(2

22.1

sin21..

AN

dPR !!! .

Resolving Power OfA Prism:Resolving power of prism is

defined as its ability to form separate images of two

neighbouring wavelengths.

aA

d

S

L1

L2

I1

I2

t


62/73

62


y Let S is a source of light; L1 is a collimating lens and L2 inthe telescope objective.

yAs the two wavelengths, and + d are very close, if the prism is set in the minimum deviation position it wouldhold good for both the wavelengths.

y The final image I1 corresponds to the principal maximumfor wavelength and image I2 corresponds to the principal

maximum for wavelength +d.

yI1 and I2 are formed at the focal plane of the telescopeobjective L2.

y The face of the prism limits the incident beam to arectangular section of width a.

yHence, the Rayleigh criterion can be applied in the case ofa rectangular aperture.

y In the case of diffraction at a rectangular aperture, theposition ofI2 will corresponds to the first minimum of the

image I1 for wavelength providedPH !ad

ora

dP

H ! ......................(1)

Here is the angle of minimum deviation for wavelength .

y From fig. THEE ! A

! 22

HTE

A

!@22

sinsinHT

EA


63/73

63

or

!2

cossinH

EA

But

!l

aEsin

l

aA!

@2

cos H ............(2)

Alsol

tA

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

yIn the case of a prism 2sin

2sin

A

A H

Q

!

2sin

2sin

AAQ

H!

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

Here and are dependent on wavelength of light .

y Differentiating eq. (4) w.r.t.

!

2

sin

2

cos

2

1 A

d

d

d

dA

P

Q

P

HH

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

!

l

t

d

d

d

d

l

a

22

1

P

Q

P

H

orP

Q

P

Hd

dt

d

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

substituting the values ofd from eq. (1)

P

QP

P

d

dt

d.!

y The expressionP

P

dmeasures the resolving power of the

prism.


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64

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 polarisedwaves and the phenomenon is aspolarisation.

Unpolarised Light: The light having vibrations along all

possible straight lines perpendicular to the direction of

propagation of light is known as UnpolarisedLight.

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

one-sidedness, is calledpolarisedlight. 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 polarisedwhen 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 circularlypolarised when the vibrations in transverse plane are circular


65/73

65

having constant period. Here the amplitude of vibrations

remains constant but the direction changes only.

3.Elliptically Polarised Light: Light is said to be elliptically polarised when the vibrations are elliptical and having 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 ofvibration and a

plane perpendicular to plane of vibration in which no vibration

occurs is known asplane ofpolarisation.


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Plane of Vibration

Plane of Polarisation

Brewsters law:In 1811, Brewster discovered a simple relation

between angle of polarisation and refractive index of themedium. He found that the refractive index of the material

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

pitan!Q ........(1)

ABCDEFGH

ipip

r

X YO

A

C

B

A

B C

D

E F

G H


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Thus when angle of incidence becomes equal to the angle of

polarisation, then from Snells law,

r

ip

sin

sin!Q ........(2)

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

or rip sin)90sin( !

or0

90! pir Also from fig. if be the angle between reflected and refracted

rays then,0

9090180)(180 !!!p

irUTherefore, 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.

re

ro

E-Ray

O-Ray

Incident

Ray


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2.When Unpolarised light is passed through uniaxial crystal, itsplits up into ordinary and extraordinary light.

3.If by some optical method one of the two rays is eliminatedthe ray, emerging through the crystal will be plane polarised.

4.In Nicol prism O- ray is eliminated by total internal reflectionand E- ray is transmitted through the crystal.

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 theacute angle (71

0) 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 backwith a transparent substance known as Canada balsam, therefractive index of which is midway forO and Eray.

5.The two end faces of the crystal are left open while the otherfaces are coated with lampblack.

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

68071

0

140

140

OOAA

DP

S0

SE

S


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Action: When Nicol crystal is used for, producing polarised

light is known aspolariserand 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 thecrystal 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.

3.Since this layer offers a rarer medium to O-ray comingfrom a denser medium of calcite, the O- ray suffers total

internal reflection.

4.This totally reflected ray is absorbed by the lampblackcoating.

POLARISAER

O- Ray

E- Ray

E-

POLARISAERANALYSER

ANALYSER

E- Ray

O- Ray

E- Ray


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5.Here the E-ray emerges out of the crystal and in this waythe crystal produces plane polarised E-ray and thus known

aspolariser.(b).Nicol as a Analyser:

1.When the two Nicol prisms are arranged coaxially thenthe first Nicol act as a polariser and produces plane

polarised E-ray.

2.The ray emerging from polariser falls on the second nicolprism known as analyser.

3.When the principal section of the two nicols are parallelto 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 intensityof transmitted ray from the second nicol decreases.

5.The intensity becomes zero when principal axis of boththe nicol become perpendicular to each other. Hence thesecond 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 /2 and path difference of /4 between

emerging ordinary and extraordinary rays is known as Quarter

Wave Plate.If 0 and e 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

)(P

QQ !! teo


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or)(4 eo

tQQ

P

! (3 for negative crystal eo QQ " )

and for positive crystal (quartz)4

)(P

QQ !! toe

or)(4 oe

tQQ

P

! (3 for positive crystal oe QQ " ).

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 and path difference of /2

between emerging ordinary and extraordinary rays is known asHalfWave Plate.

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

is

2)(

PQQ !! teo

or )(2 eot QQ

P

!

and for negative crystals like quartz ( eo QQ )

)(2 oet

QQP

!


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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.

END

Unit2 Wave Optics

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Transcript of Unit2 Wave Optics