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PHASE DIFFERENCE BETWEEN e-RAY AND o-RAY
Natural light incident on the surface of an
anisotropic crystal undergoes double refraction
and produces two plane polarized waves namely eray and o-ray.
The two waves travel along the same direction in
the crystal but with different velocities. v0 < ve
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PHASE DIFFERENCE BETWEEN e-RAY AND o-RAY
2
PHASE DIFFERENCE BETWEEN e-RAY AND o-RAY
As a result, when the waves emerge from the rear face of the
crystal, an optical path difference would have developed between
them.
 The optical path difference can be calculated as follows
Let d be the thickness of the crystal.
The optical path for o-ray
within the crystal
The optical path for e-ray
within the crystal
The optical path difference
between e-ray an o -ray
= µod
= µed
∆ = = (µe - µo) d
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PHASE DIFFERENCE BETWEEN e-RAY AND o-RAY
Consequently, a phase difference arises between the two waves
δ
2π
μe  μo d
λ
As the two component waves are derived from the same incident
wave, the two waves are in phase at the front face and have
emerged from the crystal with a constant phase difference
Hence, it may be expected that the waves are in a position to
interfere with each other
However, as the planes of polarization of o-ray and e-ray are
perpendicular to each other, interference cannot take place
between e- and o-rays
The waves instead combine with each other to give an elliptically
polarized wave.
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Let two light waves traveling in the same direction, x; one wave is
polarized in the xy -plane and the other is polarized in xz- plane .
We are interested to know the state of polarization of the resultant
wave.
Let the two orthogonal waves be represented by
Ey = E1 cos (kx-ωt)
EZ = E2 cos (kx-ωt + δ)
(1)
(2)
The waves are of the same frequency ν= ω/2π. δ is the phase
difference between the waves.
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eq.(2) can be expanded as
EZ = E2 cos (kx-ωt) cosδ - E2 sin (kx-ωt) sin δ
(4)
= E2 cos (kx-ωt) cosδ ± [1- cos2 (kx-ωt) ]1/2 E2 sin δ
We find from eqn (1)
coskx  t  
Ey = E1 cos (kx-ωt)
Ey
E1
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
Put this value in Ez
EZ = E2 cos (kx-ωt) cosδ ± [1- cos2 (kx-ωt) ]1/2 E2 sin δ
EY
E Z  E2
cos δ 
E1
E 2y
1  2 E2 sin δ
E1
5 
2


 Ey 
E2
E y cos δ    1    E2 sin δ
E Z 
E1


 E1 
On squaring both sides, we obtain
E z2 
E 2y E22
2
cos
δ
2
E1
2E y E z E2
E1
cos δ  E22 sin2 δ 
E 2y E22
2
sin
δ
2
E1
7
Rearranging the terms, we get
2 2
E
2E y E z E2
y E2
2
2
2
E z  2 (cos δ  sin δ ) 
cos δ  E22 sin2 δ
E1
E1
E2
z 
E 2y E22

2
E1
2E y E z E2
E1
6 
cos δ  E22 sin2 δ
Dividing both the sides E22 by and rearranging the terms , we obtain
E2
y
E z2
2E y E z
2


cos
δ

sin
δ
2
2
E1 E2
E1 E2
Equation (7) is the general
equation of an ellipse.
7
 Hence , the tip of the resultant vector
traces an ellipse in the yz- plane.
Ellipse is constrained within a rectangle having sides 2E1 and 2E2.
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Special cases :1. When δ= 0, or ±2mπ, the two waves are in phase, cos δ =1 and
sin δ = 0, and the eqn.(7) reduces to
E2
y
2
E1

2
EZ
2
E2

2E y E z
E1 E2
0
2
 E y Ez 
E  E   0
2
 1
Ey
Ez

0
E1 E2
E2
 Ez 
Ey
E1
8 
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The above equation represents a straight line, having a slope  E2 
E 
 1
Therefore , the equation represents a wave having its plane of
polarization making an angle α with respect to the y-axis .
 It means that the resultant of two plane-polarised waves, which are
in phase(i.e., coherent waves), is again a plane-polarised wave.
2.When δ= π, or ± (2m + 1) π ,the two waves are in opposite phase,
then cos δ = -1 and sin δ = 0 and the eqn (7) reduces to
E y2
Ez2 2 E y Ez
 2 
0
1
E2 E2
E1 E2
2
 E y Ez 


 0
 E1 E2 
10
Ey
Ez

0
E1 E2
E2
9

Ez  
Ey
E1
This equation represents a straight line of a slope   E 2 
 E 
 1
Therefore, the equation represents a wave having its plane of
polarization making an angle tan1  E2  with respect to the y-axis
E 
 1
It means that the resultant of two plane-polarised waves, which are
in opposite phase (i.e, coherent waves), is again a plane-polarised
waves.
3. When δ= π/2, or ± (2m + 1) π/2 ,
then cos δ = 0 and sin δ = 1 and the eqn (7) reduces to
2
E2
E
y
z 1
10 

2
2
E1 E2
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This is the equation of an ellipse whose major axis and minor axis
coincide with y – and z-co-ordinate axes.
Therefore, when the two plane polarized waves are out of phase by
90o , their resultant is an elliptically polarized wave.
4. In the particular case, δ= π/2, or E1 = E2 =E0 , eqn. (7) reduces to
EY2  E z2  E02
This is the equation of circle. Hence the resultant light is circularly
polarized.
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when e-ray and o- ray overlap on each other after emerging from an
anisotropic crystal plate, they cannot produce interference fringes
as in a double slit experiment
On the other hand, they combine to produce different states of
polarisation depending upon their optical path difference
When the optical path difference is 0 or an even or odd multiple of
λ/2, the resultant light wave is linearly polarized.
When the optical path difference is λ/4, the resultant light wave is
elliptically polarised
In the particular instance when the wave amplitudes are equal and
the optical path difference is λ/4, the resultant light wave is
circularly polarised.
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QUARTER WAVE PLATE
It is a thin plate of birefringent crystal having the optic axis
parallel to its refracting faces and its thickness adjusted in
such a way that :
when a plane polarized light propagates through it , a path
difference of quarter-wave (λ/4) is introduced between the
e-ray and o-ray.
Thus, for a quarter wave plate
λ
μe  μo d 
4
λ
d
4μ e  μ o
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So a phase difference δ between e-ray and o-ray will be
2π
π
δ
   90 
λ
2
A quarter-wave plate is used in producing elliptically or
circularly polarized light.
It converts plane polarized light into elliptically or
circularly polarized light depending upon the angle that
the incident light makes with the optic axis of the quarter
wave plate.
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Action of quarter wave plate on elliptically and circularly polarised
light
consider elliptically polarised light being incident on a quarter wave
plate
Elliptically polarised light may be viewed as made up of two
coherent plane polarised waves of different amplitudes, and
differing in phase by 90°.
the quarter wave plate introduces an additional phase difference of
90° leading to a total phase difference of 180° between the two
component waves
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When they emerge out of the plate, they combine to form linearly
polarised wave, the action of the quarter wave plate on circularly
polarised light wave is similar.
Circularly polarised light incident on a quarter wave plate is
converted into linearly polarised light.
HALF WAVE PLATE
A half wave plate is a thin plate of birefringent crystal having the
optic axis parallel to its refracting faces and its thickness chosen
such that it introduces a half-wave (λ/2) path difference (or a phase
difference of 180°) between e-ray and o-ray.
17
As a result, when they emerge from the rear face of the crystal, an
optical path difference developed between them:
λ
μe  μo d 
2
λ
d
2μ e  μ o
A half wave plate introduces between e-ray and o-ray a phase
difference δ given by
2π 

δ      π  180 
 λ 
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The half wave plate rotates the plane of the incident
light through an angle 2θ.
This inverts the handedness of elliptical or circular
polarized light changing right to left and vice versa.
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PRODUCTION AND ANALYSIS OF PLANE,
CIRCULAR AND ELLIPTICALLY POLARIZED
LIGHT
 PLANE POLARIZED LIGHT
When a beam of monochromatic light is allowed to
pass through a Nicol prism, it splits up into two
parts: ordinary and extraordinary rays. During the
transmission of these rays through the prism, the
ordinary ray is totally reflected back by a Canada
balsam layer, whereas the extraordinary ray is
transmitted through the Canada balsam. Hence, the
emergent light is only the extraordinary ray, which is
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plane-polarized.
ANALYSIS OF POLARIZED LIGHT
The unknown polarized light is allowed to fall
normally on a polarizer. The polarizer is slowly
rotated through a full circle and observe the
intensity of the transmitted light. If the intensity of
the transmitted light is extinguished twice in one full
rotation of the polarizer, then the incident light is
plane polarized.
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Circularly Polarized Light
When two mutual perpendicular coherent linear
vibrations of same amplitude and period are
superimposed with phase difference π/2, the
resultant motion will be circular motion.
These can easily be obtained by allowing a beam of
monochromatic plane-polarized light to be normally
incident on a quarter wave plate.
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ANALYSIS OF CIRCULARLY POLARIZED LIGHT
If the intensity of the transmitted light remains
constant on rotation of the polarizer, then the
incident light is either circularly polarized or
unpolarized.
To distinguish between circularly polarized and
unpolarized light, we take the help of a quarter wave
plate.
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The light is first made to be incident on the quarter
wave plate and then it passes through the polarizer.
If the incident light is circularly polarized, the quarter
wave plate converts it into plane polarized light.
When this linearly plane polarized light passes
through the polarizer, it would be completely
extinguished twice in one full rotation of the
polarizer.
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If the intensity of the transmitted light stays
constant, then the incident light is unpolarized.
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 ELLIPTICALLY POLARIZED LIGHT
(Production)
Unpolarized light is first converted to plane
polarized ( Nicol prism). The plane polarized light is
then made incident on a quarter wave plate.
When this emerges out of the crystal
produces elliptically polarized light.
ANALYSIS
If the intensity of the transmitted light varies
between a maximum and a minimum value but does
not become extinguished in any position of the
polarizer, then the incident light is either elliptically
polarized or partially polarized
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To distinguish between elliptically polarized and
unpolarized light, we take the help of a quarter wave
plate
If the incident light is elliptically polarized, the
quarter wave plate converts it into a plane polarized
beam. When this linearly polarized light passes
through the polarizer, it would be extinguished twice
in one full rotation of the polarizer.
if the transmitted light intensity varies between a
maximum and a minimum without becoming zero,
then the incident light is partially polarized
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OPTICAL ACTIVITY
For a crossed Nicol position in which the polarizer
and analyzer are perpendicular no light comes out of
the analyzer. If we place quartz plate cut with faces
perpendicular to the optic axis; inbetween polarizer
and analyzer, light comes out of the analyzer.
Quartz has the property to turn the plane of
vibration. When plane polarized light enters the
quartz, its plane of vibration is rotated gradually.
The amount of rotation by which the plane of
vibration is turned depends upon the thickness of
the quartz plate and the wavelength of incident light.
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polarizer
 

Quartz plate
Two Crossed Nicol
analyser
I


Optic Axis
Rotation of vibration
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The property of turning the plane of vibration is
known as optical activity and substances that show
optical activity are known as optically active
substances.
.
Calcite does not have the property to rotate the
plane of vibration, therefore, it is not an optically
active substance.
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Substances such as quartz, sugar solution,
turpentine and cinnabar, etc. are optically
active substances.
Substances which rotate the plane of vibration in
clockwise direction or to the right are called
dextrorotatory or right-handed rotation. Righthanded rotation means that it is right-handed for an
observer.
31
A substance that rotates the plane of vibration in the
anticlockwise direction or to the left is called
levorotatory substance.
It is observed that some quartz crystals are
dextrorotatory while others are levorotatory.
The amount of rotation in a solution depends upon
the concentration of the solution.
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SPECIFIC ROTATION
The angle by which the plane of vibration rotates
depends on

the thickness of the substance

density of the material or concentration of the
solution
wavelength of light
the temperature


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The specific rotation of a solution for a given
temperature and wavelength of light is the rotation
of plane of vibration produced by one decimeter
length of solution when the concentration of
solution is one gram per c.c.
The specific rotation S is
When length is taken in cm, then
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FRESNEL’S THEORY OF OPTICAL ROTATION

Fresnel’s theory of optical rotation is based on the fact
that any plane polarized vibration may be regarded as the
resultant of two circularly polarized vibrations rotating in
opposite direction with the same velocity or frequency.
35
He assumed that a plane polarized light on entering
a crystal along optic axis is resolved into two
circularly polarized vibrations rotating in opposite
directions with the same angular velocity.
36
2.
In an optically inactive substance these two
circular components travel with the same speed
along the optic axis. Hence at emergence they give
rise to a plane polarized light without any rotation
of the plane of polarization.
37
3. In an optically active crystal, like quartz , two
circular components travel with different speeds so
that relative phase difference is developed between
them.
In dextro-rotatory substance vr>vl
and in leavo rotatory substance vl>vr
38

4. On emergence from an optically active
substance the two circular vibrations recombine to
give plane polarized light whose plane of vibration
has been rotated w.r.t that of incident light through
a certain angle.
39
So This explanation was based on the following assumptions:
1. A plane polarized light falling on an optically active medium along its
optic axis splits up into two circularly polarized vibrations of equal
amplitudes and rotating in opposite directions –one clockwise and
other anticlockwise.
2. In an optically inactive substance these two circular components
travel with the same speed along the optic axis. Hence at emergence
they give rise to a plane polarized light without any rotation of the
plane of polarization.
3. In an optically active crystal, like quartz , two circular components
travel with different speeds so that relative phase difference is developed
between them.
4. In dextro-rotatory substance vR>vL and in leavo rotatory substance
vL>vR..
5. On emergence from an optically active substance the two circular
vibrations recombine to give plane polarized light whose plane of
vibration has been rotated w.r.t that of incident light through a certain
angle depends on the phase diff between the two vibrations.
Plane polarized means resultant of R and L.
MATHEMATICAL TREATMENT


Consider a plane polarized light beam of amplitude a,
incident on a material. According to Fresnel, it splits
up in to two components ‘R’ and ‘L’.
Their components in the direction of X and Y are
•For clockwise circular vibrations ‘R’:
If angle between A and R is ωt;
a
x1  sin(t ),
2
a
y1  cos(t )
2
A
R
L
•For anti clockwise circular vibrations ‘L’:
If angle between A and L is also ωt (same angular speedO
for optically inactive material)
a
x 2   sin(t ),
2
a
y 2  cos(t )
2
y
B
x
42
Therefore the resultant vibration be
along the x axis : x  x1  x 2  0
along the y axis : y  y1  y 2  a cos(t )
Plane of vibration is along original direction.
The result shows that two oppositely directed
circular motions of equal velocity combine to give
linear motion along the direction of motion (optically
inactive material)
43
FOR OPTICALLY ACTIVE SUBSTANCES

According to Fresnel the two circular components are
propagated through the plate with different angular
speeds. So when they emerges out of the crystal there is a
phase difference  between them.

Suppose clockwise component moves faster compared to
anticlockwise component.
44
•FOR CLOCKWISE VIBRATIONS:
a
x1  sin(t   )
2
a
y1  cos(t   ) ........(1 )
2
A
For anti clockwise vibration:

2
A’
L
angle is (ωt)
R
O
a
x 2   sin t
2
a
y 2  cos t ........(2 )
B
2
The resultant displacement along the two axes are,
y  y1  y 2
x  x1  x 2
a
a
a
a
 cost     cos t
 sin(t   )  sin t
2
2
2
2



 a sin cos t   ....(i)  a cos  cos t    .......(ii )
2
2
2
2


45
•So the vibrations represented by equation (i) and (ii) are
perpendicular to each other. They have the same period
and phase.
•Therefore the resultant vibration is plane polarized and it
makes an angle /2 with the original direction.
•So plane of vibration has rotated through an angle /2 on
passing through the crystal.
•Dividing (i) by (ii) we get
sin

x

2

 tan
y cos 
2
2
This is equation of straight line inclined at /2 with
y-axis. That is with the vibrations of incident light.
46

Let refractive indices of the clockwise and anticlockwise vibrations be
R and L , ‘t’ be the thickness of the quartz plate, then the optical
path difference between the two components will be
   L  R  t
•Corresponding phase difference will be

2
t L  R 

•Angle of rotation of plane of vibration will be
c c
 
  t
    L   R  t or
2 
  vL vR 
If velocities vR vL
Right handed optical active material.
•In case of left handed optically active crystals
 
    R   L  t or   c  c t
and vL  vR


2 
  vR vL 
47