APPLIED PHYSICS(R20)

1. Define principle of superposition? Explain phenomenon of interference?

According to Young's principle of superposition, if two or more waves are

traveling and overlap on each other at any point then the resultant displacement
of wave is sum of the displacements of individual waves. If two waves are
represented by

=> Y1 = A1sinωt & Y2 = A2sin(ωt + Φ) or Y2 = A2sin(ωt+δ)

According to superposition principle, the resultant wave amplitude is

Y = Y1 + Y2 => Y = A1sinωt + A2sin(ωt+δ)

The resultant amplitude is (A)2 = (A1)2 + (A2)2 + 2A1A2Cosδ ------> (1)

Interference: when two light waves of some frequency, nearly same amplitude
and having constant phase difference travel and overlap on each other, there is a
modification in the intensity of light in the region of overlapping. This
phenomenon is called interference. The resultant wave amplitude depends on
phase difference between the individual waves.

Condition for Maxima or bright fringes:

From (1) => (A)2 = (A1)2 + (A2)2 + 2A1A2Cosδ

If Cosδ = +1 then δ = 2nπ where n = 0, 2π, 3π ..........

Therefore => (A)2 = (A1)2 + (A2)2 + 2A1A2

=> (A)2 = (A1 + A2)2

=> A = (A1 + A2)

In terms of path difference,

=> ∆ = (λ/2π)×δ = nλ

Where n = 1, 2, 3, ............

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

Condition for minima or dark fringes:

If Cosδ = -1 then δ = (2n--1)π where n = π, 3π, 5π, ...........

From (1) => (A)2 = (A1)2 + (A2)2 -- 2A1A2

=> (A)2 = (A1 -- A2)2

=> A = (A1 -- A2)

In terms of path difference,

=> ∆ = (λ/2π)×δ = (2n--1)λ/2

Where n = 1, 2, 3, ............

2. Explain the formation of colours in thin film?

Everyone is familiar with the brilliant colours exhibited by a thin oil film spread
on the surface of water and also by a soap bubble. These colours are due to
interference between light waves reflected from the top and the bottom surfaces
of thin films. When white light is incident on a thin film, the film appears
coloured and the colour depends upon the thickness of the film and also the
angle of incidence of the light.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

3. Explain the phenomenon of interference in thin films? (Or) derive the

equation of cosine law?

Consider a transparent thin film of uniform thickness t and its refractive index µ
bounded by two plane surfaces K and K′ as shown in figure.

A ray of monochromatic light AB incident

on the surface K of the film is partly
reflected along BC and partly refracted
into the film along BD. At the point D on
the surface K′, the ray of light is partly
reflected along DE and partly transmitted
out of the film along DG. The reflected
light then emerges into air along EF which
is parallel to BC. The ray EH after
refraction at H, finally emerges along HJ.

BC and EF are reflected rays parallel to each other and DG and HJ are
transmitted rays parallel to each other. Rays BC and EF interfere and similarly
the rays DG and HJ interfere.

Interference due to the Reflected Beam:

EM is drawn normal to BC from E. Now the path difference between the waves
BC and EF is

=> ∆ = (BD+DE)in film – (BM)in air

We know, that a distance in air is numerically equal to µ times the distance in

medium

=> ∆ = µ (BD + DE) – BM

From the figure, it is clear that BD = DE

=> ∆ = (2µ . BD) – BM ------->(1)

From the figure ∆ BME, Sin(i) = BM/BE => BM = BE.Sin(i)

We know that µ = (Sin i / Sin r), BM = µ.BE sin r -------->(2)

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

From the figure ∆ BDL, Sin(r) = BL/BD = ((1/2)BE)/(BD)

=> BE = 2(BD)sin(r)

From (2) =>BM = 2µ(BD)(Sin r)2

From (1) => ∆ = (2µ . BD) – 2µ(BD)(Sin r)2

=> ∆ = (2µ.BD)(1–Sin2r) = (2µ.BD)Cos2r

From figure, ∆ BDL, Cos r = DL/BD = t/BD

Therefore, ∆ = 2µtCos(r)

This is known as cosine law.

A ray of light travelling in air and getting reflected at the surface of a denser
medium, undergoes an automatic phase change of π (or) an additional path
difference of λ/2.

Since the reflection at B is at the surface of a denser medium, there is an

additional path difference λ/2 .

The effective path difference in this case, ∆ = 2µt cos r + λ/2

(i) For the constructive interference, path difference ∆ = nλ, where n = 0,1,2,3
and the film appears bright

=> 2µt cos r + λ/2 = nλ

So, 2µt cos r = (2n-1) λ/2

(ii) For the destructive interference, path difference ∆ = (2n+1) λ/2 where n = 0,
1, 2, 3 … and the film appears dark.

=> 2µt cos r + λ/2 = (2n+1)λ/2

=> 2µt cos r = nλ

If light is incident normally i = 0 and hence r = 0. Therefore the condition for

bright fringe is 2µt = (2n–1)λ/2 and for dark fringe is 2µt = nλ.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

Interference due to the Transmitted Light:

The path difference between the transmitted rays DG and HJ is, in a similar way,
∆ = 2µt cos r. In this case there is no additional path difference introduced
because both reflections at the point D and E take place backed by rarer medium.

Hence, condition for brightness is 2µt cos r = nλ and condition for darkness is
2µt cos r = (2n – 1) λ/2.

Conditions for interference:

1) For producing stable pattern, the two sources must have nearly the same
frequency.
2) For clear pattern, the two sources must have similar amplitude.
3) For producing interference pattern, coherent sources are required.

4. Define coherence and types?

Two sources are said to be coherent sources if they produce two waves having
same frequency, same waveform and have constant phase different between
them which does not change with time.

Coherence is an ideal property of waves that enables stationary (i.e. temporally

and spatially constant) interference.

There are two types of coherence - spatial and temporal coherence.

Spatial coherence is a measure of the correlation between the phases of a light

wave at different points at right angles to the direction of propagation. Spatial
coherence tells us how uniform the phase of the wave front is.

Temporal coherence is a measure of the correlation between the phases of a

wave at different points along the direction of propagation or the predictable
relationship between signals observed at different moments in time. Temporal
coherence tells us how monochromatic a source is.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

5. Describe with theory of Newton rings experiment to determine wave

length of monochromatic source of radiation or light?

When a parallel beam of monochromatic light is incident normally on a

combination of a Plano-convex lens L and a glass plate G, as shown in Fig.1, a
part of each incident ray is reflected from the lower surface of the lens, and a
part, after refraction through the air film between the lens and the plate, is
reflected back from the plate surface. These two reflected rays are coherent;
hence they will interfere and produce a
system of alternate dark and bright rings
with the point of contact between the lens
and the plate as the center. These rings are
known as Newton’s ring for a normal
incidence of monochromatic light, the path
difference between the reflected rays (see
Fig.1) is very nearly equal to 2µt where µ
and t are the refractive index and thickness
of the air-film respectively. The fact that
the wave is reflected from air to glass surface
introduces a phase shift of π.

Therefore, for bright fringe

=> 2µt = (2n+1) λ/2 ; n = 0, 1, 2, .......------->(1)

and for dark fringe => 2µt = nλ ; n = 0, 1,

2, .......-------->(2)

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

From Equation (10), we will find wavelength of monochromatic light.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

6. Define diffraction& Compare Fresnel and Fraunhofer classes of

diffraction?

Diffraction: When waves encounter obstacles or openings like slits, they bend
round the edges. This bending of wave is called diffraction. Diffraction is the
effect produced by the limiting part of the wavefront. Smaller is the width of the
slit, more will be diffraction for given wavelength. It is also found that if the
wavelength and the width of the slit are so changed that ratio (λ/d) remains
constant, amount of bending or diffraction does not change. If ratio λ/d is more,
then more is the diffraction.

Compare Fresnel and Fraunhofer classes of diffraction:

Fresnel Diffraction Fraunhofer Diffraction

1. If the source of light and screen are 1. If the source of light and screen are
at finite distance from the obstacle, at infinite distance from the obstacle
then the diffraction is referred to as then the diffraction is referred to as
Fresnel Diffraction. Fraunhofer diffraction.

2. Fresnel diffraction patterns occur on 2. Fraunhofer diffraction patterns

flat surfaces. occur on spherical surfaces.

3. To obtain Fresnel diffraction, zone 3. To obtain Fraunhofer diffraction,

plates are used. diffraction grafting is used.

4. In Fresnel diffraction, incident wave 4. In Fraunhofer diffraction, incident

fronts are spherical. wavefronts are plane.

5. In Fresnel diffraction, the convex 5. In Fraunhofer diffraction, the

lens is not required to converge the convex lens is required to converge the
spherical wave fronts. wave fronts.

6. The intensity of Fresnel diffraction 6. The intensity of Fraunhofer

is varying. diffraction remains constant.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

7. Describe Fraunhofer diffraction due to single slit? Draw intensity

distribution graph?

The adjacent figure represents a narrow-slit AB of width ‘e’. Let a plane wave
front of monochromatic light of wavelength 'λ' is incident on the slit. Let the
diffracted light be focused by means of a convex lens on a screen. According to
Huygens Fresnel, every point of the wave front in the plane of the slit is a
source of secondary wavelets. The secondary wavelets traveling normally to the
slit i.e., along OPo are brought to focus at Po by the lens. Thus, Po is a bright
central image. The secondary wavelets traveling at an angle ‘Θ’ are focused at a
point P1 on the screen.

The intensity at the point P1 is either minimum or maximum and depends upon
the path difference between the secondary waves originating from the
corresponding points of the wave front.

In order to find out the intensity at P1, draw a perpendicular AC on BR.

The path difference between secondary wavelets from A and B in direction ‘Θ’
is BC i.e.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

∆ = BC = ABsinΘ = esinΘ

So, the phase difference δ = (2π/λ)∆ = (2π/λ)(esinΘ)

Let us consider that the width of the slit is divided into ‘n’ equal parts and the
amplitude of the wave from each part is ‘a’.

So, the phase difference between two consecutive points

δ = (2π/λn)(esinΘ)

Then the resultant amplitude R is calculated by using the method of vector

addition of amplitudes.

The resultant amplitude of 'n' number of waves having same amplitude 'a' and
having common phase difference of 'δ' is

R = a.sin(nδ/2)/sin(δ/2)

Substituting the value of δ in above equation

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

R = a.sin((2π/λ)(esinΘ)/2)

sin((2π/λn)(esinΘ)/2)

Substituting α = (π/λ)(esinΘ) in above equation

R = a.[sinα/sin(α/n)]

As α/n is small, sin(α/n) = α/n

Now R = na(sinα/α)

Put na = A in above equation

Therefore R = A(sinα/α) -------->(1)

Therefore, the Intensity is given by

I = R2 = [A(sinα/α)]2 --------->(2)

Case (i): Principal Maximum:

Eq(1) takes maximum value for α = 0

α = (π/λ)(esinΘ) = 0

sinΘ = 0 => Θ = 0

The condition Θ = 0 means that this maximum is formed by the secondary

wavelets which travel normally to the slit along OPo and focus at Po. This
maximum is known as “Principal maximum”.

Case (ii): Minimum Intensity positions:

Eqn (1) takes minimum values for sinα = 0 .

sinα = sin(+nπ)

α = + nπ

(π/λ)(esinΘ) = + nπ

esinΘ = + nλ -----------(3)

Where n = 1, 2, 3,………..

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

In eqn(3) n =0 is not applicable because corresponds to principal maximum.

Therefore, the positions according to eqn (3) are on either side of the principal
maximum.

Case (iii): Secondary maximum:

In addition to principal maximum at α = 0, there are weak secondary maxima

between minima positions. The positions of these weak secondary maxima can
be obtained with the rule of finding maxima and minima of a given function in
calculus. So, differentiating eqn (2) and equating to zero, we have

(dI/dα) = 0

We get α = tanα ---------(4)

The values of 'α' satisfying the eqn (4) are obtained graphically by plotting the
curves y = tanα & y = α on the same graph. The points of intersection of the
two curves gives the values of α is

α = 0, +3π/2, +5π/2, .................., +(2n+1)π/2

But α = 0, gives principal maximum, substituting the values of α in eq(2), we

get I1 = ((A)^2)/22, I2 = ((A)^2)/62, I3 = ((A)^2)/125, .................

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

Intensity distribution graph:

8. Describe Fraunhofer diffraction due to double slit? Draw the intensity

distribution graph?

The double slits have been represented as A1B1 and A2B2 in Fig... The slits are
narrow and rectangular in shape. The plane of the slits is perpendicular to plane
of the paper. Let the width of both the slits be equal and it is ‘e’ and they are
separated by opaque length ‘d’. A monochromatic plane wave front of wave
length ‘λ’ is incident normally on both the slits.

Light is made incident on arrangement of double slit. The secondary wavelets

travelling in the direction of OP0 are brought to focus at P0 on the screen SS′ by

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

using a converging lens L. P0 corresponds to the position of the central bright

maximum. The intensity distribution on the screen is the combined effect of
interference of diffracted secondary waves from the slits.

The intensity at point P1 on the screen is obtained by applying the Fraunhofer

diffraction theory at single slit and interference of diffracted waves from the two
slits. From the single slit diffraction,

2α = (2π/λ) (esinΘ) or α = (π/λ) (esinΘ)

The diffracted wave amplitudes, A(sinα/α) at the two slits combine to produce
interference. The path difference between the rays coming from corresponding
points in the slits A1B1 and A2B2 can be found by drawing a normal from A1 to
A2R. A2D is the path difference between the waves from corresponding points
of the slits.

In the triangle A1A2D,

sinθ = (A2D/A1A2)

or the path difference A2D = A1A2 sin θ = (e + d) sinθ

The corresponding phase difference is

2β = (2π/λ)(e+d)sinΘ

Applying the theory of interference on the wave amplitudes A(sinα/α) at the

two slits gives the resultant wave amplitude (R).

R = 2A(sinα/α)cosβ

The intensity at P1 is

I = R2 = 4(A)2[(sinα/α)cosβ]2

Put Io = 4A2

I = 4I0(sinα/α)2cos2β ---------->(1)

This equation represents the intensity distribution on the screen. The intensity at
any point on the screen depends on α and β. The intensity of central maximum
is 4I0.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

In equation (1) the term cos2β corresponds to interference and (sinα/α)2

corresponds to diffraction. Now, we will shall look at the conditions for
interference and diffraction maxima and minima.

Interference maxima and minima: If the path difference A2D = (e + d) sinθn

= ± nλ where n = 1, 2, 3… then ‘θn’ gives the directions of the maxima due to
interference of light waves coming from the two slits. The ± sign indicates
maxima on both sides with respect to the central maximum. On the other hand if
the path difference is odd multiples of λ/2.

i.e., A2D = (e + d) sinθn = ± (2n--1)(λ/2), then θn gives the directions of minima

due to interference of the secondary waves from the two slits on both sides with
respect to central maximum.

Diffraction maxima and minima: If the path difference B1C = esinθn = ± nλ,
where n = 1, 2, 3… then θn gives the directions of diffraction minima. The ±
sign indicates minima on both sides with respect to central maximum. For
diffraction maxima esinθn = ± (2n--1)(λ/2) is the condition. The ± sign indicates
maxima on both sides with respect to central maximum. The intensity
distribution on the screen due to double slit diffraction is shown in Fig. Fig (a)
represents the graph for interference term, Fig (b) shows the graph for
diffraction term and Fig(c) represents the resultant distribution.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

9. Differences between Diffraction and Interference?

INTERFERENCE DIFFRACTION

1. Interference is due to the interaction 1. Diffraction is due to the interaction

of light coming from the two different of light coming from different parts of
wave fronts originating from the same the same wave front.
source.
2. Diffraction fringes are not of the
2. Interference fringes are of the same same width.
width.
3. All bright fringes are not of the
3. All bright fringes are of the same same intensity.
intensity.
4. All points of minimum intensity are
4. All points of minimum intensity are not perfectly dark.
perfectly dark.
5. The spacing between the fringes is
5. The spacing between the fringes is not uniform.
uniform.
6. Diffraction phenomenon requires an
6. Interference phenomenon doesn't obstacle.
require an obstacle.
7. Diffraction changes the path of the
7. Path of the wave stays intact after incident wave.
interference.

10. Define the phenomenon of polarization?

Polarization: The phenomena of reflection, refraction, interference, diffraction

is common to both transverse waves and longitudinal waves. But the transverse
nature of light waves is demonstrated only by the phenomenon of polarization.

Unpolarized light: “In a beam of light, if the oscillations of E vectors are in all
direction in a plane perpendicular to the direction of propagation, then the light
is called unpolarized light”.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

If in beam of light all electric vector (E) are coplanar and parallel to each other
is plane polarized light.
Process by which getting the plane polarized light from unpolarized light is
called polarization.

“The plane containing the direction of the beam and the direction of oscillation
of E vectors is called the plane of oscillation. In figure abcd is the plane of
oscillation".

“A plane perpendicular to the plane of oscillation and passing through the beam
of light is called the plane of polarization".

In figure lmno is the plane of polarization.

When light passes through tourmaline crystal, freely transmit the light
components which are polarized to a definite direction. While crystal absorbs

light strongly whose polarization is perpendicular to this definite direction. Thus

emergent beam of light only coplanar and parallel E vectors are found. This
definite direction in a crystal is known as an optic axis.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

11. State and prove law of malus?

According to malus, when completely plane polarized light is incident on the

analyzer, the intensity I of the light transmitted by the analyzer is directly
proportional to the square of the cosine of angle between the transmission axes
of the analyzer and the polarizer.

i.e I α cos2θ

Suppose the angle between the transmission axes of the analyzer and the
polarizer is θ. The completely plane polarized light form the polarizer is
incident on the analyzer. If E0 is the amplitude of the electric vector transmitted
by the polarizer, then intensity I0 of the light incident on the analyzer is

I α E02

The electric field vector E0 can be resolved into two rectangular components i.e.
E0 cosθ and E0 sinθ. The analyzer will transmit only the component (i.e E0 cosθ)
which is parallel to its transmission axis. However, the component E0sinθ will
be absorbed by the analyzer. Therefore, the intensity I of light transmitted by
the analyzer is,

I α ( E0 x cosθ )2

I / I0 = ( E0 x cosθ )2 / E02 = cos2θ

I = I0cos2θ
K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

Therefore, I α cos2θ. This proves law of malus.

When θ = 0° (or 180°), I = I0 cos20° = I0 That is the intensity of light

transmitted by the analyzer is maximum when the transmission axes of the
analyzer and the polarizer are parallel.

When θ = 90°, I = I0 cos290° = 0 that is the intensity of light transmitted by the

analyzer is minimum when the transmission axes of the analyzer and polarizer
are perpendicular to each other.

12. State and prove Brewster’s law?

Polarizing angle: When the light is allowed to be incident at a particular angle,

(for glass it is 57.5°) the reflected beam is completely plane polarized. The
angle of incidence at which the reflected beam is completely plane polarized is
called the polarizing angle (iP).

Brewster’s Law: Sir David Brewster conducted a series of experiments with

different reflectors and found a simple relation between the angle of polarisation
and the refractive index of the medium. It has been observed experimentally
that the reflected and refracted rays are at right angles to each other, when the
light is incident at polarizing angle.

From Fig, iP + r = 90º

r = 90° – iP

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

From Snell’s law, (sin iP/sin r) = μ -----------> (1)

Here μ is the refractive index of the medium (glass)

Substituting r in equation (1), we get,

sin iP / sin (90° – iP) = μ

So, sin iP / cos iP = μ

Thus, μ = tan(iP)

The tangent of the polarizing angle is numerically equal to the refractive index
of the medium.

13. Explain the phenomenon of double refraction?

Bartholinus discovered that when a ray of unpolarised light is incident on a

calcite crystal, two refracted rays are produced. This phenomenon is called
double refraction as shown in figure (a).

Hence, two Double Refraction images of a single object are formed. This
phenomenon is exhibited by several other crystals like quartz, mica etc.

When an ink dot on a sheet of paper is viewed through a calcite crystal, two
images will be seen in figure. On rotating the crystal, one image remains
stationary, while the other rotates around the first. The stationary image is
known as the ordinary image (O), produced by the refracted rays which obey

the laws of refraction. These rays are known as ordinary rays. The other image
is extraordinary image (E), produced by the refracted rays which do not obey
the laws of refraction. These rays are known as extraordinary rays.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

Inside a double refracting crystal, the ordinary ray travels with same velocity in
all directions and the extra ordinary ray travels with different velocities along
different directions.

A point source inside a refracting crystal produces spherical wavefront

corresponding to ordinary ray and elliptical wave front corresponding to
extraordinary ray.

Inside the crystal there is a particular direction in which both the rays travel
with same velocity. This direction is called optic axis. The refractive index is
same for both rays and there is no double refraction along this direction.

14. Explain the construction and working of Nicol prism? How it works to
analyze plane polarized light?

Geometry of Nicol prism:

Optic Axis: It is a direction inside the crystal such that the single ray of light
does not split into O-ray and E-ray as their velocities are equal. Double
refraction is not observed along optic axis.

Principal plane and Principal section: A plane containing the optic axis and
perpendicular to the opposite faces of the crystal called principal plane. Section
of a crystal along principal plane is called principal section of the crystal.

Construction of Nicol Prism: A calcite crystal (CaCO3) of length 3 times its

breadth is cut along the proper direction. Its principal section is a parallelogram
of 71° and 109° angles. The two pieces are ground so that the angles of
principal section are changed to 68° and 112°. As shown in the figure.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

The two pieces are joined together along diagonal AD by sticky liquid known as
Canada balsam. The side faces are blackened.

Working: Its working is based on a phenomenon of double refraction. In

double refraction an incident ray splits into two component rays called as
ordinary ray (O- ray) and extra ordinary ray (e-ray).

O-ray has vibrations perpendicular to principal section. Refractive index of

calcite for O-ray is 1.658 and Refractive index of Canada balsam is 1.55. Thus
for O-ray the calcite crystal is denser medium. It is totally reflected by Canada
balsam and absorbed by blackened surfaces.

E ray has vibration in the plane of principal section. Refractive index of the
crystal for E-ray is 1.486. Thus, for E-ray the Calcite behaves as rarer medium.
So, it is transmitted by Canada balsam and finally emerges out of the crystal as
polarized beam of light.

Application:

Nicol as Analyzer: Two nicols N1 and N2 are arranged in the line of incident
ray. N1 is polarizer and gives out a plane polarized light.

(i) Two nicols are parallel, i.e. their diagonals of principal sections are parallel.
Thus, intensity is maximum for emerging light.

(ii) N1 and N2 are crossed. The principal section of N1 and N2 are

perpendicular. E-ray becomes O-ray as its vibration are perpendicular to
principal plane of N2. So the ray is totally reflected and absorbed by side faces.
Thus, intensity is zero for emerging light.

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.

 APPLIED PHYSICS(R20)

15. What are wave plates or retardation plates? Types?

A retardation plate, also known as plate retarders or wave plates, is an

optically transparent birefringent crystal which resolves a beam of
unpolarized light into two orthogonal components (ordinary light rays and
extra ordinary light rays); change the relative phase difference between the
components; then recombines the components into a single beam with new
polarization characteristics. These plates are very useful to produce different
kind of polarized light and convert one type of polarized light to other. The
retardation plates are mainly of two types as follows.

Quarter Wave Plate:

A retardation plate of such a thickness that it produces a path difference of
λ/4 (quarter of the wavelength of incident light) or a phase difference π/2
between the two components (ordinary light rays and extra ordinary light rays)
of incident light beam passing through it, is known as quarter wave plate.
Hence, for a quarter waveplate

Path difference = (µo~µe) x thickness

λ/4 = (µo~µe) x t

t = λ/[4(µo~µe)t]

Half Wave Plates:

If the thickness of the retardation plate is such that it produces a path

difference of λ/2 (half of the wavelength of incident light) or a phase
difference π between the two components (E-ray and O-ray) of incident light
beam passing through it, then this retardation plate is known as half wave
plate. Hence, for a half wave plate

Path difference = (µo~µe) x thickness

λ/2 = (µo~µe) x t

t = λ/[2(µo~µe)t]

K LOKESWARARAO, MSc, CSIR-UGC NET, Lecturer in physics,

RCE, Eluru - 534007.