Interference of Light

 Light Waves:

Fig.1

A light is a harmonic electromagnetic wave consisting of periodically

varying electric and magnetic fields, oscillating at right angles to each other and
also the direction of its propagation (Fig.1). Electric field is represented by E
and magnetic field is represented by B. As the electric vector ⃑E has much
optical effect when it passes through materials, so ⃑E vector will be discussed
here. The electric field E is represented by the equation:

E=Eo sin ( ωt−kx ) or E=Eo cos (ωt−kx) or E=Eo sin ⁡( kx−ωt )

 Phase difference and Coherence

Crest

1
No constant
phase diff.
2 Phase diff.=0
or 2π
A
3 6
Trough
(a)

4
7
Phase diff.=π

5 (c)
(b)
Fig.2
 Consider that two waves are passing through space. In Fig.2, two waves
have the same wavelength and at the same time the crest of one wave
corresponds to the crest of the other (wave 1 and 2).Then, we say that two
waves are coherent and with zero phase difference. If the wave 3 starts at point
A, then the phase difference between waves 2 and 3 is 2π. In both cases, the
waves are said to be in phase.

In Fig.2, when one wave reaches to its crest while the other falls to its
trough, the phase difference between them is π. The waves are said to be out of
phase.

In both cases, the waves maintain a constant phase difference. “If two
waves maintain a constant phase difference in their passage with time, then they
are said to be coherent.”

In Fig.2, two waves of different frequencies change their phase with time
and can never maintain a constant phase difference. These waves are said to be
incoherent.

“Sources, which produce coherent waves, are called Coherent sources.”

 Optical path and Geometrical path

Optical path length indicates the distance traversed by light in a medium at time
‘t’ as compared to that in vacuum.

Liquid
Vacuum
Consider, c= speed of light in vacuum v
c v= speed of light in medium
µ = refractive index of medium

L
L

Fig.3a Fig.3b
At time t, let the distance travelled by light in vacuum be Δ (Fig.3a), then
Δ=ct -------------------------------------------(1)
At time t, light passes a distance ‘L’ called geometrical distance, in the medium
(Fig.3b.
So, L=vt -------------------------------------------(2)
Deviding equ(1) by equ(2), we get,

Δ c
= =μ
L v
Δ=μL

Here, Δ is called optical path length and L is called geometrical path length.

 Path difference and phase difference

Consider a wave of wavelength λ. From the beginning to the end of a wave

cycle, the phase is 2π λ

2π

Fig.4

Let, L=geometrical distance in the medium

Δ=optical distance in the medium

The path λ corresponds to phase 2Δ

2Δ
,, ,, 1 ,, ,, λ
2Δ
×L
,, ,, L ,, ,, λ
 2Δ
× pathdifference
So, Phase difference= λ
λ
× phase difference
And, path difference= 2 Δ

 Superposition of waves

“When two or more waves overlap, the resultant displacement at any point and
at any instant may be found by adding the instantaneous displacements that
would be produced at any point by the individual waves if each were present
alone”.
E⃗ = ⃗E sin (ωt−kx ) E⃗ = ⃗E o2 sin(ωt−kx )
Let, 1 o1 and 2 , then the resultant wave is
⃗ =E
E ⃗ +E⃗
the sum of the individual waves, i.e. 1 2

 Interference

When two or more waves of the same frequency but different amplitudes and
phases are superimposed, the resultant wave pattern will be different from the
individual waves. The phenomenon of redistribution of light wave due to
superposition of two or more waves from coherent sources is known as
interference of light.

=
+

+ =

Fig.5
 Constructive and destructive Interferences

When two waves of the same frequency and in phase are superimposed, the
amplitude of the resultant wave will be equal to the sum of the amplitudes of the
two waves. This phenomenon is known as constructive interference (Fig.5). If
the amplitude of the resultant wave is zero, the phenomenon is known as
destructive interference (Fig.5).

For constructive interference:

E R =E+ E=2 E
E R =E−E=0
For destructive interference:

--------------------------------------------------------------------------------------------
 Huygens’s principle

Fig.6

All points on a wave front can be considered as point sources for the production
of spherical secondary wavelets. After a time t, the new position of the wave
front will be the surface of tangency to these secondary wavelets (Fig.6).
----------------------------------------------------------------------------------------------
 Theory of Interference

Consider two light waves:

E A =E1 sin (ωt−kx ) -----------------------(1)

And E B =E2 sin (ωt−kx + δ ) -------------------(2)

Where, δ is the phase angle between EA and EB. The two waves are
superimposed. According to the principle of superposition, the resultant wave’s
amplitude ER will be:
E R=E A +E B=E1 sin( ωt−kx )+E 2 sin(ωt−kx +δ )
=E1 sin(ωt−kx )+E2 [sin (ωt−kx )cos δ+cos(ωt−kx )sin δ ]
=( E1 + E 2 cos δ )sin(ωt −kx )+ E 2 sin δ cos(ωt−kx ) ----------------------(3)

Let, after superposition, the resultant wave gets new amplitude E and new phase
angle φ. Then, the wave can be represented by:
E R=E sin (ωt−kx + ϕ )
=E cos ϕ sin(ωt−kx )+E sin ϕ cos(ωt −kx ) -----------------------------(4)

Equations (3) and (4) are same. Comparing the coefficients of sin( ωt−kx ) and
cos(ωt−kx ) , we get:
E sin ϕ=E 2 sin δ -------------------------------------(5)
E cos ϕ=E1 + E 2 cos δ ------------------------------(6)
Squaring and adding equations (5) and (6), we get,
E2 ( sin2 ϕ +cos 2 ϕ )=( E1 + E2 cos δ )2 + E22 sin 2 δ
E2 =E21 +2 E 1 E2 cos δ + E 22 cos2 + E22 sin2 δ
E2 =E21 + E22 + 2 E 1 E 2 cos δ -----------------------------------------------(7)
Also, from equ(5) and (6):
E 2 sin δ
tan ϕ=
E 1 + E2 cos δ -------------------------------------(8)
From equ(8), φ can be obtained. Moreover, the intensity of light is related to the

amplitude by the relation: I ∞ E2 or I =kE 2 -------------------------------(9)

Applying this for I1 and I2 we can write:
I 1=kE12
--------------------------------(10)
I 2=kE 22
-------------------------------(11)

2 2
Introducing the value of E , E1 and E2 , in equ(7), we get:
2

I=I 1 +I 2 +2 √ I 1 I 2 cos δ ----------------------------------

(12) where cosδ is the interference term. When δ=0, I will be maximum, i.e.
I max =I 1 +I 2 +2 √ I 1 I 2 --------------------------------------(13)
If I1=I2=Io, then Imax=4Io. When δ=180, I will be minimum, i.e.
I min=I 1 +I 2 −2 √ I 1 I 2 --------------------------------------(14)
If I1=I2=Io, then Imin=0. At points that lie between the Imax and Imin, when I1=I2=Io,
I will be:
I=I o +I o +2 √ I o I o cosδ
I =I o + I o +2 I o cos δ
=2 I o (1+cos δ )
or
=4 I o cos 2 δ

δ
-5π -4π -3π -π π 3π 4π 5π

Fig.7
Fig.7 shows I versus δ graph. For incoherent waves:
I ave=I 1 +I 2 +2 √ I 1 I 2 ⟨cosδ ⟩ -----------------------------(15)
But, ⟨cosδ ⟩=0 I ave=I 1 +I 2
, So . Phase shift occurs at a rate of 108/s. Eye,
photo-film etc. cannot detect it. So we do not see any interference pattern. We
see uniform illumination.

Condition for maximum intensity: Maximum value of

δ δ
cos 2 =1 cos =±1
2 or 2
δ
cos =cos0 , cos π , cos2 π . .. .. . .. .. .. . .cos nπ
∴ 2
δ
=nπ
∴ 2 or δ =2 nπ
λ λ
x= ×δ x= ×2 nπ
Path difference, 2π or 2π
∴ x=nλ

Condition for minimum intensity: Minimum value of

δ δ
cos 2 =0 cos =0
2 or 2
δ π 3π 5π π
cos =cos , cos
, cos . .. .. . .. .. .. . .cos(2 n+1)
∴ 2 2 2 2 2
∴ δ=(2n+ 1)π
λ λ
x= ×(2 n+1)π x=(2n+ 1)
Path difference, 2π or 2
1
x=(n+ )λ
∴ 2
 Young’s double slit experiment

x
S1 E
θ θ d/2
d M O
N d/2
S2 F

D
θ

Fig.8

Fig.8 shows the basic arrangement of Young’s double slit experiment.

Monochromatic source of wavelength λ=5893Ǻ is used. S is a source,
producing cylindrical wave. Two slits S 1 and S2 are at equal distances from S.
Waves are in phase. Screen is at a distance D from the slits. M is the midpoint
of S1S2. P is an arbitrary point on the screen. θ is the angle between MP and MP.
x is the distance from O to P. The two waves from S 1 and S2 will interfere
constructively if S2N contains integral multiple of λ. The interference will result
in alternate bright and dark fringes.
d d
( x− ) ( x+ )
Now, from the figure, PE= 2 and PF= 2 .

[ d
2 ][ d
S2 P2 −S1 P2 = D2 +( x + )2 − D2 +( x− )2
2 ]
d2 d2
=D 2 + x 2 + xd + −D2 −x 2 + xd−
4 4
=2 xd
 2 xd
S2 P−S 1 P=
S 2 P+ S1 P

If S2P≈S1P=D (i.e. the point P is very near to O), then

2 xd
S2 P−S 1 P=
2D
xd
=
D
But, S2P‒S1P is the path difference between two light waves.

Bright Fringe:
The condition for bright fringe is:
S2 P−S 1 P=nλ ( where n=0, 1, 2 etc .)
xd
or =nλ
D
nλD
x=
d

Dark Fringe:
The condition for dark fringe is:
λ
S2 P−S 1 P=(2n+1 ) (where n=0, 1, 2 etc. )
2
xd λ
or =(2 n+1 )
D 2

Fringe Separation:
The nth fringe occurs when:
nλD
x n=
d
The (n+1)th fringe occurs:
( n+1) λD
x n+1=
d
λD
β=x n+1 −x n=
So, Fringe separation: d . Fringe width is β/2.
Also, the path difference: S2 N =S 2 P−S1 P
or d sin θ=nλ ( bright fringe)
1
and d sin θ=(n+ )λ (dark fringe )
2

 Fresnel bi-prism
S1
S

S
S

(a)

S 2
30´ (b)

179°

(c)

30´
Fig.9

In Fig.9 (a), a light source S is placed in front of a prism. The light ray will pass
through the prism and will be bent at an angle called deviation angle as shown.
'
If we look the emergent beam, it seems that light comes from the source S not
'
from S. The source S is called virtual source. In Fig.9 (b), the source will
' '
appear at S1 and S2 . One source turns into two virtual sources. If we cut two
prisms from the apex to the base and join them base to base, then Fig.9 (c) will
be produced. Fig.9 (c) is called a biprism. Fresnel used a biprism to show
interference phenomenon. The biprism consists of 2 prisms of very small
 refracting angles joined base to base. A thin glass plate can be cut such that
Fig.9 (c) can be obtained.

 Fresnel bi-prism experiment

S1

d S θ O

S 2

D
Screen
Fig.10

Theory: The two light rays will interfere and will produce interference fringe as
' '
shown in Fig.10. As O is equidistance from S1 and S2 , the central fringe will be
bright and of maximum intensity. Alternate bright and dark fringes will be
observed. The fringe separation or fringe with can be obtained from the relation:

λD
β=
d
βd
or λ=
D
where β is the fringe separation, d is the distance between virtual sources, and D
is the distance from the source to the screen.

Determination of wavelength:
Measuring β, d and D, λ can be determined.

β-measurement: An eyepiece is set at the field of view (Fig.11).

Cross-wire is used to measure β. The vertical line of cross-wire
 is set at the middle of a fringe. Micrometer is used to get reading,
xO. Then, the micrometer is moved to N number of fringe. Thus,
Fig.11
a reading xN is taken. Then β can be obtained from the following:
x N −xO
β=
N

d-measurement:

S1 S1
'

d d
S S

S 2 S2
'

u v v u

d1 d2

Fig.12
A convex lens of short focal length is placed in between the prism and the eye-
piece (Fig.12). The lens is moved back and forth near the biprism till sharp
images of the slit are obtained in the field of view. The distance between the
images as seen in the field of view is d1. The distance of the slit is u and the
distance of eye-piece is v. Then,
v d
magnification= = 1
u d
The lens is moved to near the eye-piece and sharp images of the slits are
focused. The distance between the images as seen in the field of view is d2. In
this case, the object distance u will be same as the image distance v of the
previous case. And the image distance v is the same as the object distance u of
the previous case. So,
u d
magnification= = 2
v d
 d1 d2
=1
From the above two relation: d2
d= √ d1 d 2

Alternative method:

S1 α

δ
d S

θ
S 2
a

Fig.13

Consider a thin prism (Fig.13). Let the refractive index of the material of prism
be μ. In case of a thin prism, the deviation angle: δ=(μ−1)α , where α is the
θ θ d /2 d θ d
δ= tan = = or =
angle of the prism. If d is small, 2 . So, 2 a 2a 2 2 a , where a
d
=( μ−1)α or d=2 a (μ−1)α .
is the distance as shown in the figure. 2 a

 Thickness measurement of a thin film

P
Film

S1
d O
S2
D

Fig.14
Let O be at equal distance from S1 and S2 (Fig.14). Without film, the optical
distance is S1O=S2O. So, a central bright fringe will appear at O. Introducing the
film of refractive index μ and thickness t, the central bright fringe is shifted to P.
∇ =∇ S P .
So, optical path difference: S 1 P 2

or [ 1
S P+ μt −t ]=S2 P

or S2 P−S 1 P=(μ−1)t
xd
=( μ−1 )t
or D
xd
t=
So, D(μ−1 )

 Lloyd’s experiment

Q
S1
B
R P O
C

Mirror
Fig.15
S1 is a light source placed in air (Fig.15). RP is a mirror. S 1R is the incident ray
on the mirror. RA is the reflected ray from the mirror. S 1P is the incident ray
partially passing at corner P. PB is the partially reflected ray. Interference
occurs in PQAB by original rays and reflected rays from R and P. O is the point
on the screen along RPO. Point O should be bright, because incident ray from
S1 falls on O. Now, move the point O to the point O. At this point, it is
experimentally found darkness rather than brightness. What is the reason? The
reason is that there is a phase difference between incoming and reflected rays at
the point P. The phase change is π or a path difference of λ/2. Otherwise, we
could not have destructive interference.

Lloyd’s experimental results confirm us that when light ray from rarer medium
is reflected from a denser medium, there is a phase change of π occurring of the
reflected rays.
 Stokes theorem

“When light ray from denser medium is reflected from rarer medium, no phase
change occurs of the reflected light”.

Err
E Er
Er Ett´
Air

Water

Ert
Et Et
Etr´

Fig.16

Let, E be the amplitude of the incident ray (Fig.16).

r be the fraction of amplitude reflected for light from air to water
t ,, ,, ,, ,, ,, transmitted ,, ,, ,, ,, ,, ,,
r´ ,, ,, ,, ,, ,, reflected ,, ,, ,, water to air
t´ ,, ,, ,, ,, ,, transmitted ,, ,, ,, ,, ,, ,,
Er and Et amplitudes of reflected and transmitted wave
Err and Etr´ amplitudes of two sets of reflected waves
Ett´ and Ert amplitudes of two sets of transmitted waves
According to the principle of optical reversibility:
Ett´+Err = E
or tt´+r2 =1
or tt´= 1‒ r2 -------------------------------(1)
Let, I= original light intensity, I 1= reflected light intensity and I 2= transmitted
light intensity. If we apply conservation principle for light intensity, then
I1 + I2 = I or E2r2+ E2t2 = E2 or r2 +t2 =1
r2= 1‒t2. So, from equ(1), tt´= 1‒ (1‒t2) or tt´= t2
So, t=t´
From this result we can say that due to the transmission of light, no phase
change occurs. However, this result is not correct for two reasons:
(i) Although using E, we can derive t=t´ through conservation principle
for intensity, an additional factor of the index of refraction will have to
be introduced in determining the intensity of new medium. (Because,
1 1
I O= ε O cE 2 I = ε cE 2
in vacuum, 2 but in medium, 2 , where ε O andε
refer to permittivity of two media related to the refractive index of the
2 2
media. Therefore, for I2, we cannot simply write I =E t . We have to
introduce refractive index.)
(ii) Law of conservation cannot be applied to intensity but to energy,
momentum etc. Again we can write:
'
Ert+ Et { r =0 ¿
'
or r =−r --------------------------------(2)
'
or r =r cos 180
The meaning of equation (2) is
If the phase of Er is 0, the phase of Err is 180°. Then the phase of Et is 180°.
Now according to equ (2) if Er has 0 phase, then Ert has 180° phase. So the
phase difference between phase Et and Etr´ is 0. That is, no phase change occurs
when light from denser medium is reflected from the rarer medium.

Problem1: A biprism is placed 5 cm from a slit illuminated by sodium light

(λ=5890Ǻ). The separation of the fringes as viewed on a screen 75 cm from the
biprism is 9.424×10‒2 cm. What is the distance between two coherent sources?

Solution: Here, λ=5890×10‒8cm, β=9.424×10‒2 cm, D=(5+75) cm=80 cm

 λD λD
β= or d=
d β
−8
5890×10 ×80
d=
9 . 494×10−2
= 0 . 05 cm

Problem 2: In a biprism experiment with sodium light, bands of width 0.0195

cm are observed at 100 cm from the slit. On introducing a convex lens 30 cm
away from the slit, two images of the slit are seen 0.7 cm apart, at 100 cm
distance from the slit. Calculate the wavelength of sodium light.

Solution: Here, β=0.0195 cm, D=100 cm

λD βd
β= or λ=
d D
I v
= and u+v=100 cm
For convex lens, O u
Image size, I=0.7 cm, slit distance from lens, u= 30 cm
0 .7 70
=
So, v=(100‒30) cm= 70 cm, Now O 30
O=0. 3 cm That is, d= 0.3 cm
0 . 0195×0 . 3
λ=
100
= 5850×10 -8 cm
o
=5850 A

Problem3: Interference fringes are observed with a biprism of refracting angle

1º and refractive index 1.5 on a screen 80 cm away from it. If the distance
between the source and the biprism is 20 cm, calculate the fringe width when
wavelength of light used is 6900Å.

Solution: Here, μ=1.5, α=1º =π/180º, a=20 cm, b=80 cm, So

D=(20+80)cm=100cm and λ=6900×10‒8 cm.
 λD
β=
We know, d=2(μ−1)αa , and d
λD
=
2( μ−1 )αa
6900×10−8 ×100
=
π
2(1 .5−1)× ×20
180 =0.02 cm.
Problem 4: The Young’s double slit interference experiment is performed with
light from a mercury vapor lamp so filtered that only the strong green line of
wavelength 5460Å is effective. The slits are 0.10 mm apart, and the screen on
which the interference pattern appears is 20 cm away. What is the angular
position of the first minimum and of the tenth minimum?

Solution: Here, λ=5460×10‒10 m, d=0.10×10‒3 m, and n=0

1
( n+ ) λ
2
sin θ=
We know, d
1
×5460×10−10
2
= =0 . 0027
0 . 10×10−3
θ=0 .15 o
At 10th maximum, n=10, so
1
( 10+ )×5460×10−10
2
sin θ= =0 . 05733
0 . 10×10−3
θ=3. 3 o

What is the linear distance on screen C between adjacent maxima?

Solution: Since θ is very small, then sin θ≃tan θ≃θ

y nλ
tanθ= and sinθ=
D d
Combining the above two formula, we get:
 nλD
y=
d , n=0, 1, 2,............... for maxima
nλD (n+1) λD
y n= y n+1 =
So, d , and d
λD
Δy=
d , and D=20×10‒10 m
5460×10−10×20×10−2
Δy= −3
=1. 09 mm
0 .10×10
Interference of Light Due to
Thin Films
 Thin Film
A thin film is an optical medium with thickness of the order of 1 wavelength of
light in the visible region. Thus, a thickness of the film ranging from 0.5 mm t0
10 mm may be considered as a thin film. Example: Thin sheet of glass, mica, air
film enclosed between two transparent plates, soap bubble etc.

Characteristics:
When light is incident on a thin film, a small part of light gets reflected from the
top surface and a major part is transmitted into the film. Again, a small part of
transmitted light is reflected back into the film by its bottom surface and the rest
of light emerges out of the film.

 Plane parallel film

Fig.17

A transparent thin film of uniform thickness bounded by two parallel surfaces is

known as a plane parallel thin film (Fig.17). Thin Films transmit incident light
strongly and reflect weekly. After two reflections, the intensity of reflected rays
drops to a negligible strength.
1
Reflected rays
2

Film

S1
S2 Emerging rays

Fig.18
We consider two reflected rays 1 and 2 (Fig.18). These rays are derived from
the same source of light but seem to come from two sources S 1 and S2. They are
virtual coherent sources. Rays 1 and 2 will interfere by a lens and the case is
called two-beam interference.

 Wedge-shaped film

θ=constant θ=varying

Fig.19

Wedge-shaped film is such a film which has is zero thickness at one end and
progressively increasing thickness at other end. It is of two types. One type has
constant wedge angle and the other has varying wedge-angle.

 Interference due to reflected light

L
i C
A i H
F
G i
B E
r
t
Film r r

K
Fig.20
Let, in Fig.20
t= Thickness of the film; µ= Refractive index of the film,
AB= A ray of monochromatic incident light, BL= Normal to the top surface,
i= Angle of incidence; BC= Reflected ray from the top surface,
BD= Transmitted ray through the film;
DE= Reflected ray from the bottom surface;
EF= Emerging ray from the top surface; DK= Emerging ray from the bottom
surface,
r= Angle of refraction; EH = Normal to BC ray.
The film is surrounded by the air. BC and DE rays will interfere if they are
overlapped by a convex lens. The rays HC and EF travel equal path.
The geometrical path difference between ABDEF and ABHC rays is: BD+DE‒
BH.
λ
The optical path difference between them is, Δ= µ(BD+DE) ‒1(BH+ 2 ) ------(1)
DG t
BD= =
In ΔBDE, ∠ BDG =∠GDE =∠r ; So, cos r cos r , ∵ DG=t
ΔBDG and ΔGDE are congruent, ∴ BD =DE
t t 2t
∴ BD+ DE= + =
cos r cos r cos r -------------------(2)

Also, BG=GE=DG tan r=t tan r

∴ BE =2 BG=2 t tan r
o o
Again, ∠ LBH +∠ HBE=90 and ∠ HEB +∠ HBE=90

So, ∠ LBH =∠ HBE=i , Then, BH=BE sin i=2 t tan r sini ------------------(3)

sini
μ=
From Snell’s law,
sin r ∴ sin i=μ sin r
2 μt sin 2 r
∴ BH =2t tan r ( μ sin r )=
cos r -------------------(4)
Using equations (2) and (4) in equation (1),
2 μt 2 μt sin2 r λ
Δ= −( + )
cos r cos r 2
2 μt λ
= (1−sin2 r )−
cos r 2
 λ
=2 μt cos r−
2
Condition for maxima:

If the difference in optical path between two rays is equal to an integral number
of full wave, i.e. Δ=nλ, the rays meet each other in phase.
Thus, constructive interference occurs when,
λ
2 μt cos r − =nλ
2
λ
2 μt cos r =(2 n+1)
or 2 , A condition for Bright Fringe

Condition for minima:

If the difference in optical path between two rays is equal to an odd integral
number of full wave, i.e. Δ=(2n+1)λ/2, the rays meet each other in opposite
phase.

Thus, constructive interference occurs when,

λ λ
2 μt cos r − =( 2 n+1 )
2 2
or 2 μt cos r =(n+1 ) λ

The phase relationship of interfering waves does not change if a full wave, λ, is
added or subtracted from interfering waves. So, ((n+1)λ can be replaced by nλ.
Thus,

2 μt cosr =nλ , A condition for Dark Fringe

 Interference due to transmitted light

P
L
i r
S

C
A
r
t
Film r r M

D
B
Q
N R
Fig.21

Let, in Fig.21,
t= Thickness of the film; µ= Refractive index of the film,
SA= Incident ray; AB= Refracted ray; BC= Reflected ray from lower surface;
BR and DQ are emerging rays; BM=Normal to CD; DN=Normal to BR,
The optical path difference between BCDQ and BR is :
Δ= µ(BC+CD) ‒1(BN) -----------------------(1)
sini BN
μ= =
Also, from Snell’s law, sin r MD or
BN=μ . MD

In Fig.21, ∠ BPC=r and PC=BC=CD ∴ BC +CD=PD

∴ Δ=μ . PD−μ . MD=μ (PD−MD )=μ . PM
PM
cos r=
In ΔBPM,
BP or PM =BP cos r

But BP=2 t ∴ PM =2t cos r

∴ Δ=2 μt cosr
When Δ=nλ, Bright fringe occurs
2 μt cosr=nλ
When Δ=(2n+1)λ/2, Dark fringe occurs
λ
2 μt cosr=(2 n+1)
2
 Interference in wedge-shaped film due to transmitted light

Microscope

45° L
Half-
silvered S
mirror

Fringe

θ Thin Film

Fig.22

The arrangement of observing interference pattern in a wedge-shaped film is

shown in Fig.22. The wedge angle, θ, is small. When parallel monochromatic
beam illuminates the wedge from above, the rays reflected from its two
bounding surfaces will not be parallel. The path difference between rays
reflected from the upper and the lower surfaces of the air film varies along its
length due to thickness variation. Thus, alternate dark and bright fringes
(Fig.22) are observed on its top surface. The thickness of the glass plates is
large compared to the wavelength of incident light. Hence observed interference
effects are entirely from wedge-shaped film.
The optical path difference between two reflected rays (from upper and lower
surfaces of the film) is:
 λ
Δ=2 μt cos r −
2
λ
2 μt cos r =(2 n+1)
Maxima occur when 2
Minima occur when 2 μt cos r =nλ

E
C
A D
B t3
θ t1 t2
O
K L M
Dark Dark Dark

Fig.23

Let us say a dark fringe occurs at A (Fig.23), where 2 μt cosr =nλ is

satisfied. For normal incidence, r=0 and cos r=1 . If the thickness of air film at
A is t1, then
2 μt 1 =nλ
------------------------------(1)
The next dark fringe will occur, say at C, where the thickness is t2. So, at C,
2 μt 2 =( n+ 1) λ
-------------------------(2)
Subtracting equ(1) from equ(2), we get:
2 μ( t 2−t 1 )=λ
2 μ. BC= λ ∵( t 2−t 1 )=BC
λ
BC=
2μ
Now, from Δ ABC , ∠ BAC=θ , BC=AB tanθ
λ
∴ AB tan θ=
2μ
Where AB= the distance between successive dark fringes, also equal to the
distance between successive bright fringe. This distance is called fringe
separation, β. That is AB=β. We may write,
λ
∴ β=
2 μ tan θ

λ
∴ β=
For small values of θ, tan θ≈θ , 2 μθ
Thus, it is clear that β decreases with the increase in θ. In Newton’s rings
experiment, the width of Newton’s ring becomes smaller with the increase of
wedge angle θ.

Characteristics of Fringes:
(i) Fringe at the apex is dark,
(ii) Fringes are straight and parallel.
(iii) Fringes are equidistant,
(iv) Fringes are localized

Problem 1: A parallel beam of light (λ=5890×10‒8 cm) is incident on a thin glass

plate (μ=1.5) such that the angle of refraction into the plate is 60 o. Calculate the
smallest thickness of the glass plate which will appear dark by reflection.

Solution: Here, μ=1.5, r=60o, n=1, λ=5890×10‒8 cm, we know,

2 μt cosr =nλ
nλ 1×5890×10−8
t= =
2 μ cos r 2×1. 5×cos 60o
=3.926×10‒5 cm

Problem 2: A soap film of refractive index 4/3 and of thickness 1.5×10 ‒4 cm is

illuminated by white light with incident angle of 60 o. The light reflected by it is
examined by a spectroscope in which it is found a dark band corresponding to a
wavelength of 5×10‒5 cm. Calculate the order of the dark band.
Solution: Here, μ=4/3, , λ=5×10‒5 cm, t=1.5×10‒4 cm, i=60o
sin i 0. 866×3
sin r = =
μ 4 or r=40.5o
2 μt cosr =nλ
n=6 . 0832≈6

Problem 3: A soap film 5×10 ‒5 cm thick is viewed at an angle of 35º to the

normal. Find the wavelength of light in the visible spectrum which are absent
from the reflected light. Refractive index of the soap film is 1.33.

Solution: Here, t=5×10‒5 cm, i=35º and µ=1.33

sin i sin 35 °
sin r = = =0 . 431
We know, μ 1. 33
r =25. 5 ° ∴ cos r=0. 90

The condition for destructive interference is:

2 μt cosr =nλ
−5
If n=1, then λ 1=2×1. 33×5×10 ×0 . 9=120 μm
o

If n=2, then λ 2=6000 A

o

If n=3, then λ 3=4000 A

o

If n=4, then λ 4 =3000 A

λ1 and λ2 lie in the visible region and are absent in the reflected light.
Problem 3: A glass wedge of angle 0.01 radian is illuminated by
monochromatic light of wavelength 6000Å falling normally on it. At what
distance from the edge of the wedge will the 10 th fringe be observed by reflected
light.

Solution: Here, θ=0.01 rad, m=10, λ=6000×10‒10 cm. For dark fringe,
2 t =mλ
t
θ= or t=θx
Wedge-angle, x

∴ 2θx=mλ

mλ
∴ x=
2θ
10×6000×10−8
=
2×0 . 01
¿ 3 mm

Problem 4: A beam of monochromatic light of wavelength 5.82×10 ‒7 m falls

normally on a glass wedge with the wedge angle of 20 seconds of an arc. If the
refractive index of glass is 1.5, find the number of dark fringes per centimeter of
the wedge length.

'' 20×π
θ=2 0 =
Solution: Here, 60×60×180 , λ= 5.82×10‒7 m, μ=1.5
 λ 5. 82×10 -7×60×60×180
θ= =
2 μθ 2×1. 5×20×π
=2 mm
Number of fringes per centimeter=1/0.2
=5 per cm.

 Newton’s rings

Microscope

45° L
Half-silvered
mirror S

Plano
Convex Circular fringe
lens P
r Air Film
A B
O
Fig.24

Newton’s rings are examples of fringes of equal thickness. Newton’s rings are
formed when a plano-convex lens P of large radius of curvature placed on a
sheet of plane glass AB is illuminated from the top with monochromatic light
(Fig.24). Air film is formed in between P and AB of variable thickness in all
directions around the point of contact (O) of the lens and the plate. The locus of
all points at a distance r from O will make a circle. The thickness of air film at
this circle is same. When light rays reflected from the curved surface of the lens
and top surface of AB interfere, interference fringes are observed. Interference
fringes in the form of a series of concentric rings with their center at O are
observed in the microscope. Newton observed the rings first and hence the rings
are called Newton’s rings.

After interference, consider dark fringes are produced. For reflected light, dark
fringes will be produced if,
2 μt cosr =nλ
Where μ= refractive index of the air=1, angle of refraction, r=0 for normal
incidence, n=order of the fringe, and λ= wavelength of light. So, we write the
above equation as:
2 t =nλ -------------------------------------(1)

R R

P t rn N
A B
Q O

Fig.25

Consider that a dark fringe is located at P. Let, R=radius of curvature of the

lens, t=thickness of the air film at PQ, i.e. PQ=t, and r n= radius of the circular
fringe.
Using Pythagoras theorem,
R2 =r 2n +( r −t )2
r 2n =2 rt−t 2

if R >> t , then 2 Rt >>t 2

∴ r 2n =2 tR --------------------------------------(2)
Using the value of 2t from equation (1), we can write:
 r 2n =nλR

So,
r n =√ nλR

Diameter of the ring:

D n =2 r n =2 √ nλR ---------------------------------(3)

Hence,
D 1 =2 √ λR , D 2 =2×1 . 4×√ λR , D3 =2×1 . 7× √ λR
D 4 =2×2× √ λR . Thus the ring gets closer as n increases.
Why is the ring circular?
At a fixed distance r from the contact point O of the lens to outward, the air film
thickness is constant. The locus of all points at r will make a circle of radius r.
Thus, this constant thickness will make a circular ring, because light rays from
this thickness will produce a circular ring after interference.

Why is the central spot Dark?

λ
Δ=2 t −
We know the optical path difference, 2 , so at point O, t ≃0
λ
|Δ|=
2
λ
A path difference 2 (corresponding to a phase difference π) between two rays
(one from the curved surface and another one from the top surface of AB) will
interfere destructively, resulting in a dark spot.

Alternative explanation:
No phase change occurs for the rays reflected back from the curved surface
(denser to rarer medium: according to Stokes’s theorem) at O. A phase change
of π occurs for the rays reflected back from the top surface of AB (rarer to
denser medium: according to Lloyd’s experiment). So the phase difference of
these two types of rays is π, giving rise to destructive interference, and hence
dark spot is produced.

Determination of wavelength of light:

A plano-convex lens of radius of curvature about 100 cm
and a flat glass plate are cleaned. The lens is kept with its
convex face on the glass plate and they are held in position
by same means. The system is held under a travelling
microscope kept in front of a sodium lamp (Fig.24). The
light from the lamp falls on a glass plate held at 45° with
light beam. The light is incident on lens normally. Fig.26
The microscope is adjusted till circular rings are observed. By turning the screw
the microscope can be moved. The microscope cross-wire is set at the centre of
the ring. The cross-wire is moved along the diameter of the rings to 12 th ring.
Now, the vertical line of the microscope cross wire is tangentially set at one of
the rings, say 12th ring from the center (Fig.26). The readings are taken from the
scale. The cross-wire is moved and set the vertical line tangentially as before to
the 11, 10, 9.......... 5th ring. The readings are taken; say these readings are x12,
x11, x10, x9,........x5. The cross-wire is moved to the opposite sides along the
diameter and set at 5, 6, 7,8,........12 th rings. The readings are taken; say these
' ' ' ' ' '
readings are x 5 , x 6 , x 7 , x 8 ,........ x 12 . The difference of readings (x5~ x 5 ) for each
2
ring gives the diameter, Dn, of the ring. A graph of D n versus ring number (n) is
drawn. A straight line will be obtained as shown in Fig.27.

Now, from equation (3), we have D2n+ p

D 2n
D 2n =4 nλR

or D2n+ p =4(n+ p ) λR
D 2n
D 2n+ p −D2n p
λ=
or 4 pR --------------(4)
p
D2n+ p −D2n
Fig.27
The slop of the straight line is: p
slop
λ=
So, 4R
And R can be measured by a spectrometer. If we want to measure R by
Newton’s rings, similar formula of equation (4) will be applicable; slope will be
measured and λ will be given.

Formula for measuring R:

P
A
h
x O
x
R‒h
x/2
R

B a/2
C
D Q
a x
BD= sin30 o =
OD=xFig.28
In Fig.28, 2, 2,

=x cos30o = √ x 2
a 3
2
2 2 , a =3 x

2
Now, ( R−h) + x2 =R2 , R2 −2 Rh+ h2 + x 2 =R2
2 Rh=x 2 +h2
x2 h a2 h
R= + R= +
2h 2 , 6h 2

 Determination of Refractive index by Newton’s rings method

The gap between the lens and the plane glass plate is filled by a liquid, whose
refractive index is to be determined. Now, the liquid film substitutes the air
film. For dark fringe,
2 μt cosr =nλ
Where μ= refractive index of the liquid, angle of refraction, r=0 for normal
incidence, n=order of the fringe, and λ= wavelength of light. Then, we have
2 μt=nλ
 nλ
2 t=
or, μ

nλR
r 2n =2 tR=
We know, μ
Now, D 2n =4 r 2n
4 nλR
D2n =
μ , Fringe diameter decreases with μ.

4 nλR
For nth ring in the liquid,
[ D2n ] Liquid =
μ

4( n+p ) λR
For (n+p)th ring in the liquid,
[ D2n+ p ] Liquid =
μ

4 pλR
∴ [ D2n+ p ] Liquid− [ D2n ]Liquid =
μ -------------------(1)

∴ [ D2n+ p ] Air −[ D2n ] Air =4 pλR

We know, for air medium, ----------------(2)

μ= 2
[ D 2n+ p ] Air −[ D 2n ] Air

From equations (1) and (2), [ D n+ p ] Liquid −[ D 2n ] Liquid

Problem 1: In a Newton’s ring experiment the diameter of the 15 th ring was

found to be 0.59 cm and that of 5 th ring was 0.336cm. If the radius of the plano-
convex lens is 100 cm, calculate the wavelength of light used.

Solution: Here,
D215=0 . 59 cm, D 25=0. 336 cm, R=100 cm and p=15-5=10
 D 2n+ p −D2n
λ=
We know, 4 pR
D215−D25
=
4 pR
5 . 92−3. 362 ×10−6
=
4×10×1
o
=5880 A

Problem 2: Newton’s rings are observed in reflected light of wavelength

5.9×10‒5 cm. The diameter of the 10 th dark ring is 0.5 cm. Find the radius of
curvature of the lens and the thickness of the air film.

Solution: Here, λ=5.9×10‒5 cm, r10=0.5 cm and m=10

r 210
R=
We know, mλ
(0. 5 )2
= =106 cm
10×5. 9×10−5

mλ
t=
Again, 2

10×5. 9×10−5
= =295 μm
2
Problem 3: A thin convex lens of focal length 4 m and refractive index 1.50
rests on a glass plate, and using light of wavelength 5460Å, Newton’s rings are
viewed normally by reflection. What is the diameter of the 5th bright ring?

Solution: Here, f=4 m, µ=1.5, λ=5460×10‒10 m, and m=5

1
f [
1 1
=( μ−1 ) −
R1 R 2 ] Here, R1 =R and R2 =−R

1
f
=( μ−1 )
2
R [ ]
1
4
=(1 .5−1)
2
R [] or R=4 m

The diameter of the mth bright ring:

Dm= √2(2 m−1 )λR

D6 =√ 2(2×6−1 )×5460×10−10×4=6 . 2 mm

Diffraction of Light
 Diffraction

When light waves encounter obstacles (or openings), they bend round the edges
of the obstacles if the dimension of the obstacles are comparable to the
wavelength of the waves. The bending of waves around the edges of an obstacle
is called diffraction.

Obstacle

Opening

d d d

λ<<d
λ≈d
λ>d
 (a) (b) (c)
Fig.1

In Fig.1, when the opening (width d) is large compared to the wavelength, λ, the
waves do not bend round the edges. When, λ≈d, the light bending is noticeable.
When d<λ, the waves spread over the entire surface behind the opening.
If we consider a point source at the edge of the hole, it produces spherical wave
front and the bending of light can be explained.

The diffraction phenomenon is classified into two categories: (i) Fresnel

diffraction and (ii) Fraunhofer diffraction.

 Fresnel diffraction
A
In the Fresnel class of diffraction, the source of light and
Point source B D
the screen are, in general, a finite distance from the Aperture

diffracting aperture (Fig.2).Incident wave front is not C

plane wave. The phase of the secondary wavelets on
the aperture is not the same. A,B and C has same phase but Fig.2
B and D on the aperture has not same phase and hence
not coherent waves.

 Fraunhofer diffraction

In the Fraunhofer class of diffraction, the

source and the screen are at infinite distance
Point source
Aperture
from the aperture. To obtain plane waves, the
source is kept an infinite distance. However,
this is easily achieved by placing the source
on the focal plane of a convex lens and placing
the screen on the focal plane of another convex
lens. The analysis of diffraction pattern is much
easier in Fraunhofer diffraction pattern, because
the phase of the secondary waves is the same.

 Fraunhofer diffraction by a single slit

Slit zsinθ
P´
A dz
θ x
P´
z b
b
O P O P
B

Fig.4

Consider that a monochromatic parallel beam of light be incident on the single

slit AB of width b (Fig.4). The secondary waves travelling in the same direction
as the incident wave come to focus at point P. The secondary wavelets
travelling at an angle θ come to focus at P´. A lens is used for converging the
rays.

Let, x=the distance of the screen from the slit

O=the centre of the slit and the origin of coordinate system.
dz= a small element of the slit or wave front with coordinates (0,z).
We know that the equation of displacement of a travelling wave propagating
along x-axis can be written as:

y= A sin (ωt−kx ) --------------------------------(1)

where A=amplitude of the wave, ω=angular frequency=2πυ and k=wave
vector=2π/λ. For small element of slit dz at a distance z from o, let us consider
the displacement of the wave be dy, then

dy =dA sin (ωt−kx ) ------------------------------(2)

where dA=amplitude element. Here,
1 cdz
dA ∝dz and dA ∝ . So, dA=
x x

cdz
dy o = sin(ωt−kx )
At P, x ----------------------------(3)
At P´, the path difference between two rays coming from O and dz is Δ=zsinθ.
So, at P´,
cdz
dy = sin [ ωt−k (x +z sin θ) ]
x
cdz
¿ sin [ ωt−kx−z sin θ ]
x -----------------(4)
The total displacement, y, can be obtained by integrating the above equation (4).
b
+
2
c
y=∫ dy= ∫ sin [ ωt−kx −kz sin θ)] dz
x b
−
2
 [ ]
b
c −cos(ωt −kx−z sin θ) +
2
¿
x −k sin θ −
b
2

¿
c
xk sinθ [
cos (ωt−kx−
kb sin θ
2
)−cos(ωt −kx +
kb sin θ
2
)
]
c kb sin θ
¿ ×2×sin( ωt−kx )sin (C + D) (D−C )
xk sinθ 2 2×sin
2
sin
2

kb sin θ
sin
cb 2
¿ × ×sin( ωt−kx )
x kb sin θ
2

kb sin θ cb sin β
=β y= sin( ωt−kx )
Let, 2 , then x β
cb sin β
A=
So, the amplitude at P´ is x β , The intensity at P´ is I≈ A .
2

cb 2 sin2 β sin 2 β cb 2
I=( ) =I o 2 I o =( )
So, x β2 β , where x

sin2 β
I=I o
Therefore, β2

Intensity graph for a single-slit:

I

β
-4π -3π -2π -π π 2π 3π 4π
 y=β

y=tanβ

Fig.5

sin2 β
I=I o
We have derived the relation: β2 -----------------------(1)
If we draw I versus β, the above graph (Fig.5) will result. For β=0,
sin β sin β
= =1
β Indeterminate, But, if β →0, then sinβ→β and hence β ,
π 4I
β=± I = 2o
thus I=Io=Principal maximum. If 2 , then π . If β=±π

[ sin(±π ) ]2
I =I o 2
=0
then, (±π ) . But if β increases from ± π , then I increases

[ sin(±2 π )]2
I =I o 2
=0
and at β=±2 π , (±2 π ) . So, I increases, becomes

maximum and then decreases in between ± π and ±2 π . Again, if β

increases from ±2 π , then I increases, becomes maximum and decreases to 0

at ±3 π . And so on. All these maxima are called secondary maxima.

The question is where these maxima are? The maxima can be obtained from the
following condition:
 dI
‖ ‖ =0
dβ max -------------------------------(2)
Using equations (1) and (2), We can write:

β 2 . 2 sin β cos β−sin2 β . 2 β

Io[ ]=0
β4
sin β
sin β=β cos β or =β
cos β
So, tan β =β
So, Secondary maxima will be obtained for tan β =β
Let, y=tanβ and y=β and let us draw y=tanβ and y=β. Then, the intersection of
these two curves will give the value of β, where we can find the secondary

maximum. From tan β =β , one root is β=0. The other roots are obtained from
the intersection points that give, β=1.43π, 2.46π, 3.47π, etc. The intensity will

[sin(1 . 43 π )]2 [sin (2 . 46 π )]2

I 1=I o 2
=4 . 96 %I o I 2=I o 2
=1. 68 %I o
be (1. 43 π ) , (2 . 46 π )

[sin (3. 47 π )]2

I 3=I o 2
=0 . 83 %I o
, (3 . 47 π )

Calculation of approximate intensity of the secondary maxima:

sin2 β 3π 5π 7 π
I=I o β= , , ,. .. . .. .. . .. .
We know, β2 . if 2 2 2
I o×4 I o ×4 I o ×4
I= , , ,. . .. .. .. . .. .. . .
then, 9 π2 25 π 2 49 π 2
Io Io Io
¿ , , ,. .. . .. .. . .. .. ..
22 .2 61 . 1 121
 ¿4 .50 %I o , 1.62 %I o , 0.82 %I o ,...............

Condition for secondary minima and maxima:

We consider that at P´, the intensity is 0. For I=0,
πb sin θ
β=± pπ or =± pπ
λ
bsin θ=± pλ
π
β=±(2 p+1 )
The intensity I will be maximum, when 2
πb sin θ π λ
=±(2 p +1) or bsin θ=±(2p+1)
λ 2 2

 Diffraction at a circular aperture

Slit
θ L
A D P´´
θ x
C P
d
B

Fig.6
In Fig.6, AB is a circular aperture of diameter d. C is the centre of the aperture.
P is a point on the screen. CP is perpendicular to the screen. A plane wave front
is incident on AB. The secondary wave along CP comes to focus at P by the
lens L. Thus, P corresponds to the position of central maxima. Now, Consider
secondary waves travelling along a direction inclined at an angle θ with CP.
They meet at P´. Let PP´=x. The path difference between waves emanating from
A and B and reaching at P, is AD. From ΔABD,
AD=dsinθ

Condition for minimum intensity at P´ is: dsin θ=pλ

λ
dsin θ=(2 p +1)
Condition for maximum intensity is: 2
Thus, on the screen, at a distance x from P, we get a circle of radius x with
minimum intensity. So, we get diffraction pattern due to circular aperture. The
pattern consists of a central disc, called Airy’s disc, surrounded by alternate
dark and bright concentric rings. The intensity of dark ring is zero and that of
bright ring gradually decreases outwards from P. When the screen is at a large
distance from the lens,
x
sin θ=θ=
f
where f is the focal length of the lens. For the first secondary minima,
dsin θ=λ
λ
sin θ=θ=
So, d
x λ λf
= or x=
This implies, f d d
 Diffraction at a double slit
X M

L
A
b θ P´
B
d a P

C
bO θ
D dz

zsinθ

Y N

In Fig.7, AB and CD are two rectangular slits in line with one another. The
Fig.7
width of the slit is AB=BC=b. The width of opaque portion BC=a. The distance
between midpoints of two slits is d. That is d=a+b. O is the centre of the
coordinate and is located at the middle of BC. A very small element of the slit is
dz, located at a distance z from O. θ is the diffraction angle. So, zsinθ is the
path difference between to waves coming from O and dz. L is a convex lens to
converge the rays and MN is a screen. Let a plane wave front be incident on
XY. All the secondary wave fronts parallel to aP will meet at P. So, P
corresponds to the position of central maximum. Rays inclined at an angle θ
will meet at point P´. Diffraction pattern has two parts:
(i) The interference of the secondary waves of the two slits
(ii) The diffraction of secondary waves from individual slits.
Now for a single slit, we know.
cdz
dy = sin [ ωt−k (x +z sin θ) ]
x
So, for double slit,
b b
+ d+
2 2
c c
y=
x
∫ sin [ ωt−k ( x + z sin θ )] dz+ x ∫ sin [ ωt−k ( x + z sin θ ) ] dz
b b
− d−
2 2

y +
= 1
y2
Here, y1 is for a single slit and can be written as:
cb sin β
y 1= sin(ωt −kx )
x β
kb sin θ
β=
where 2
And y2 can be written as:
b
d+
2
c
y 2=
x
∫ sin(ωt−kx−kz sin θ )dz
b
d−
2
b
c d+
= [ cos ( ωt−kx−kz sin θ ) ] 2b
xk sin θ d−
2

=
c
[
xk sin θ
cos(ωt −kx−kd sin θ−
kb sin θ
2
)−cos(ωt −kx−kd sin θ+
kb sinθ
2
)
]
xk sin θ [ ]
c kb sin θ
= 2. sin(ωt −kx−kd sin θ ). sin( )
2
kb sin θ
sin
cb 2
= sin(ωt−kx−kd sin θ )
x kb sin θ
2
 cb sin β
= sin(ωt−kx−kd sin θ )
x β
Therefore,

y= y 1 + y 2
cb sin β cb sin β
= sin(ωt−kx)+ sin( ωt−kx−kd sin θ )
x β x β
cb sin β
= [ sin(ωt−kx )+sin(ωt −kx−kd sin θ )]
x β
cb sin β kd sin θ kd sin θ (C + D) (C−D)
2×sin cos
= . 2 sin(ωt −kx− ). cos 2 2
x β 2 2
kd sinθ
=γ
Let , 2 . Then,

cb sin β
y=2 . cos γ sin(ωt−kx−γ )
x β
cb sin β
A=2 . cos γ
Here, amplitude x β
cb 2 sin 2 β 2
I =4( ) 2
. cos γ
Therefore,
x β
sin2 β
I =4 I o 2 . cos2 γ
β

If β→0 and γ→0 then

I=I max =4 I o . The intensity of central maximum of
a double-slit is 4 times the intensity of central maximum of a single slit.

Intensity
 θ
Fig.8

Above Figure 8 shows the intensity distribution due to double slit The dotted
curve represents the intensity distribution due to diffraction for double slit. The
solid line represents the intensity due to interference of light from both slit. The
intensity, I, is not completely zero for double slit diffraction. The reason for the
presence of interference is due to the superposition of waves coming from two
slits.

 Mising order
Let, b=slit width and a=slit separation. If b is constant, diffraction pattern
remains the same. Keeping b constant, if the spacing a is altered, the spacing
between interference maxima changes. Depending on the relative values of b
and a, certain order of interference maxima will be missing in the diffraction
pattern.
Now the condition of interference maxima is:
dsin θ=nλ
or (a+ b)sin θ=nλ , where n is an integer
The condition for diffraction minima is:

bsin θ= pλ , where p is an integer

If the values of b and a are such that both equations are satisfied simultaneously
for the same value of θ, then the positions of certain interference maxima
correspond to the diffraction minima at the same position on the screen.
From above two equations:
(b+a)sin θ n
=
b sin θ p ---------------------------(1)
n
=2
(i) Let b=a, then from equation (1): p
If p=1, 2, 3, ........................, then n=2, 4, 6,.......................
So, 2, 4, 6,........... interference pattern will lie at the minima of diffraction. Thus,
the orders 2, 4, 6,.... of interference will be missing in the diffraction pattern.
So, there will be 3 interference maxima in the central diffraction maximum. The
graphs are shown in Figure 9:

Intensity

θ
Fig.9
 n
=3
(ii) If 2b=a, then: p
For p=1, 2, 3,....................; n=3, 6, 9,....................

The orders 3, 6, 9,... of interference maxima will be missing in the

diffraction pattern. . So, there will be 5 interference maxima in the central
diffraction maximum.

(iii) If b+a=b, then a=0

The interference maxima will be missing, because there is only one slit.
Only diffraction pattern will appear.

 Diffraction grating
Definition:
A diffraction grating is an optical device, consisting of a large number of
narrow slits placed side by side that are separated by opaque space or obstacle.

Types of diffraction grating:

There are two types of diffraction grating.
(i) Plane transmission grating
(ii) Plane or Concave reflection grating
Transmission grating: When a wave front of light is incident on a grating
surface, light is transmitted through narrow slits and obstructed by the opaque
portion. Such a grating is called transmission grating.

Reflection grating: When a wave front of light is incident on a grating surface,

light is reflected from the position of mirror in between lines. Such a grating is
called reflection grating.

Line
Fabrication of grating:
Gratings are fabricated by ruling equidistant
parallel lines on a transparent material (Fig.10), like
glass, quartz, with the help of a fine diamond point.
The space in between any two lines is transparent Slit
to light and ruled lines are opaque to light. Fig.10
This type of grating acts as transmission grating.
On the other hand, if the lines are drawn on a silvered surface (Plane or
Concave) of a mirror, then light is reflected from the positions of the mirror in
between any two lines. Such a grating acts like a plane or concave reflection
grating.
Number of lines and their spacing:
Usually, study for visible region 104 lines/cm or 106 lines/m are drawn in a
grating. The spacing of a grating is from µm to wavelength of light. When the
line width is 4000-8000Å, deviation of light is produced.

 Intensity distribution due to Diffraction grating

X M

L
θ
P´
dz
θ
 Fig.11

Figure 11 shows an arrangement of a grating (AB) where plane wave fronts of

light are incident and the diffracted rays are viewed on XY-screen. Let a= width
of each line, b= width of each slit. Here, d=(a+b) is called grating element or
grating constant and is denoted by 1/N´. So, N´=1/(a+b)= number of lines per
unit length.
By Huygens’s principle of secondary wavelets, each of the infinite points in the
plane of the slit sends out secondary wavelets in all direction. The secondary
waves travelling in the same direction (θ=0) are focused by a convex lens L at a
point P on XY. At P, we get central maximum. To know the intensity pattern at
other points on XY, we consider waves whose travelled direction makes θ with
the incident beam at a point P´. The waves reaching P´ have different phases.
Thus we will get dark and bright bands at two sides of P.

cdz
dy = sin [ ωt−kx −kz sinθ ) ]
For a single slit, x
If we let, φ ( z )=sin [ ωt−kx−kz sin θ ) ] , then for N slits:
b b b
+ d+ 2 d+
2 2 2
c
y=∫ dy= [ ∫ φ( z)dz+ ∫ φ ( z)dz+ ∫ φ ( z)dz+.. .. .. . .. .. . .. .. . .. .
x b b b
− d− 2d−
2 2 2
b
( N+1) d+
2
+ ∫ φ( z)dz
b
( n−1 )d−
2 ............................................................(1)
On simplification,
cb sin β
y= [ sin(ωt−kx)+sin(ωt −kx−kd sin θ )+
x β
sin( ωt−kx−2kd sin θ )+.. .. . .. .+sin {ωt−kx−( N−1 )kd sin θ}]
-------------------------(2)

kb sin θ kd sin θ
β= ' γ=
where 2 . If y =ωt −kx , n=N-1 and 2 , then
'
cb sin β
y= [ sin { y +sin ( y ' −2 γ )+sin( y ' −2 . 2 γ )+. . .+sin( y ' −n . 2 γ )]¿
x β
------------------------------(3)
p=n
cb sin β
=
x β
∑ sin( y ' − pm)
p=0 ----------------------------------(4)
where γ=m/2.
' nm n+1
p=n sin( y − )sin( )m
2 2
∑ sin( y ' − pm)= m
p=0 sin
But, 2 --------------------(5)
-----------------------------------------------------------------------------------------
Formula verification: Taking left side,
p=1
∑ sin( y ' − pm)=sin { y +sin( y ' −m)=2 sin( y ' − m2 )cos m2 ¿
'

p=0

' m m m
2 sin( y − )cos sin
2 2 2
=
m
sin
2
' m
sin( y − )sin m
2
=
m
sin
2
Taking right side,
nm (n+1 )m
sin( y ' − )sin
2 2
=
m
sin
2
m
sin( y ' − )sin m
2
= for n=1
m
sin
2
------------------------------------------------------------------------------------------

From equations (5) and (6):

' nm n+1
sin( y − )sin( )m
cb sin β 2 2
y=
x β m
sin
2

N −1 Nm
sin {y ' −( )m}sin
cb sin β 2 2
= ∵ n=N −1
x β m
sin
2
 cb sin β sin Nγ m
= sin {ωt−kx−( N −1)γ } ∵ γ=
x β sin γ 2
cb sin β sin Nγ
A=
Here, amplitude, x β sin γ
cb 2 sin2 β sin2 Nγ
I =( )
Intensity, x β 2 sin2 γ
cb 2
I =( ) =I o
If γ→0 and β→0, then, x
sin 2 β sin 2 Nγ
I =I o
So, β 2 sin 2 γ --------------------------------------(6)

sin2 β
Here, β2 is diffraction terms due to a single slit and

sin2 Nγ
sin2 γ is interference terms due to coherent waves from N slits

Principal maxima:

If sin γ=0 , then γ=±nπ , where n=0, 1, 2, ......

Again, N=1, 2, 3, ....., then sin Nγ=0 ,
sin Nγ
=
Hence, sin γ indeterminate
sin Nγ
=
The value of sin γ when γ =±nπ can be obtained by differentiating
numerator and denominator using L. Hospital’s rule:
 d
sin Nγ
sin Nγ dγ N cos Nγ
Lt γ →±nπ =Lt γ →±nπ = Lt γ →±nπ =±N
sin γ d cos γ
sin γ
dγ
sin 2 β 2
I =I o 2
N
Hence, β

2sin 2 β
I =N I o
β2 ------------------------------------------(7)

This equation shows principal maxima when

γ ≈±nπ
kd sinθ
=±nπ
or, 2
2 π ( a+b )sin θ
=±nπ
or, λ 2

or, (a+ b)sin θ=±nλ ---------------------------(8)

n=0 corresponds to zero order (central) maximum. We obtain 1 st, 2nd,

3rd,............. order principal maxima for n=1, 2, 3, ........... Positive and negative
signs are used to show same order maximum on both sides

1st Central 1st

n=-1 n=0 n=1

Fig.12
Minima:

If sin Nγ=0 but sin γ ≠0 , then I=0. Therefore, for minima,

sin Nγ=0
or Nγ=± pπ
kd sin θ
N =±pπ
or, 2
2 π ( a+ b )sin θ
N =± pπ
or, λ 2

or, N (a+ b )sin θ=± pλ ------------------------------((9)

Where p has all integral values except 0, N, 2N, ....nN, because for these values
we get similar conditions of equation (8), which show maxima instead of
minima. This is contradictory. So, p=1, 2, 3, ...(N-1). Since for these values of p
we get minima, then there must have maxima in between these minima. These
maxima are called secondary maxima.

Secondary maxima:
Since (N-1) minima occur between two adjacent principal maxima, so there
must be (N-2) other maxima between two principal maxima [Let N-1=4, so
there will be (N-1)-1=3 other maxima, as can be obvious from figure 14]. These
maxima can be obtained from the condition as follows:

sin 2 β sin 2 Nγ
I =I o
We know, β 2 sin 2 γ
dI max sin2 β sin2 γ . 2 .sin Nγ . cos Nγ . N−sin2 Nγ . 2 sin γ . cos γ
=0=I o 2 .[ 2 ]
dγ β sin γ

dI max sin2 β N cos Nγ sin γ −sin Nγ cos γ

=0=I o 2 .2 . sin Nγ . sin γ [ 4 ]
dγ β sin γ

N cos Nγ sin γ −sin Nγ cos γ

4
=0
sin γ
or,

or N cos Nγ sin γ−sin Nγ cosγ=0

or N cos Nγ sin γ=sin Nγ cosγ
or N tan γ=tan Nγ --------------------------------(10)
The positions of secondary maxima are obtained from the roots of equation

(10), other than those for which γ =±nπ (that correspond to the principal

sin2 Nγ
2
maxima). To evaluate the value of sin γ from N tan γ=tan Nγ , we can
use the right-angled triangle (Fig.13).
This gives: √ N 2+cot 2 γ
N
sin Nγ=
√ N 2+cot 2 γ N
2 2
sin Nγ N Nγ
=
or sin2 γ ( N 2 + cot2 γ ) . sin2 γ cotγ

Fig.13
 N2
= 2 2
N sin γ + cos2 γ
N2
=
1+( N 2−1 )sin 2 γ
So, Intensity of secondary maxima:

sin 2 β N2
I =I o 2
β 1+( N 2 −1) sin2 γ

I (seconadry max ) 1
=
I ( principal max ) 1+(N 2−1 )sin2 γ
--------------------(11)

The ratio decreases as N increases and zero when N→α. Graph of secondary

sin2 β
maxima is shown in Fig.14 β2

sin2 Nγ
sin2 γ

I
Numbers of maximum orders availableFig.14
with a grating:

The maximum orders can be obtained from:

(a+ b)sin θ=nλ

( a+b ) sin θ
n=
λ -----------------------------(12)

So, λ will decide the maximum orders of diffraction for a given (a+b). For n max,
θmax=90°.

(a+b )
n max =
So, λ
If (a+b)<3λ, then nmax<3.

Absent spectra with a diffraction grating:

Sometimes 1st order is clearly visible, 2nd order not visible and 3rd order is again
visible and so on. This occurs for a given θ if the path difference between two
extreme ends is equal to nλ.

The principal maxima are given by: (a+ b)sin θ=nλ

The minima for a single slit are: b sin θ= pλ

(a+ b) n
=
So, b p
This is the condition for absent spectra. To suppress the 2 nd order n=2=2p, i.e.
( a+ b)
=2
p=1. So, b
a=b
The width of ruling is equal to width of the sheet.
Effect of increase in the width of a ruled surface:
For a transmission grating, the principal maximum of wavelength λ in the
direction θ is given by:

(a+ b)sin θ=nλ --------------------------------------(1)

If N is the total number of slits in the grating, then equ (1) becomes:

N (a+b )sin θ=nN λ --------------------------------------(2)

Now, for the first minimum adjacent to the
nth maximum in the increasing direction of
θ, let its direction be along θ+dθ (Fig.15). Here,
θ
dθ is the angular width of the nth maximum. dθ

Fig.15
Now, the directions of minima are given by:

N (a+ b )sin θ=pλ --------------------------------------(3)

Where p can take any value except, 0, N, 2N, .....nN. As a result, if nth principal
maximum occurs at nN, then 1st minimum after nth maximum will occur at
nN+1. So,

N (a+b )sin (θ +dθ )=(nN +1) λ --------------------(3)

or, N (a+b )[sin θ cos dθ+cosθ sin dθ ]=(nN +1) λ -------------(3)

Since dθ is very small, cosdθ≈1 and sindθ≈dθ. So,

N (a+b )[sin θ+cos θdθ ]=(nN +1) λ -------------(3)

or, N (a+b )sin θ+N (a+b )cosθdθ=(nN +1 )λ -------------(3)

Using equation (2):

nN λ+ N (a+b )cosdθ=(nN +1) λ -------------(3)

or, N (a+ b )cos dθ= λ

λ
dθ=
or, N (a+b )cos θ
1 1 λ 1 1
dθ= =
2 2 nλ 2 nN cot θ
N cos θ
or, sin θ
So, larger width (a+b) of ruled surface, smaller is the angular half width and
sharper is the maxima. Half width is inversely proportional to N and cotθ. This
means, if N and cotθ are larger, sharper is the diffraction maxima.

Dispersive power of a grating:

Definition: Dispersive power of a grating is defined as the ratio of the difference

in the angle of diffraction of any two neighboring spectral lines to the difference
dθ
in wavelength between two spectral lines. It is defined as dλ .
dθ
dλ = difference in angle of diffraction per unit change in wavelength. For a

given λ, the diffraction condition of nth principal maxima is:

(a+ b)sin θ=nλ ----------------------------------(1)

or, (a+b)cosθdθ=nd λ
dθ n
=
dλ (a+ b )cos θ
 '
nN
=
cos θ
dθ
where, N´=Number of lines per unit length. So, dλ is proportional to n,
dθ
proportional to N´ and inversely proportional to cosθ. If θ increases, dλ
increases, It is minimum at θ=0. If θ is small, dθ is proportional dλ. This is
called normal dispersion.

Resolving power of an instrument:

A fortune teller observes our palm by a magnifying glass to see lines on palm
very clearly and distinctly. We use best quality binocular to see a cricket match
in a stadium. An appropriate telescope is necessary to see very nearby stars at a
large distance. A camera lens of superior quality is required to have good
photograph. So, optical instruments like microscope, telescope, lens, prism and
gratings that aid our vision depend on physical phenomenon, known as
diffraction of two nearby points. The condition to obtain a best quality images,
also known as resolution, was given by Rayleigh.
We know magnification or magnifying power of a microscope or a telescope
depends on focal length of the lenses used in the instrument. Image size can be
increased by changing focal length of a lens or a proper choice of a lens. But
there is a certain limit of magnification of an optical instrument. This is due to
the fact that for a wave surface, the laws of geometrical optics do not hold good.
From the knowledge of diffraction of light, it is clear that the image of a point
source is basically diffraction instead of a point. When a light is kept in front of
a circular aperture, the diffraction pattern of the point source of light consists of
a central bright disc surrounded by alternately dark and bright diffraction rings.
When aperture diameter (of a lens or holes) is larger, the diffraction effect of a
point source is small. If we keep two point sources very close to each other,
then the diffraction disc of one may be overlapped by the other and hence image
of two point sources cannot be seen clearly. Consequently, optical instruments
are used to resolve the image of two point sources when their diffraction
patterns are well separated. For example, consider two points very close to each
other. Our eyes cannot separate them, because the diffraction of the two points
occurs by the aperture of eye lens overlap each other, whereas a magnifying
glass can separate them.

Definition of a resolving power:

The ability of an optical instrument, expressed in numerical measure, to resolve
the images of two closely spaced points is termed as its resolving power.

Criterion of Lord Rayleigh:

Rayleigh’s criterion states that “two nearby images are said to be resolved if the
position of the principal maximum in the diffraction pattern of one coincides
with the first minimum of the diffraction pattern of the other and vice versa.

A B
Explanation of Rayleigh’s criterion:
A B
principal
maxima
Consider Fig.16 consisting of two principal
maxima A and B having wavelength λ1
and λ2, respectively. In this case, the
λ1 Fig.16 λ2 θ
diffraction is large enough and two images
A B
are thus separate. This means θ1 corresponding
C
to principal maximum B is greater than θ2 principal
maxima
corresponding to1st minimum to the right of A. A B increased

So, spectral lines are well resolved

In Fig.17, the central maxima have
λ λ+dλ θ
wavelengths λ and λ+dλ. The angle of diffraction
Fig.17
for 1st minimum A is greater than that for
A B
principal maxima of B. The two maxima Dip in the
A B middle
overlap and the images cannot be separated. principal C principal
maxima maxima
The maximum intensity is C. If we go from
A to B, we find an increase in intensity at C.
λ λ+dλ θ
In Fig.18, the position of the principal maximum
Fig.18
of A (of λ) coincides with the position of the first
minimum of B (of λ+dλ). Similarly, the position of the principal
maximum of B coincides with the position of the first minimum of A. The
resultant intensity shows a dip at C (in the middle), where the intensity is about
20% less than that at A or B. Thus, we can say that A and B are distinguishable.
In another words of Rayleigh's criteria, "Two images are said to be just resolved
if the radius of the central disc of either patterns is equal to the distance between
the centers of the two patterns."

Resolving power of a grating:

The important property of a diffraction grating is its ability to resolve two or
more spectral lines which have nearly the same wavelength.
"The resolving power of a diffraction grating is defined as the capacity to form
separate diffraction maxima of two wavelengths which are very closed to each
λ
other". This is measured by the ratio dλ , where d λ is the smallest difference in
two wavelengths which are just resolved by grating and is the wavelength of
either of them, or the mean wavelength λ.

Expression for resolving power:

Consider the surface of a A X
plane transmission grating AB.
λ+dλ P2
N is the number of slits; nth
a+b is the grating element. dθn λ P1 order
A beam of light consisting θn central
of wavelength λ and λ+dλ image
is incident on the grating.
XY is the field of view of
a telescope. B Y

P1 is nth primary maximum

of a spectral line for λ, diffracted at θn.
P2 is nth primary maximum of a spectral line for λ+dλ, diffracted at θn+dθn.
To see clearly the two wavelengths, the Rayleigh's criterion (Two lines will be
resolved if the principal maximum of λ+dλ (in nth order) in a direction θ n+dθn
falls over the first minimum of λ in the same direction θ n+dθn. We shall consider
the 1st minimum of λ in a direction θn+dθn as follows:
The principal maximum of λ in the direction of θn is given by:

(a+ b)sin θn =nλ ----------------------------------(1)

or,
N (a+b )sin θ n=nN λ
The equation of minima is:

N (a+b )sin θ n= pλ ----------------------------------(2)

where p has all integral values except 0, N, 2N,...........nN, because for these
values of p, the condition for maxima is satisfied and we get diffraction maxim.
Thus the 1st minimum adjacent to nth principal maximum in the direction
θn+dθn can be obtained by substituting the value of p as nN+1. Therefore, the 1st
minimum in the direction θn+dθn is:

N (a+b )sin (θ n +dθn )=(nN +1) λ --------------------------(3)

Now, the principal maximum of wavelength λ+dλ in the direction θn+dθn is:
 (a+b)sin(θ n +dθ n )=n( λ+dλ ) --------------------------(4)

or,
N (a+b )sin (θ n +dθn )=nN ( λ+dλ ) -----------------------(5)
From equ (3) and (5), we get:

(nN +1 )λ=nN ( λ+dλ )

or, nN λ+ λ=nN λ+nNd λ
or, λ=nNd λ
λ
=nN
So, dλ --------------------------------------(6)
This is the equation of resolving power of a grating of N slits for diffraction
order n.
Conclusions:
λ
∞n
(i) dλ
λ
∞N
(ii) dλ
( a+b ) sin θn
n=
Again from equation (1): λ

λ N ( a+ b) sin θn
=
So, dλ λ ---------------------------(7)
dθ λ
Distinction between dλ (dispersive power) and dλ (resolving power) of a
grating:
λ
1. dλ indicates the limit of resolution of two close objects to be just
resolved.
 dθ
dλ indicates an idea of angular separation between two spectral lines.
λ
2. dλ is measured by nN.

dθ n
dλ is measured by
(a+ b)sin θn
λ
3. dλ increases with N
dθ
dλ remains unchanged with N

λ
4. dλ remains unchanged with grating element (a+b)
dθ
dλ increases with (a+b)

Problem 1: A slit of width b is illuminated by a white light. For what value of b

will the first minimum for red light (λ=6500 Å) fall at θ=30°?

Solution: At first minimum, n=1

nλ 1×6500
b= = =13 , 000 Å
We know, sin θ sin 30

What is the wavelength of light whose first diffraction maximum (except central
maximum) falls at θ=30°, thus coinciding with first minimum for red light?

Solution: The maximum is half way between 1st and 2nd minima. Then let n=1.5,
 So, b sin θ≃1. 5 λ and b sin θ≃λ ,
' λ 6500
λ= = =4300 Å
Therefore 1 .5 1. 5

The 2nd maximum of light of λ=4300 Å will coincide with 1 st minimum of

λ=6500 Å.

Poblem 2: In Fraunhofer diffraction pattern due to a narrow slit, a screen is

placed 2 m away from the lens to obtain a pattern. If the slit width is 0.2 mm
and the first minima lie 5 mm on either side of the central maximum, find the
wavelength of light.

Solution: Here, b=0.02 cm, x=0.5 cm, D=200 cm, and n=1
x
sin θ=
We know, b sin θ=λ but D
bx bx 0 . 02×0 . 5
=λ λ= = =5000 Å
Therefore, D So, D 200

Problem 3: Diffraction pattern of a single slit of width 0.5 cm is formed by a

lens of focal length 40 cm. Calculate the distance between first dark and the
next bright fringe from the axis. Given λ=4890 Å.

Solution: Here b=5×10‒3 m, f =0.4 m, λ=4890×10‒10 Å

y1
sin θ=
For minima, b sin θ=nλ , where n=1 and f
 by 1
=λ
Therefore, f so,

fλ 0 . 4×4890×10−10 −5
y 1= = =3 . 912×10 m
b 5×10−3
λ y2
b sin θ=(2 n+1 ) sin θ=
For maxima, 2 , where f and n=1
3 fλ
∴ y 2= =5. 868×10−5 m
2 b
∴ y 2 − y 1=1 . 9×10−5 m=1 .9×10−2 mm

Problem 4: In double-slit Fraunhofer diffraction, what is the fringe spacing on a

screen 50 cm away from the slits if they are illuminated with blue light (λ=4800
Å). If slit separation is 0.10 mm and slit width is 0.02 mm, what is the linear
distance from the central maximum to the first minimum of the fringe envelop?

Solution: Here, D=50cm, d=0.10 mm, b=0.02 mm and λ=4800 Å

y
d =λ
We know, d sin θ= λ or D
λD 4800×10−10×50×10−2
∴ y= = −3
=2 . 4×10−3 =2 . 4 mm
d 0 . 10×10
The distance to the first minimum of the envelop is determined by the
sin β 2
( ).
diffraction factor β The first minimum occurs for β=π.
kb sin θ πb sin θ
β= ∴ =β
We know 2 . λ
 β λ β
sin θ= n= =1
So, π b, since β=nπ , so π

λ 4800×10−10
sin θ= = −3
=0. 024
Therefore, b 0. 02×10

The value is very small so, θ=tan θ≃sin θ

y=D tan θ=D sin θ=50×0 . 024=1. 2 cm

Problem 5: Deduce the missing orders for a double-slit Fraunhofer diffraction

pattern, if the slit widths are 0.16 mm and they are 0.8 mm apart.

Solution: Here b=0.016 cm and a=0.08 cm

For maxima, (a+ b)sin θ=nλ and for minima b sin θ= pλ

n (0 . 016+0. 080 )
∴ = =6
p 0 . 016
n=6 p , if p= 1, 2, 3 ....... etc. then n=6, 12, 18,......etc.

So, 6, 12, 18, ..... etc. orders are missing.

Problem 6: A parallel beam of light of wavelength 5460 Å is incident at an

angle of 30° on a plane transmission grating which has 6000 lines/cm. Find the

highest order spectrum that can be observed.

(a+b)[sin θn +sin i]=nλ

Solution: We know,

1
(a+ b)= 5
m
Here, θ=30°, λ=5460×10‒6 m, θn=i and 6×10
 2( a+b ) sinθ n 2 1
n= = × =3
So, λ 5460×10−10×6×105 2

Problem 7: Calculate the possible order of spectra with a plane transmission

grating having 8000 lines/cm when light of wavelength 4500 Å is used.

( a+b ) sin θn
n=
Solution: We know, λ
Here, possible order=highest order, so sin90°=1
(a+b ) 1
n= = =2 .77≈3
So, λ 8000×4500×10−8

Problem 8: A diffraction grating has 6500 lines/cm and is used with sodium
light at normal incidence. Calculate the dispersive power of the grating for the
2nd order.

dθ n
=
Solution: We know, dλ (a+ b )cos θ , but (a+ b)sin θ=nλ
Given n=2, λ=589.3 nm,

nλ 2×5893×10−9 6500×102
sin θ= = × =0 . 766
So, (a+b ) 1 1

cosθ=√ 1−sin 2 θ=0 . 41

dθ 6500×102 ×2
= =1585365 . 9
dλ 0 . 41
Problem 8: To resolve 589.6 nm and 589.0 nm in its first order diffraction with
a plane transmission grating, calculate the minimum number of lines required
per centimeter in a 1.5 cm width grating.

Solution: Here, n=1, λ1=589.6 nm, λ2=589.0 nm,

dλ=λ 1−λ 2 =0 .6×10−9 m

λ 1 + λ2
λ= =589 . 3 nm
Mean 2
1 λ 5893×10−9
N= × = −9
≈983
Therefore, n dλ 1×0 .6×10
So, to resolve the wavelength separation of 0.6×10 ‒9 nm, the minimum number
of lines is 983. These numbers of lines are located in a 1.5 cm width of grating.
Therefore, Number of lines in 1 cm is 983/1.5=655.
 Polarization of Light

Natural Light: E

Fig.1
Light is emitted randomly oriented excited atoms. Each excited atom radiates a
polarized wave train for 10‒8 s. If there are a lot of atoms which coherently emit
(frequency same) light waves, their combinations will give a single resultant
polarized wave which persists not greater than 10 ‒8 s. However, new wave
trains are constantly emitted and these emissions might not be coherent i.e. if
we consider the propagation direction k of all waves, the E vectors of them will
be perpendicular to k, but along various directions about k (Fig.1). Thus the
overall polarization changes in a random fashion, as the rate of emission is very
rapid. The resultant wave thus produced is called natural light or unpolarized
light.
A light beam whose E fields are symmetrically distributed about the
propagation direction of the wave is known as unpolarized light. A perfectly
monochromatic plane wave is always polarized

Polarization of light:

Plane

E
k

Fig.2

Natural light has E vectors random about the direction of propagation k. But if
the E fields or the vibration directions of E are confined to a plane, then this
light is called polarized light. The above plane is called plane of vibration and a
plane perpendicular to the above plane is called plane of polarization.

Processes of polarization:
An unpolarized light can be polarized (linearly) by the following ways:
 (i) Reflection, (ii) Refraction, (iii) Double refarction, (iv) Scattering and (v)
Selective absorption.

Polarization by reflection (Discovered by Malus in 1809)

θ θ
A C
A C
θp θp
Rarer medium θ θ Partially polarized Completely polarize

B B
Partially polarized
Denser medium
θr θr

D D

E
N´
N´ E

Let AB be an unpolarized incident beam, which can be polarized into two

components. ABN is a plane of incidence. ↕ is a resolved component of E-
vector parallel to ABN i.e. in the plane ABN (also called π-component) and ●
(dot) is another component perpendicular to ABN (also called σ-component).
When ray AB is reflected from the surface it is found that the reflected ray BC
has a predominance of dot component and is said to be partially polarized.
When incident angle θ varies, the predominance of dot component also varies,
and at a particular angle θp, called polarizing angle, BC does not contain any
parallel component. In this way, unpolarized light is completely polarized.

Explanation of polarization by reflection:

A C
θp θp
Completely polarized
An atom in denser medium oscillates according to E ║ and E┴ fields. From
classical electrodynamics, an oscillating atom does not radiate wave whose
propagation direction k is parallel to oscillating direction. According to the
Figure, E║ is nearly parallel to BC. So, E ║ field will not propagate along BC.
But E┴ is perpendicular to BC and it can propagate along BC. Thus, E ║ is
missing in BC. When θ=θp, E║ is completely parallel to BC. So, this component
is completely missing in BC ray. On the other hand, the refracted ray BD is
partially polarized due to the action of reflecting surface. Because, of the E ║
fields vibrating in the plane of incidence, parts are reflected and parts are
transmitted at all angle θ. Thus, we always get E║ and E┴ fields in the BD rays.

Explanation of Polarization by elastic-solid theory:

Consider that E┴ is the perpendicular component of electric field of the incident
wave. E║ is the parallel component of electric field of the incident wave. Et┴ is
perpendicular component of electric field of the transmitted wave. Et║ is parallel
component of electric field of the transmitted wave. Er┴ is perpendicular
component of electric field of the reflected wave. Er║ is parallel component of
electric field of the reflected wave. θ=θp is the polarizing angle of the incident
beam. θr is the angle of refraction. Using the elastic-solid theory on polarization
of light, Fresnel derived the following relation:
Et┴ 2 sin θr cosθ
=
E┴ sin ⁡(θ+θr )
..........................................(1)
t
E║ 2 sin θr cosθ
=
E║ sin ⁡(θ+θ r)cos ⁡(θ−θr )
...........................(2)

Here, E┴= E║ and θ+θr=90°, so from equ(1), we can write

t
E┴
E┴
=2 sinθ r cosθ ..................................(3)
t t
t E┴ 1 E┴
and from equ(2), E ║= E║ =
E┴ cos ⁡(θ−θ r ) cos ⁡(θ−θr )

Since, θ‒θr <90°, then, Et║> Et┴

He also derived the following equations:
r
E┴ sin ⁡(θ−θ r)
E┴
=‒
sin ⁡(θ+θr )
..........................................(4)

Er║ tan ⁡(θ−θr )

=
E║ tan ⁡( θ+θr )
.......................................(5)
Since, θ+θr=90°, then from equ(5), Er║=0 and from equ(4),
r
E ┴=‒ E ┴ sin ⁡(θ−θr ) ...........................................(6)

Here, ‒ve sign indicates the phase of the reflected wave. Since, θ‒θr <90°, then
from equ(6), , |E t║|<|E ┴|. But, since Er║=0, all components are Er┴ in the reflected
light.

Polarization due to refraction

θp

33º
We know that light is partially polarized by reflection. In transmitted, Et║ light
component is larger. Therefore, if the natural light is passed through a stack of
glass plate, the transmitted light will be nearly polarized by Et║ component. It
has been proved that if I║ is intensity of parallel component of light, I ┴ is the
intensity of perpendicular component, m is the plate number and µ is the
refractive index of the plate, then the degree of polarization is
I ║−¿ I
P= ┴
¿
m
I ║+¿ I = 2
¿
┴
2μ
m+( 2
)
1−μ
About 15 glass plate and 33º inclination of plate with beam can produce parallel
component of light.

Brewster’s law
Sir David Brewster found that the polarizing angle θp depends upon the
refractive index µ of the medium when he performed experiments of reflected
from various surfaces. He proved:
μ=tan θ p

In statement, “The tangent of the angle at which polarization is obtained by

reflection is numerically equal to the refractive index of the medium.”

A C
θp θp
Completely polarized
AB is incident ray, BC is reflected ray, BD is refracted ray, θp is polarizing
angle θr is refracted angle. Brewster found that maximum polarization occurs
when BD is perpendicular to BC.
Here, BD┴BC, so ∟CBD=90º. Therefore, θp + θr =90º or θr =90º‒ θp
According to Snell’s law, refractive index of the medium:
sin θ p sin θ p sin θ p
μ= = =
sin θr sin ⁡(90−θ p) cos ⁡θ p

or, μ=tan θ p

Polarization by dichroic crystals (Selective absorption):

(i) Polarizing sheet

Polarizing sheet

Figure shows unpolarized light falling on a sheet of commercial polarizing

material, called polaroid. These sheets consist of a certain chracteristic
polarizing direction, shown by parallel lines. The sheet passes the wave train
components whose electric vectors vibrate parallel to this direction and absorb
wave whose electric vector is perpendicular to this direction.

Construction of polaroid:
A clear plastic sheet of long chain molecules of polyvinylalcohol (PVA) is
heated and stretched in a direction to many times than its original length. Then,
PVA molecules are aligned along the stretching direction. The stretched PVA
sheet is then exposed to iodine vapour. The iodine atoms are attached to the
long PVA molecules (replacing OH) and form conducting parallel chains.

Mechanism of polarization: Molecules

Transmission
axis
Ey

Ex

Polaroid

Natural light can be resolved into Ex and Ey components. When Ex and Ey appear
on the polaroid sheet, Ey components will be absorbed by the electrons of iodine
and current will flow along the direction of chain or electron collides with
lattice atoms, imparting energy to them and thereby heating the sheet. In
contrast, electrons do not move along x-direction as there is no conduction path
between parallel molecules. Thus Ex component is not absorbed, creating a
transmission axis along x, which we call polarizing direction.
Moreover, accelerating electrons radiate in forward and backward direction.
The forward wave is cancelled out with incident wave. Only backward or
reflected wave is wave is obtained.
 (ii) Tourmaline:

T1 Parallel T2

T1 Crossed T2

Tourmaline crystals have the property of selectivity absorbing one of the two
components Ex or Ey of light. Dichroism exhibited by a number of minerals or
organic compounds. When a beam of unpolaraized light is sent through a
tourmaline crystal T1, the transmitted light is plane-polarized that can be
verified by a second slab T 2. When T1 and T2 are parallel, the light transmits to
T2 is same as that transmitted by T 1. When T1 and T2 are perpendicular, no light
passes to T2.

Malus’ law:
Ey θ
Ey

Ex
I
Iy
Polarizer, P1
Analyzer, P2
θ Ey
Consider that unpolarized light is incident on a polarizer P 1. Ex component of
unpolarized light is blocked and Ey component passes through P1. The intensity
of Ey is Iy. Let us place a second polaroid (analyzer) P 2 whose transmission axis
makes an angle θ with the direction of Ey.
Then, perpendicular component to transmission axis, E´x=Eysinθ will be blocked,
and parallel component E´y=Eycosθ will be transmitted through P2. If I is the
finally transmitted light, then,

´2 2 2
I ∝ E y or I =k E y cos θ
2
and I y =k E y
2 2
I k E y cos θ
Therefore, = 2
Iy k Ey

2
I =I y cos θ

I is maximum when θ=0º or 180º and I is minimum when θ=90º or 270º. The
above equation describes a lack of symmetry about the propagation direction.
Thus, if rotate P2, we will see a variation of light intensity through the analyzer.
This type of behavior is exhibited by plane polarized transverse wave.
Longitudinal wave does not show it.

Note:
For unpolarized light θ=0 to 2π are equally probable except transmission axis.
In this case, we have to take average of cos2θ.
 2π 2π
1 1 1+ cos 2 θ
¿ cos θ≥ ∫ cos θdθ= ∫
2 2
dθ
2π 0 2π 0 2

[ ]
2π
1 sin 2 θ 2π 1
¿ cos 2 θ≥ θ+ = =
2π 2 0 4π 2

So, I=Io/2, The intensity of a plane polarized light is half the original intensity.

Problem 1: Find the angle of refraction of light which is incident on a crown

glass slab of refractive index 1.52 at the polarizing angle.

Solution: Here, µ=1.52

We know, μ=tan θ p θ p =tan−1 μ=tan−1 ( 1.52 )=56.7 °

Now, θp+θr =90°, therefore, θr =90°‒56.7°=33.3°

Problem 2: Two polarizing sheets have their polarizing directions parallel so

that the intensity of the transmitted light is a maximum. Through what angle
must either sheet be turned if the intensity is to drop by one half?

Solution: Suppose, Iy=maximum intensity, then I=Iy/2

Iy
I =I y cos θ
2
or 2
2
=I y cos θ

2
cos θ=
1
2 or cosθ=±
√ 1
2
θ=± 45 ° , ± 135 °

Problem 3: How will you orient a polarizer and an analyzer so that a beam of
natural light is reduced to half of its original intensity after passing through the
analyzer?

Solution: We know,
 Io Io
cos θ . Here,
2 2
I =I y cos θ= I=
2 2
Io Io
= cos θ . or or θ=0 °
2
So, cosθ=1
2 2

Problem 4: A beam of light is passed through a calcite crystal. If the plane of

vibration of the incident beam makes an angle of 30º with the optic axis,
compare the intensities of the extra-ordinary and ordinary light.

Solution: Here,
2
I e =I y cos θ
2
and I o=I y sin θ
I e cos 2 θ
= 2 =cot θ . So,
2
I e =3 I o
I o sin θ

Problem 5: Prove that two plane polarized light wave of equal amplitude, their
planes of vibration being at right angles to each other, cannot produce
interference effects.

Solution: Let the equation of two light weaves be:

E1=E 01 cos ⁡(kx−ωt +∅ 1)

E2= E02 cos ⁡(kx−ωt +∅ 2)

Here, the electric field vector, ⃗E =⃗E 1+⃗E 2 and the intensity of light, I=¿ E2 >¿ ,
E =⃗
E.⃗
E =( ⃗
E 1+ ⃗
E 2) . ( ⃗
E 1+ ⃗
E2 )
2
where
E = E1+ E2 +2 ⃗
E1. ⃗
2 2 2
E2

I =I 1+ I 2 + I 12=¿ E 1> +¿ E12> ¿+¿ 2 ⃗

E1 . ⃗
2 2
and E2 >¿

The last term of the above equation is the interference term, because it contains
the phase angle. Since, ⃗E1 and ⃗E2 are perpendicular to each other, then ⃗E1 . ⃗
E 2=0.

So, I =I 1+ I 2, where,
 2 T 2
E 01 E01
I 1=
T 0
∫ 2
cos ( ks−ωt+∅ 1 ) dt=
2
2 T 2
E 02 E02
Similarly, I 1=
T 0
∫ 2
cos ( ks−ωt+∅ 2 ) dt=
2

2 2
E01 E02
So, I= + , This term is independent on phase angle. Hence
2 2
interference does not occur.
Double refraction:
In some materials, the speed of light or index of refraction is independent of the
direction of propagation in the medium and also independent of the state of
polarization of light. Liquids, amorphous solids (glass), polycrystals and cubic
crystals (cubic symmetry) normally show this behavior and are said to be
optically isotropic. But many other crystals are optically anisotropic. Examples,
(i) Mica cleaves in one plane only, (ii) A cube of graphite crystal shows
different resistances between pairs of opposite faces, (iii) A cube of nickel
crystal shows different magnetization in different directions, and (iv) Calcite or
quartz as discussed earlier. Other examples, Ice (H 2O), Quartz (SiO2), Wurtzite
(ZnS), Dolomite (CaO.MgO,2CO2) and Siderite (FeO.CO2).

e-ray

o-ray

e-ray

o-ray

A B
 109º θe e-ray

71º o-ray
θo

Figure shows a beam of unpolarized light falling on a CaCO3 (calcite) crystal at

right angles to the face. The single beam splits into two at the face. This double
bending of one beam through the crystal is called double refraction. Here, o-ray
is called ordinary-ray and e-ray is called extraordinary ray. Huygens, in 1678,
discovered that o- and e-rays have their planes of vibration at right-angle to
each other. He also found that o-ray obeys Snell’s law e-ray does not.
According to Figure A, θ=0, but θr ≠0 for e-ray. Similar reasons occur in
Figures B and C. e-ray does not lie in the plane of incidence. The reason that e-
ray does not follow Snell’s law but o-ray does is as follows: (i) For o-ray, the
c
speed of light vo is same in all directions, so μo = v =constant. (ii) For e-ray, the
o

c
speed varies from vo to a larger value ve with directions, thus μe = v ≠constant,
e

but changing. Minimum value of ve is vo.

Following is a table of μo and μe :

Crystal type Sample Formula μo μe μo −μ e

name
Uniaxial positive Ice H2O 1.309 1.313 +0.004
crystal Quartz SiO2 1.544 1.553 +0.009
Wurtzite ZnS 2.356 2.378 +0.022
Uniaxial negative Calcite CaCO3 1.658 1.486 ‒0.172
crystal Siderite FeO.CO2 1.875 1.635 ‒0.240
Origin of double refraction:
Consider an anisotropic crystalline substance. Its binding force between positive
charge and electrons will be anisotropic. Thus we consider 3 different binding
forces along x, y and z axes that are represented by springs of 3 stiffness as
shown in Figure (here, spring constant k varies).

z
- y
-
- + - x

-
-
An electron placed at the end of each set of spring is displaced from its
equilibrium position. The, it will oscillate with different frequency, as

represented by ν= 1
2π √ k
m
. Now light propagates in a substance by exciting the

atoms of it. The electrons are driven by oscillating E-fields and the atoms
radiate. These secondary wavelets recombine and the resultant waves moves on.
c
The speed of wave (v) or refractive index ( μ= v ¿is determined by the frequency

( v=νλ ) of E and the natural frequency of atoms. Due to the anisotropy of

binding force, an anisotropy of µ occurs. For example, the force on an electron

√
is F=eE and v= F . v changes if F changes and hence µ changes. Besides, the
m
speed of wave is governed by the direction of E. If E is parallel to stiff spring
 F
(x-axis), i.e. along strong binding (if radius of spring R is large, then stress σ= A
F
=less; since σ is proportional to strain, so Δx small. hence k= Δx =large), the

electron’s natural frequency will be high along x axis (as, ν o ∝ √ k ¿. In contrast,

with E along y-axis, where the binding force is smaller, ν o is weaker. If the
binding force along y and z axes same and x-axis different, then two indices of
refraction (µ) occurs. A material of this sort, which displays two µs, is said to be
birefringent. If the ω of E is nearly equal or slightly larger than ω o of oscillation
along y-axis, then E component along y-axis is absorbed but along x-axis
transmitted. A birefringent material that absorbs either E-perpendicular or E-
parallel is called dichroic. Suppose that the crystal symmetry is such that the
binding force along y and z-axes are identical, then x-axis defines the optic axis
in the crystal (the optical properties w.r.t. optic axis along other directions
same). Thus, if light moves along optic axis (E in yz-palne), it is strongly
absorbed as ω = ωo. But if it moves along other axes, it comes out as linearly
polarized light.

Optic Axis:
The optic axis is the direction in uniaxial crystal along which the e-ray and o-
ray travel with the same speed and hence double refraction does not take place
if light passes along this direction.

Theory of Production of Plane, Circularly and elliptically polarized light:

Optic axis

e-ray

Eo

o-ray e-ray
o-ray
Plane Polarized Light: When the light has unideriectional vibration, it is known
as plane or linearly polarized light.
Circularly Polarized Light: When two plane polarized light waves are allowed
to superimpose and the tip of the resultant electric vectors traces a circle, this
light is known as circularly polarized light.
Elliptically Polarized Light: If the magnitude of the resulting electric vector
varies periodically during its rotation and its tip traces an ellipse, the resultant
light is called elliptically polarized light.
Consider a calcite crystal inside which light is passing. Its refracting faces are
cut parallel to optic axis. The incident light has linear vibrations of amplitude
Eo, making angle θ with optic axis. This light can be resolved into two
components Eocosθ (e-ray) and Eosinθ (o-ray). As per Hugen’s theory, both ray
travel in the same direction with different velocities. Calcite is a negative crystal
where e-ray is faster than o-ray. So, a path difference occurs between them and
hence a phase difference occurs. Let e-ray has ϕ phase difference than o-ray.

Now the incident light wave= E o sin ⁡(ωt−kx)

So, the e-ray along optic axis is E x =Eo sinθsin(ωt−kx +ϕ ¿
= E1sin(ωt−kx ¿ ....................................(1)
and the o-ray along y-axis is E y =Eo cosθ sin(ωt−kx ¿
= E2sin(ωt−kx ¿ ........................................(2)

Ey
From eq(2), sin(ωt−kx ¿= E ....................................(3)
2

√
2
Ey
and cos(ωt−kx ¿= 1− 2 ...............................(4)
E2
 Ex
From eq(1), E =sin ( ωt −kx ) cosϕ+cos ( ωt −kx ) sinϕ
1

√
2
E E
= y cosϕ+ 1− 2y sinϕ
E2 E2

or,
Ex E y E2
− cosϕ=¿ 1− y2 sinϕ
E1 E 2 E2 √
( )
2 2
E E E
or ( x − y cosϕ) = 1− 2y sin 2 ϕ
E1 E 2 E2

( )
2 2 2
Ex E y 2 Ex E y Ey 2
or, 2
+ 2 cos ϕ−2
E E
cosϕ= 1− 2 sin ϕ
E1 E2 1 2 E2
2 2
Ex E y Ex E y 2
or, 2
+ 2 −2 cosϕ=sin ϕ.......................................(5)
E1 E2 E1 E2

This is a general equation of an ellipse.

Special cases: ϕ depends upon the crystal thickness “t” of the crystal plate.
(i) If the thickness of the plate is such that ϕ=0, 2π, 4π etc. between o-
and e-rays, then sinϕ=0 and cosϕ=1. So, eq(5) becomes
2 2
Ex E y Ex E y
2
+ 2 −2 =0
E1 E2 E1 E2
2
Ex Ey
or, ( − ) =0
E1 E 2
E2
E y= E ..................................................(6)
E1 x

This is a equation of a straight line. This concludes that the emerging

light is plane or linearly polarized.
(ii) If the thickness of the plate is such that ϕ=π, 3π, 5π etc., then sinϕ=0
and cosϕ=-1, So eq(5) becomes
−E 2
E y= E ..................................................(7)
E1 x

This equation also shows plane polarized light.

 π 3π 5π
(iii) If the thickness of the plate is such that ϕ= 2 , 2 , 2 etc., then sinϕ=1

and cosϕ=0, So eq(5) becomes

2 2
Ex E y
2
+ 2 =1
E1 E2

This is an equation of an ellipse. The emergent light will be elliptically

polarized.
π 3π 5π
(iv) If the thickness of the plate is such that ϕ= 2 , 2 , 2 etc., and E1=E2,

then
2 2
E x + E y =1

This is an equation of a circle. The emergent light will be circularly polarized.

Problem 6: A beam of circularly polarized light falls on a polarizing sheet.

Describe the emerging beam.
y

Em
Ey
θ
x
Em Ex Em

Em
Consider that a circularly polarized light falls on a sheet. As it enters the sheet,
the circularly polarized light can be represented by:
E x =Em sinωt

E x =Em cosωt

Here, Em is the amplitude of light wave; x and y are arbitrary perpendicular

axes. The resultant of circular polarization:
Ecp =¿ √ E + E =√ E
2
x
2
y
2
m ¿¿¿

2
The resultant intensity is: I cp ∝ Em

Let the polarizing direction of the sheet make an arbitrary angle θ with x-axis.
The instantaneous value of the plane-polarized wave transmitted the sheet is:
E=E y sinθ+ E xy cosθ

¿ Em cosωt sinθ+ Em sinωt cosθ

¿ Em sin ⁡(ωt +θ)

2 2
Transmitted intensity, I trans ∝ Em sin (ωt +θ)

Average value of sin2 ( ωt +θ ) is:

T
⟨ sin (ωt+θ) ⟩= 1 ∫ sin2 ( ωt +θ ) dt= 1
2
T 0 2
2
1
So, I trans ∝ E
2 m

Our eyes respond to this intensity. Now,

2
1
I trans= kE
2 m
1
¿ I
2 cp
Since I trans does not contain θ, so a single polarizer cannot be used to distinguish
unpolarized or circularly polarized light.
Retardation Plates
Retardation plates or retarders are a class of optical elements that serve to
change the state of polarization of an incident wave.
When plane polarized light is incident on a retardation plate, it splits the light
into two plane polarized waves (ordinary and extra-ordinary waves) and one of
the waves lags behind the other by a known amount. Upon emerging from the
retarder, two waves superpose on each other to produce a wave, which is of
different state of polarization.

Types of retardation plates

There are two types of retardation plates
(i) Quarter wave plate
(ii) Half wave plate
Examples: Quartz crystals and biaxial mica

Quarter wave plate

A quarter wave plate is a thin plate of birefringent crystal having the optic axis
parallel to its refracting faces and its thickness are so adjusted that it introduces
a quarter wave (λ/4) path difference (or 90º phase difference) between the e-ray
and o-ray propagating through it.
Let,
µe is the refractive index of e-ray, µo is the refractive index of o-ray, and d is the
thickness of the crystal. Then, the optical path difference= µed‒ µod
λ
By definition, (µe‒ µo)d= λ/4=∆ , So, d= 4(μ ‒ μ ) .
e o

2π 2π λ
Phase difference, δ= λ ×∆= λ × 4 = 90º

Quarter wave plate produces elliptically and circularly polarized light.

Half wave plate
A half wave plate is a thin plate of birefringent crystal having optic axis parallel
to its faces and its thickness are so adjusted that it introduces a half wave (λ/2)
path difference (or 180º phase difference) between e-ray and o-ray.
λ
By definition, (µe‒ µo)d= λ/2=∆ , So, d= 2(μ ‒ μ ) .
e o

2π 2π λ
Phase difference, δ= λ ×∆= λ × 2 = 180º

Example 7: A quartz quarter wave plate is to be used with sodium light of

wavelength 5490 Å. What must be its thickness? Refractive index for o-ray is
1.544 and that for e-ray is 1.553.

Solution: Let the crystal thickness be d, then

λ
by definition, d=
4(μ e ‒ μ o)
5890
¿
4 (1.553 ‒ 1.544)

= 0.016 mm
Problem 8: How may a circularly polarized light be verified?

Solution: Insert a quarter-wave plate. If the beam is circularly polarized, the two
components (e- and o-rays) will have a phase difference of 90º between them.
The quarter wave plate will introduce a further phase difference of ± 90 ° so that
the emerging light will have a phase difference of either 0º or 180º. In either
case, the light will now be plane polarized and can be made to suffer complete
extinction by rotating a polarizer in its path.

Problem 9: A plane polarized light wave of amplitude Eo falls on a calcite

quarter-wave plate with its plane of vibration at 45º to the optic axis of the plate,
which is taken as y-axis (Figure). The emerging light will be circularly
polarized. In what direction will the electric vector appear to rotate? The
direction of propagation is out of the page.

y Optic axis y

Em Em

x x
Em Em Em ωt1 Em

45º 45º
t=0 t=t1
Em Em

Plane of vibration

The wave component whose vibrations are parallel to the optic axis (the e-
wave) can be represented as it emerges from the plate as
1
E y =( Eo cos 45 ° ) sinωt= E sinωt=Em sinωt
√2 o
The wave component whose vibrations are at right angles to the optic axis (the
o-wave) can be represented as it
−1
E x =( E o sin 45° ) sin ( ωt −90 ° )= E cosωt =−Em cosωt
√2 o
The 90º phase shift is due to the action of the quarter-wave plate. Note that Ex
reaches its maximum value ¼ of a cycle later than Ey does, for, in calcite, wave
Ex travels more slowly than Ey.
To decide the direction, consider the tip of E at (a) t=0 and (b) t=t1 time and
choose that ωt1 is small angle. At t=0,
E y =0 and E x =−Em (as shown in left Figure)
At t=t1, E y =Em sinω t 1 ≅ E m ω t 1 (as shown in right Figure)
E x =−Em cosω t 1 ≅ −E m
The E vector rotates clockwise direction.

Analysis of polarized light:

Light may exhibit (i) unpolarized, (ii) polarized and (iii) mixed type of waves.
Our eye cannot distinguish these. Using a polarizer and a quarter wave plate, the
type of polarization of light can be understood. The following steps are used:
(i)
Polarizer
Imax
Rotation

I=0. I=0.
Unknown light
Tel. Eye.
Imax

Unknown light is allowed to fall on a polarizer. The polarizer is then rotated

through a full circle and the intensity of the transmitted light is observed. If the
intensity is extinguished twice in one full rotation of the polarizer, then the
unknown light is plane polarized

(ii)
Polarizer
Imax
Rotation

Imin Imin
Unknown light
Tel. Eye
.
Imax
If intensity of transmitted light varies between a maximum and a minimum
value but does not become extinguished in any position of the polarizer, then
the unknown light is either eliptically polarized or partially polarized.

(iii)

Polarizer
I
Rotation

I I.
Unknown light
Tel. Eye.
I
If intensity of transmitted light remains constant on rotation of the polarizer,
then the unknown light is either circularly polarized or unpolarized.

(iv)
Polarizer
I
QWP Rotation

I=0 I=0

Tel. Eye.
Partially or elliptically polarized
I

If the incident light is eliptically polarized, the quarter wave plate converts it
into a plane polarized beam. When this plane polarized light is passed through a
polarizer, it will be extinguished twice in one full rotation of the polarizer. On
the other hand, if intensity becomes maximum and minimum without being
zero, then the incident light is partially polarized.
On the other hand, if intensity becomes maximum and minimum without being
zero, then the incident light is partially polarized.

(v)
Polarizer
Imax
QWP Rotation

I=0 I=0

Tel. Eye
Unpolarized or circularly .
Imax
elliptically polarized

If the incident light is unpolarized or circularly polarized, the quarter wave plate
converts it into a plane polarized beam. When this plane polarized light is
passed through a polarizer, it will be extinguished twice in one full rotation of
the polarizer. On the other hand, if intensity stays constant, then the incident
light is unpolarized.
Optical activity:
When a linearly polarized light propagates through an “optically active”
medium, like sugar solution, its plane of polarization rotates.
“The property by which the plane of polarization of light rotates about its
propagation direction passing through an optically active substance is known as
the optical activity.”
The rotation of the plane of polarization is due to the fact that the modes of the
optically active substance are left circularly polarized (LCP) and right circularly
polarized (RCP) which propagate slightly with different velocities. Here, modes
mean that LCP light incident on the substance will propagate as an RCP;
similarly RCP will propagate as an LCP but with different velocity.
 Examples of optically active substances
(i) Sugar solution, (ii) turpentine and (iii) alcoholic solution
Optically active substances are of two types:
(i) Right-handed or dextro-rotary: The substances that rotate the plane of
polarization to the clockwise direction. Example; Cane sugar
(ii) Left-handed or laevo-rotary: The substances that rotate the plane of
polarization to the anti-clock wise direction. Example: Fruit sugar.

This rotation occurs as a result of the helical structure of sugar molecules. The
method of determining the concentration of sugar solution by measuring the
rotation of the plane of polarization is a widely used method in industry.

Laurents Half-Shade Polarimeter:

Definition: A device designed for the accurate measurement of the angle of
rotation of the plane of polarization of a plane polarized light by an optically
active medium is known as a polarimeter.
A sccharimeter is a kind of polarimeter to measure percentage of can-sugar (or
can-juice) in the solution. For this purpose, the polarimeter remains calibrated.
Polarimeter has medical application to find concentration of sugar in urine to
diagnose diabetes.

L Circular scale
N1 HSP T N2
O

Construction: Figure depicts the parts of a Laurent’s half shade polarimeter. O

is a monochromatic source of light of wavelength λ. L is a convex lens placed to
produce parallel beam. N1 and N2 are two Nichol prisms, known as polarizer
and analyzer, respectively. N2 is capable of rotating about a common axis of N1
and N2 while N1 is permanently set. Rotation of N2 is observed by a circular
scale. Vernier scale provides accuracy of reading. Light after passing through
N1 becomes plane polarized light with its vibration in the principle plane of N1.
The polarized light passes through a half-shade plate (HSP) and then through a
tube T containing sugar solution. Normally T is hollow with a large diameter,
but when it is filled with solution care must be taken to avoid air bubble in the
path of light. T is closed at both ends by cover-slips. The emergent light after
passing N2 is viewed by a telescope which is focused on the half shade.

Action of half shade:

A Optic axis Y
R P P
θ θ R θ θ

D B X´ X
O N O M

Q S
C
Y´

Glass Quartz

If an optically active substance is kept between two initially crossed Nichols N1

and N2, the field of view is not dark. To make view dark N2 is rotated. Even
after rotation, the view does not become dark. Hence the measurement becomes
inaccurate. To overcome this difficulty, a half shade plate is used. The plate
consists of semicircular glass ADC and a semicircular half wave quartz crystal
ABC (cut parallel to the optic axis) so that ABC introduces a phase change of π
between e- and o-rays. These two plates are cemented along diameter AC. The
thickness of the glass plate is so adjusted that it absorbs same amount of light as
the quartz plate.
Consider that plane of vibration of a plane polarized light incident normally on
HSP is along PQ. PQ makes an angle θ with AC. The vibrations emerge from
the glass plate without any change i.e. along PQ. In quartz plate, light is divided
into e- and o-rays. Let o-ray along XX´ and e-ray along (parallel to) optic axis
YY´. They travel in the same direction with different speeds. So, there is a path
difference and hence a phase difference between e- and o-rays. Since the wave
plate is half wave plate (i.e. λ/2), it corresponds to a phase π. The phase change
between e- and o-rays is π. Due to this phase change, the direction of o-ray is
reversed. If the original o-ray has vibration along OM, then after HSP, its
vibration is along ON. So, the resultant vibration of e-ray and o-ray is along OR
that makes θ with Y axis.
Now, if the principal plane of Nichol N2 is parallel to PQ, then light from glass
portion will be unobstructed and light from quartz will be partially obstructed.
Thus glass part will be brighter than quartz part. If principal plane of N2 is
parallel to RS, light from quartz part will be unobstructed but that from glass
part will be partially obstructed. When incident light vibration is parallel to RS,
its resultant after emerging HSP will be along OP. Thus, if the principal plane of
N2 is along PQ, the light from quartz portion will pass unobstructed while from
glass part is partly obstructed. Thus quartz becomes brighter than glass. If the
principal plane of N2 is parallel to Y-axis, it is equally inclined to two plane
polarized light (OP and OR or OS and OQ), so the field of view will be equally
bright or dark. So, the half shade serves the purpose of dividing the field of
view in two halves. When N2 is slightly rotated from the position of equal
brightness, a marked change in the intensity of two halves is easily seen.
(Calcite is a negative crystal; here v e>vo or µo>µe. Quartz is a positive crystal;
here vo>ve or µo<µe.)

Specific rotation:
If an optically active substance is placed between a polarizer and an anlyzer and
if the analyzer is rotated, then the field of view changes from bright to dark or
vice versa. The angle of rotation is equal to the angle through which the plane of
polarization is rotated by the optically active substance. This angle depends on:
(i) the thickness of the substance (l), (ii) concentration of the solution (c),
(iii) the wavelength of light, and (iv) the temperature.
For fixed λ and T, rotation angle, θ ∝l and θ ∝ c. So, θ=slc
θ
s=
lc

“The specific rotation for a given wavelength of light at a given temperature is

conventionally defined as the rotation produced by one decimeter long column
of the solution containing 1 gm of optically active material per cc of solution.”

Here, l is known, c is known and θ is determined from the apparatus.

Problem 10: Calculate the specific rotation of sugar if the plane of vibration is
turned through 26.4º, traversing 20 cm length of 20% sugar solution.

Solution: We know,
10 θ 26.4 × 10
s= =¿ =66
l(cm)c 20 × 0.2