UNIT- I PHYSICAL OPTICS FOR INSTRUMENTS

INTERFERENCE
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Syllabus: INTERFERENCE: Introduction – Interference in thin films by reflection – Newton's rings.
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1. Explain the principle of superposition of waves.

Principle of superposition of waves:-

Statement: when two or more waves travelling simultaneously in a medium, the resultant
displacement at any point is the algebraic sum of the displacements due to individual waves.

If y1 and y2 are the instantaneous displacements of two waves, then the resultant displacement is
given by

y = y1  y2

‘+’ sign has to be taken when two both waves are in same phase and ‘–‘sign when both waves are
out of phase.

Coherence: Two waves are said to be coherent, if they have same frequency and constant phase
difference.

2. Explain the phenomenon of interference.

Interference:
When two or more waves having same frequency and constant phase difference travelling in
the same direction are superimposed with each other, then there is a modification of
amplitude in the region of superposition. i.e. the energy is redistributed in the form of maxima
and minima.

If a is the amplitude of the two waves, then the resultant amplitude for constructive
interference is

i.e. y = a + a = 2a.

and for destructive interference is

i.e. y = a – a=0

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3. Write the conditions for obtaining interference Pattern.

a. Two sources must be coherent.

b. Two sources should be narrow and monochromatic.
c. The separation between the two sources should be small.
d. The distance between the sources and the screen should be large.
e. The background should be dark.

4. With ray diagram discuss the theory of thin films in the condition for constructive
and destructive interference in the case of reflected system.

INTERFERENCE IN THIN FILMS

The colors of thin films, soap bubbles and oil slicks can be explained by using the phenomenon
of interference on the basis of division of amplitude.

Interference in thin films will be discussed in two classes:

i) Interference in uniform thin films.
ii) Interference in non -uniform thin films.

INTERFERENCE IN PLANE PARALLEL FILMS DUE TO REFLECTION (UNIFORM)

Consider two plane surfaces PQ and P′Q′ are separated by a distance ’t’. Let µ be the refractive
index of the film between the surfaces. Let a ray of light OA be incident on the surface PQ at A,
then some part of the light is reflected into air as AR ray, The remaining part is transmitted into the
medium and is reflected at the lower surface P′Q′ at C, meets the upper surface at D and emerges as
DR′ ray as shown in fig.

The path difference between the two reflected rays AR and DR′
 = path[ AC+CD] in medium – [AB] in air 

  µ[AC+CD] – AB --------------(1) 

From ΔAEC, cos r = CE / AC

CE t
AC   ( CE  t)
co
cosr sr

ΔAEC and ΔCED are similar, so 

AC = CD and AE = ED

t t 2t
AC CD    - - - - - - - - - - - - - - - (2)
cos r cos r cos r

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 AB
From ΔABD, sin i 
AD

AB  AD sin i
sin i
We Know that  sin r

sin i   sin r

AB  AD  sin r

AE
From Δ AEC , tan r 
CE
AE  CE tan r  t tan r
 AD  AE  ED  2t tan r ( AE  ED)
Hence AB  2 t tan r   sin r

2  t sin2r
AB  - - - - - - - - - (3)
cos r

Substitute eqn (3) & (2) in eqn (1)

2 t 2 t
   sin 2 r
cos r cos r

⇒
2 t
 1  sin 2 r 
cos r

2 t
  cos 2 r
cos r

  2  t cos r 

According to Stoke’s law, when the light is reflected by denser medium in to rarer medium,

then the reflected ray suffers a phase change of π radians or undergoes a path change of .
2
The path difference between the reflected rays AR and DR′

  2  t cos r -
2
Condition for maxima:
Hence the condition for maxima, i.e. for an air film to appear
bright is
 
  2  t cos r -  n⇒2  t cos r  (2n  1)  where n=0, 1, 2,………
2 2

Condition for minima:

Hence the condition for minima, i.e. for an air film to appear

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  
dark is   2  t cos r -  (2n 1)
2 2
2 t cos r  (2n 1)  
2 2
2 t cos r  (2n  2) 
2
2  t cos r  (n 1) where n= 0, 1, 2, 3………..

or 2  t cos r  n where n= 0,1,2,3………..

5. Explain the colours in a thin film when exposed to sunlight.

Colours in thin films:

 When white light is incident on a thin film, it gets splits up by reflection at the top and
bottom surfaces of the film. These reflected rays interfere producing colors in thin films.
 Different points on the thin film satisfy the condition for the constructive interference i.e.,

2  t cos r  (2n  1) 
2
for different values of and t when illuminated with white light and hence appear
multicolored.
 In case of soap bubble, t is constant and varying values of  and r satisfy the above
condition give multicolor.
 In case of oil slick, r is constant and varying values of  and t satisfy the above
Condition gives multicolor.

6. Discuss the theory of Newton’s rings with relevant diagram

NEWTON’S RINGS: (INTERFERENCE IN NON-UNIFORM FILMS DUE TO

REFLECTION)
When a Plano convex lens with its convex surface is placed on a plane glass plate, an air film of
gradually increasing thickness is formed between them. The thickness of the film at the point of
contact is zero. If a monochromatic light is allowed to fall normally and the film is viewed in
reflected light, alternate dark and bright concentric circular rings are observed around the point of
contact. This phenomenon was first described by Newton, so these rings are called Newton’s
rings.

EXPERIMENTAL ARRANGEMENT:-
The experimental arrangement consists of a Plano- convex lens ‘L’ of large radius of curvature
and is placed on a plane glass plate ‘p’ as shown in fig . The light from monochromatic source is
reflected normally on to the air film by means of glass plate G inclined at 450. A part of the
incident light is reflected by the curved surface of the lens ‘L’ (Ray 1) and remaining is
transmitted. The transmitted light is reflected back from the plane surface of the glass plate ‘P’
(Ray 2) undergoes a path change of λ/2(Stoke’s principle). These two reflected rays are

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 superimposed and produce an interference pattern in the form of bright and dark circular rings.
These rings can be viewed in a microscope ‘M’ focused on the film.

The conditions for the bright and dark rings are governed by the following relations:

The condition for bright ring:


  2μt cos r   n
2

2 μ t cos r  (2n 1) where n = 0,1,2,3 …
2
For normal incidence, cos r = 1 and for the air film µ = 1

2 t  (2n 1) (Bright fringes)
2
The condition for dark ring:

 
 2μ t cos r   (2n 1)
2 2
2μ t cos r  n λ Where n=0,1,2,3…

For normal incidence, cos r = 1 and for the air film µ = 1

2 t  n λ (Dark rings)

7. Derive the expressions for the diameters of nth dark and bright rings by forming Newton’s
Rings.

Theory of Newton's rings:

In the reflected monochromatic light, Newton's rings are alternate bright and dark circles with
a central dark spot. Let R be the radius of curvature of the lens. At Q, let the thickness of the film
PQ = t satisfies the condition for a dark ring to form by interference. Let it be an n th dark ring with
a radius, SQ = rn.

Determination of radius of n th dark ring

Consider an n th dark ring with radius HE.

From the diagram

OE = R – t, HE = rn , OH = R
From the Δ OHE

OH2 = HE2 + OE2

R2 = rn 2 + (R-t) 2

rn 2 = R2 - (R-t) 2

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 rn 2 = R2 - R2 - t2 + 2Rt

rn2  2R t  t2 (as 2 R t >> t2, t2 can be neglected) 


 rn2  2R t . 

For dark rings, the condition is 2t = n λ

rn2  n R or 
rn nR
The diameter of the dark ring is therefore given by
Dn  2 rn = 2 n λ R or Dn  4n λ R
Dn  n

Thus, the radii (also diameters) of the dark rings are proportional to the square root of
natural numbers.

Determination of radius of n th bright ring

Let us suppose that a bright ring is located at the point Q. The radius of the nth bright ring is
given by rn2  2R t.

For bright rings, the condition is

λ
2 t  2n  1 2

2 Dm

r2  2n 1 R or r  2n  1λ R .

n n 0
2 2 No. of ring, m

Dn  (2n  1)2R

It is clear that the radii (also diameters) of bright rings are proportional to the square root of
the odd natural numbers.

8. Why a Dark spot is observed at the center of the Newton’s rings?


The thickness of the air film at the point of contact is zero i.e. t = 0, hence   which is the
2
condition for destructive interference. So a dark spot is observed at the center of the Newton’s rings.

9. Why Newton’s rings are circular?

In Newton’s rings arrangement, a thin air film is enclosed between a Plano-convex lens and a glass
plate. The thickness of the air film at the point of contact is zero and gradually increases as we move
outward. The locus of points where the air film has the same thickness then fall on a circle whose
centre is the point of contact. Thus, the thickness of air film is constant at points on any circle having
the point of lens-glass plate contact as the centre. The fringes are therefore circular.

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10. Determine the wavelength of light source and Radius of curvature of given convex lens in
Newton’s Rings experiment.

The wave length of incident monochromatic light can be determined by forming Newton's rings
and measuring the diameters of the dark rings using travelling microscope.

For mth dark ring, D2m = 4mλR

Similarly, for nth dark ring, D2n=4nλR

D2m - D2n = 4mλR - 4nλR
D2m - D2n = 4(m-n)λR

Dm  Dn
2 2
 This is an expression for wavelength of incident light.
4(m  n) R

D  Dn
2 2
R m This is an expression for Radius of curvatureof lens..
4(m  n)


PROBLEMS
1. A parallel beam of light of wavelength 5890 A0 is incident on a thin glass plate (μ = 1.5) such
that the angle of refraction into the plate is 600. Calculate the smallest thickness of the glass
plate which will appear dark in reflected light.

The condition is given by 2 t Cos r = m. Taking m = 1, the smallest thickness of plate that
causes destructive interference is

 5890  1010
t   0.39 μm .
2 Cos r 2  1.5  Cos 600

2. A Newton’s ring arrangement is used with a source emitting two wavelengths and
2  4.5  10 5 cm . It is found that nth dark ring due to 1 coincides with (n+1)th dark ring for

2 If the radius of curvature of the curved surface is 90cm, find the diameter of nth dark ring for
1 .

Solution : r  m R , m = 0,1,2,.... for dark ring. Hence rn  n  6 10 5  90 and

rn1  (n  1)  4.5  10 5  90 .

n x 6 x 10-5 x 90 = (n+1)4.5 x 10-5 x 90

6n = 4.5 (n+1) or 6n = 4.5n + 4.5 or 1.5n = 4.5 or n = 4.5/1.5 = 3

 rn  3  6 10 -5  90  1620 10 5  0.0162  0.127cm

Diameter of nth ring = 0.127 x 2 = 0.254 cm.

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3. In a Newton’s ring experiment, the diameter of the 4th and 12th dark rings are 0.400
cm and 0.700 cm, respectively. Find the diameter of the 20th dark ring.

Solution: We know that D2m  D2n  4(m  n) λR

Here m = 12 , n = 4, and (m-n) =12-4 = 8

Dm =0.700 cm and Dn = 0.400 cm

 D12  D 4  4  8  λR
2 2
1

For 20th and 4th dark rings

D220  D24  4  16  λR 2
Dividing (2) by (1)
D 220  D 24 4  16  λR
 2
2
D12  D 24 4  8  λR
D 220  0.4 2 D 220 - 0.16
or 2 or 2
0.7 2  0.4 2 0.33
or D 220 - 0.16  2  0.33  0.66
D 220  0.66  0.16  0.82 or D 20  0.906 cm.
th
The diameter of the 20 dark ring = 0.906 cm

4. In Newton ring’s experiment the diameter of 10 th ring changes from 1.40 cm to 1.27 cm when
a liquid is introduced between the lens and the plate. calculate the refractive index of the
liquid.

Solution: Given the diameter of 10 th ring in air D10 = 1.40 cm

the diameter of 10 th ring in liquid D’10 = 1.27 cm

the diameter of n th ring in air 𝐷𝑛2 = 4𝑛𝜆R

4𝑛𝜆R
the diameter of n th ring in liquid 𝐷𝑛′ 2 =
μ

𝐷2
From above two equations 𝜇 = 𝐷′𝑛2
𝑛

𝐷2
For 10th ring 𝜇 = 𝐷10
′2
𝑛

(1.40)2
𝜇 = (1.27)2

μ = 1.2152

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 DIFFRACTION

Syllabus: Introduction – Fraunhofer diffraction - Fraunhofer diffraction at double slit (qualitative) –

Diffraction grating – Grating spectrum – Resolving power of a grating – Rayleigh’s criterion for resolving power.
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1. What is Diffraction?
The phenomenon of bending of light waves around the edges of obstacle and
spreading of light waves into the geometrical shadow of an obstacle placed
in the path of light is called diffraction.

In general, diffraction of waves becomes noticeable only when the size of

obstacle is comparable to the wavelength.

λ≅d

Examples:
 The spaced tracks on a CD or DVD act as a diffraction grating to form the familiar rainbow
pattern.
 Sound produced inside one room reaches the other room after bending the edges of the doors.

2. Explain Huygen’s wave theory.

HUYGEN’S WAVE THEORY
Let a primary wave front XY from point source S of
monochromatic light reach the slit placed in its path. The
light passing through the slit has to cause brightness inside
AB on the screen and regions beyond A and B have to be
dark i.e. regions of geometrical shadow. According to
Huygens wave theory, every point on the primary wave
front acts as a secondary source of disturbance and
secondary wavelets are generated from these points. These
secondary wavelets enter into the geometrical shadow
region and interfere to produce diffraction pattern beyond
A and B .

3. Write the differences between Fresnel and Fraunhofer diffraction.

1) Fresnel diffraction . 2) Fraunhofer diffraction

Fresnel diffraction:

1. In Fresnel diffraction, a point source or an illuminated narrow

slit at finite distance is used.
2. The incident wave front is either spherical or cylindrical.
3. The source and screen are at finite distance from the aperture
producing diffraction.
4. Lens is not used to focus the rays.
5. The centre of diffraction pattern may be bright or dark.
6. Mathematical investigations are complicated and approximate.

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 Fraunhofer diffraction:

1. In Fraunhofer diffraction, an extended source at

infinite distance is used.
2. The incident wave front is plane wave front.
3. The source and screen are at infinite distance from
the aperture producing diffraction.
4. Converging lens is used to focus the rays.
5. The centre of the diffraction pattern is always bright.
6. Mathematical investigations are easy and accurate.

4. Discuss in detail Fraunhoffer diffraction at a double slit with suitable diagram and draw the
intensity distribution curve.

Consider AB and CD be two parallel slits of equal width e and are separated by a distance d.

The distance between the corresponding points of the two slits is (e+d). Let a parallel beam of
monochromatic light of wavelength λ be incident normally upon the two slits. The diffracted light be
focused by a convex lens L on the screen XY as shown in figure.

Explanation: By Huygens’s principle, every point in the slits AB and CD sends out secondary
wavelets in all directions. From the theory of diffraction at a single slit, the resultant amplitude due to
the wavelets diffracted from each slit in direction θ is

𝐴 𝑆𝑖𝑛∝
R=
∝

𝜋𝑒 𝑠𝑖𝑛𝜃
Where A is constant and ∝ =
𝜆

𝐴 𝑆𝑖𝑛∝
let A and C be the points of the slits sending waves of amplitude in a direction θ.
∝

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∴ The resultant amplitude at a point P on the screen will be the result of interference between the
𝐴 𝑆𝑖𝑛∝
waves of the same amplitude ∝
and having a phase difference ø

Let us draw AK perpendicular to CK. The path difference between the wave lets from A and C in the
direction ‘θ’ is (e+d) sin θ i.e. CK= (e+d) sin θ
2𝜋
Hence the corresponding phase difference, ø = 𝜆
𝑥 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓 .

2𝜋
∴ø= 𝜆
(e+d) sin θ.

The resultant amplitude R at P can be determined by the vector amplitude diagram,

OB2=OA2+AB2+2(OA)(AB) cos OAB

 A sin    A sin    A sin   A sin  
2 2
R2       2   cos 
          

 A sin  
2
R2    (2  2 cos  )
  

A 2 sin 2  
I  R2  4 cos 2
 2
2

Thus the resultant intensity at any point on the screen is given by the product of two factors i.e.

 A sin  
2

a) The diffraction term   gives a central maximum in the direction θ = 0 having

  
alternately minima and secondary maxima of decreasing intensity on either side.

The minima are obtained in the directions given by

Sin α=0

   n

e sin 
  n


e sin   n where n=1,2,3…..

The position of secondary maxima appear at

3 5 7
  , , ..............
2 2 2


b) The interference term cos 2 gives a set of equidistant dark and bright fringes.
2

The bright fringes are obtained in the directions given by


cos 2 1
2

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
 n
2

 e  d sin 
 n


e  d sin    n Where n=0, 1, 2, 3,…..

The dark fringes are obtained in the directions given by


cos 2 0
2

 
 (2n  1)
2 2

 e  d sin 
 n


e  d sin   (2n  1)  Where n=0, 1, 2, 3,…..

2

Therefore the intensity distribution of Fraunhoffer diffraction due to double slit is the product of a
 A sin  
2

constant term 4, diffraction term   and interference term cos 2 as shown in figure
   2

5. What is Diffraction Grating? Explain Fraunhofer diffraction at ‘n’ slits with necessary theory.

 A diffraction grating is an optically plane glass plate on which a large number of equidistant
parallel lines are ruled.
 The ruling region becomes opaque and the region between the rulings is transparent.
Therefore diffraction grating is nothing but closely placed multiple (N) slits.
 The combined width of a ruling and a slit is called grating element.
 A good quality of grating contains 15000 lines per inch.

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Let XY represents the grating. The plane of the grating is perpendicular to the plane of paper. In
grating AB,CD, EF etc. represents the slits ,each of width ‘e’ and these slits are separated by equal
opaque regions BC,DE,FG etc. each of width ‘d’.(e + d) is called the grating element.

Let plane waves of monochromatic light of wave length ‘λ’ incident normally on the grating. By
Huygens –principle, every point on the plane wave front acts as a source of secondary wavelets.
These secondary wavelets spread out in all directions.

The secondary wavelets travelling in the same direction of the incident light focus at point O on the
screen. Now consider the secondary waves from each slit traveling in a direction  are equivalent to
A sin 
a single wave of amplitude .


Let us consider the diffracted waves at the corresponding points A and C. AN is the normal drawn to
CN.

The path difference between the waves on reaching the point P is CN.

CN
From  ACN sin  
le
AC

CN=AC sin 𝜃

CN= (e+d) sin 𝜃

2
And the corresponding phase difference is (e  d ) sin   2 (say)


Therefore the resultant amplitude at P in the direction of θ is

A sin  sin N
R
 sin 

The Resultant intensity is

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 A2 sin 2  sin 2 N
I  R2 
2 sin 2 

The first factor gives A sin2  a diffraction pattern due to a single slit, while the second factor
2 2


sin N
2
gives the interference pattern due to N slits. The variation in intensity is due to second
sin 2 
factor and is discussed below.

Case i: Principal Maxima

When sin β =0,

i.e β = ± nπ where n = 0,1,2…….

sin N 0
We have sinN β =0 and thus  which is indeterminate
sin  0

By applying L hospital rule

sin N
Lt  n  N
sin 

Then β = ± nπ

These maxima are most intense and are called principal maxima. These are obtained in the direction

β = ± nπ


(e  d ) sin    n

(e  d)sinθ   n λ

Where n =0,1,2,……we get zero order maximum, first order maximum….

Case 2: Condition for minima

When sinNβ =0, but sinβ ≠0

sin N
 0, then I  0
sin 

These minima are obtained in the directions given by sinNβ =0

Nβ = ± m л


N (e  d ) sin   m


N (e  d ) sin   m

Where m takes all the integral values except 0, N, 2N,…, nN.

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 Case 3: Secondary maxima:

The positions are obtained by differentiating I w.r.t β and equating to 0

dI A 2 sin 2   sin N  N cos N sin   sin N cos 

 2  0
d  2
 sin   sin 2 

N cos N sin   sin N cos   0

TanN  NTan

Sin 2 N N2

sin 2  1  ( N 2  1) sin 2 

A 2 sin 2  N2
I
2 1  ( N 2  1) sin 2 

In an actual grating, N is very large. Hence these secondary maxima are not visible.

6. Analyse the Grating Spectrum obtained when a plane grating is exposed to monochromatic and
multi chromatic light.
Grating Spectrum: Theory of formation of principal maxima

The position of the principal maxima is given by equation.

(e+d) sin  n   n (where n=0, 1, 2, 3, …)

This relation is called grating equation.

The angle of diffraction depends on the values of λ and n.

For a particular wavelength  , the angle of diffraction  is different for different orders of principal
maxima.

The principal maximum occurs at   0 irrespective of the wave length  , which is known as zeroth
order.

The 1st order maxima obtained for n=1 then (e+d) sin 1  

The 2nd order maxima obtained for n=2 then (e+d) sin 2  2 and so on.

Grating spectrum with the monochromatic and white light are shown below:

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 15

 7. Derive an expression for maximum no of possible orders in a diffraction grating.

We know that the positions of the principal maxima are given by the equation

(e+d) sin   n where n= 0, 1, 2, 3,…. .

(e  d ) sin 
n


Where (e+d) is the grating element.

n is the order of maxima.
λ is the wave length of the incident light.
The maximum angle of diffraction is 90o.

Hence maximum possible orders are given by

(e  d ) sin  ed
n  
 

1 1 2.54
n ( N(e  d)  1 inch  (e  d )  inch  cm.)
N N N

1
nmax 
N

8. State and explain Rayleigh’s criterion of Resolution.

RAYLEIGH’S CRITERION OF RESOLUTION

According to Rayleigh’s criterion, two spectral lines of equal intensity should be regarded as
separate that is just resolved, if central maximum of the diffraction pattern due to one coincides
with the first minima of the other and vice-versa.

UN RESOLVED
 If the difference in wavelengths of two spectral lines
is so small, the central maxima corresponding to the
wavelengths come closed.
 Then the resultant intensity curves show a sufficient
overlapping and it is higher than the individual
intensities of both two spectral lines appear as well
resolved.

JUST RESOLVED

 If the difference in wavelengths is smaller such

that the central maximum of one coincides with
the first minimum of the other and vice versa.
 The resultant intensity curve shows a distinct dip
in the middle of the two central maxima as shown in fig.

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 16

 WELL RESOLVED

 The difference in wavelength is such that the

central maximum of two wave lengths is quite
separate.
 There is a distinct point of zero intensity in the
middle of the resultant intensity curve.

9. Define Resolving power of a Grating and derive an expression for it.

Resolving power of a Grating:

The resolving power (R) of a grating is defined as its ability to form two distinctly separate
maxima corresponding to two spectral lines which are very close to each other.

λ
R
dλ

where dλ is the minimum wavelength difference that can be distinguished by a spectrograph and λ is
the mean of the two wavelengths.

Expression for the resolving power of a grating:

Consider the diagram shown in the figure. XY is the transmission grating surface and AB is the
screen. P1 is the position of nth principal maximum of a spectral line of wavelength λ at an angle of
diffraction θ and P2 is the nth principal maximum of a neighboring spectral line of wavelength λ + dλ
at a diffraction angle θ+d𝜃 .

According to Rayleigh’s criterion, these two lines will

appear just resolved if the position of P2 (corresponding
to wave length λ+dλ) will coincide with the first
minimum (of order (nN+1)) of P1(corresponding to
wavelength λ).

The condition for minimum is

N(e+d) Sin 𝜃 = mλ where m has all integral values

except 0, N, 2N, 3N, ……

The condition for first minimum beside the nth Principal maximum at P1 of wavelength λ in the
direction of +d𝜃 is

N(e+d) Sin(𝜃+d𝜃) = (nN+1)λ -------------(1)

The condition for nth Principal maximum is (e+d) Sin 𝜃 = nλ

The condition for nth Principal maximum at P2 of wavelength λ+dλ in the direction 𝜃+d𝜃 is

(e+d) Sin (𝜃+d𝜃) = n (λ+dλ) -------------- (2)

Equating the right hand sides of eqs.1 & 2,

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 17

 λ
nλ  dλ  nλ 
N

λ λ
or n dλ  or  nN
N dλ

λ
The quantity R   nN is a measure of the resolving power of a grating.
dλ

From this equation we note the following:

i) R is independent of the grating element d

ii) R is directly proportional to n, the order of the spectrum
iii) R is directly proportional to the total number of lines (N) on the grating surface.

10. Write the applications of Diffraction:

Diffraction is used

1. To measure the wavelength of spectral lines.

2. To determine the structure of crystals.

3. To find the velocity of the sound.

4. To know size and shape of the tumors etc., inside the human body can be assessed by

Ultrasound scanning.

SOLVED PROBLEMS

1. Calculate the possible order of spectra with a plane transmission grating having 18000 lines per
0
inch when light of wavelength 4500 A is used.
0
Given : d = 18000 lines/inch = 7.09 × 10 lines/m, λ = 4500 A = 4500×10-10 m
5

d sinθ
Solution: Order of spectra, n  . The highest order occurs when θ = 900
λ

d 1 1
 n   3
λ Nλ (7.09  10 lines/m) (4500  10 10 m)
5

2. In a plane transmission grating the angle of diffraction for the second order principal maximum
is 300 for a light of wavelength 5×10-5 cm. Calculate the number of lines/cm on the grating surface

Given: λ= 5×10-5 cm, θ = 300 , m = 2

Solution: The grating equation for normal incidence is nλ  d sin θ

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 18

 nλ 2  5  10 5
 d   2  10 4 cm.
sin θ 0.5

1
Therefore number of lines/cm   5000.
d

λ
The resolving power of a grating is  n N , where N is the total number of lines on a length of 2.5
dλ
cm.
λ 5893  10 -8
 N   491
n dλ 2  6  10 -8

491
Number of lines per cm   196.4
2.5

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 19

 POLARIZATION

Introduction – Types of Polarization – Double refraction – Quarter wave plate and Half Wave plate.

1. What are the different types of polarized light?

Plane or Linearly polarized light
If electric component of light passing through the medium vibrates only along a single direction,
Perpendicular to the direction of propagation, the wave is said to be plane polarized light or
linearly polarized.

The electric vector can be resolved into two rectangular components Ex & Ey. Therefore electric
Vector may be considered as superposition of two mutually perpendicular electric fields.

Circularly polarized light

If the two electric vectors Ex & Ey having same magnitude but vibrating in two mutually
perpendicular planes at a phase difference of Π/2 radians superimpose, the magnitude of the
resultant vector E remains constant about the direction of the propagation. Such light is called
CIRCULARLY POLARIZED LIGHT.

Elliptically polarized light

If the two electric vectors Ex &Ey having unequal magnitudes vibrating in two mutually
perpendicular planes superimpose at a phase difference of Π/2 radians, the magnitude of the
resultant vector E changes with the time and the vector E sweeps a flattened helix in space.
Such light is Called ELLIPTICALLY POLARIZED LIGHT.

2. Explain Brewster's law and Malus Law.

Brewster's law:

When ordinary light is reflected from the surface of a transparent medium like glass, it becomes
partially polarized. The degree of polarization varies with the angle of incidence.

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 20

 Brewster proved that the tangent of the angle of polarization (p) is numerically equal to the
refractive index of the medium.
Unpolarized light Plane polarized light

𝝁 = 𝐭𝐚𝐧 𝒑 (Brewster's law)

From Brewster's law,
Fig. 3.3
sin p sin p p p
  tan p  . From Snell's law,   . Air 0
90
cos p sin r
Glass r

From the above two equations, we get,

sin r = cos p = sin (900 - p)

r  90  p or r  p  90 0

Malus Law:

This law states that the intensity of light transmitted from the analyzer is proportional to the square
of the cosine of the angle between the planes of transmission of analyzer and polarizer.

𝑰 = 𝑰𝒐 𝒄𝒐𝒔𝟐 𝜽

Where 𝐼𝑜 = 𝑎2 the intensity of plane is polarized light incident on the

analyzer and 𝜃 is the angle between the planes of transmission of polarizer
and analyzer.
The intensity of transmitted light is maximum when the planes of polarizer
and analyzer are parallel, i.e. when 𝜃=0, 𝐼 = 𝐼𝑜 (max).
When the two planes are perpendicular, there is no transmitted light i.e.
When 𝜃=90𝑜 , I = 0.

3. Explain the construction and working Principle of Nicol Prism.

Nicol Prism is an optical device used for producing and analyzing plane polarized light.
Principle:
When light is passed through a doubly refracting crystal, it is split up into ordinary ray and
extraordinary ray. Both these rays are plane polarized perpendicular to each other. One of these rays,
usually the 0 - ray, is cut off by total internal reflection.
Construction: A'
A D
Consider a calcite crystal whose length is three times as that
of its width. The end faces of this crystal are grounded in 900
0 S2
such a way that the angle in the principal section becomes 68
and 1120 instead of 710 and 1090.The calcite crystal is cut into S R e-ray
two pieces by a plane perpendicular to the principal section as S1
710 900
shown in fig. The cut faces are grounded and polished 680
optically flat and then cemented together by Canada balsam C
B C'
which is a transparent substance. The refractive index of it is Fig. 3.12 0-ray
in between that of a O-ray and e-ray i.e. for sodium light μ0 =
1.6584,
μcanada balsam = 1.55 and μe = 1.4864.
Working:
When a beam of light enters into Nicol Prism, it is doubly refracted in to o-ray and e-ray. Canada
balsam acts as a rarer medium for an ordinary ray and denser medium for an extra ordinary ray.
Therefore, the o-ray gets completely reflected by total internal reflection and the e-ray emerges out of
the Nicol Prism with vibrations parallel to the Principal plane.
Uses:
Nicol Prisms are used as Polarizers and Analyzers.

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 21

4. Write a short note on Double refraction or birefringence.

Polarization by Double refraction:

When a beam of unpolarized light is

allowed to fall on a calcite crystal or
quartz crystal, it is split up into two
refracted beams. The phenomenon is
called double refraction or birefringence
and such crystals are called doubly-
refracting crystals.
The two refracted rays are plane polarized. The refracted ray which obeys the laws of refraction and
having vibrations perpendicular to the principal section of the calcite crystal is known as the
ordinary ray or the 0 - ray. The other refracted ray which does not obey the laws of refraction and
having vibrations in the principal section is called the extraordinary ray or the e-ray.

5. Explain Quarter wave and Half Wave Plates.

A wave plate is a double refracting uniaxial crystal which introduces specific path difference
between the o-ray and e-ray for a particular wavelength of light.
Consider a double refracting uniaxial crystal cut with its optic axis parallel to the refracting faces.
When a plane polarized light of wavelength λ is incident normally to the surface, the light splits up
into ordinary and extraordinary rays. They travel in the same direction but with different velocities.

Quarter wave plate:

If the thickness of the crystal plate is such that it introduces a phase difference of π/2 radians or the
path difference of λ/4, then it is called a Quarter wave plate.

The thickness‘t’ of the crystal to create a path difference λ/4 is given by

λ
t ( o ~ e ) = .
4

λ
 Thickness of the quarter wave plate, t  .
4(μ 0 ~ μ e )

λ
For a positive crystal (such as Quartz), t .
4(μ e  μ 0 )

λ
For a negative crystal (such as calcite), t .
4(μ o  μ e )

A quarter wave plate is used to convert linearly polarized light into circularly polarized light. So it is
used for the production and detection of circularly polarized light.

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 22

Half - Wave plate:

If the thickness of the crystal plate is such that it introduces a phase difference of  radians or

a path difference of , then it is called a half wave plate.
2

The thickness of the crystal to create a path difference of λ/2

λ
is given by t
2(μ 0 ~ μ e )

λ
For a positive crystal, like quartz, t .
2(μ e  μ 0 )

λ
For a negative crystal (such as calcite), t
2(μ o  μ e )

A half wave plate is used to convert the left handed circularly polarized in to right handed and
vice versa.

SOLVED PROBLEMS

1. Calculate the thickness of a half wave plate of quartz for a wavelength of 5000A0. Given e =
1.553 and 0 = 1.544.

 8
t  5000  10  2.78 103 cm
2(e  0 ) 21.553  1.544 

2. Calculate the thickness of the doubly refracting plate which can produce a path difference of

between the e-ray and o-ray
4
0
Given: λ = 5890 A , μ0 =1.53, μe =1.54

0
λ 5890 A 0
Solution: t   1.45  10 5 A  14.5 μm
4 μ e  μ o  4 1.54  1.53 

3. Plane polarized light passes through a calcite plate with its optic axis parallel to the faces.
Calculate the least thickness of the plate for which the emergent light will be plane polarized.

0
Given: μ0 = 1.6584, μe = 1.4864 and λ = 5000 A

Solution: When a plane polarized light is incident on a half wave plate, the emergent light will also be
plane polarized.

λ 5000  10 -10 m
t   1.45 μm
2 μ o  μ e  2 1.6584  1.4864

Dept of Basic Science, VISHNU INSTITUTE OF TECHNOLOGY, Bhimavaram. Page 23