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Interference and Diffraction

Principle of Superposition of Waves

a. Statement :

When two or more waves arrive at a point in the

medium simultaneously then each wave produces

its own displacement independent of the other

waves. The resultant displacement at that point is

given by the vector sum of individual displacements

produced by each wave.

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Interference of Light

Redistribution of light intensity in the region of medium

is due to the physical process called as “Interference of

light”

Constructive Interference

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Destructive Interference

Path Length and Path Difference

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Condition for Constructive Interference and

Destructive Interference (Or Conditions for

Brightness & Darkness of a point)

a. Condition for Constructive Interference

(brightness):-

1. Overlapping

2. Path difference = n λ = (2n) λ /2

[n = 0, 1, 2, ...........]

3. Phase difference

4. 4. Intensity

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b. Condition for Destructive Interference

(darkness):-

1. Overlapping

2. Path difference =(n – 1/2)λ

= (2n – 1) λ/2

[n = 1, 2, 3, 4..........]

3. Phase difference

4. Intensity

M.C.QQ.1    In interference of light

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          (a.1) Light energy is created              (b.1) Light energy is destroyed          (c.1) Light energy is redistributed     (d.1) Light energy is doubledQ.2    Select the correct statement          (a.2) Only longitudinal waves produce interference.       (b.2) Only transverse waves produce interference.          (c.2) Only standing waves produce interference.   (d.2) Both longitudinal and transverse waves produce interference. Q.3    Two sources of light are said to be coherent if they emit light waves of the same          (a.3) Frequency and speed                                    (b.3) Wavelength and constant phase difference          (c.3) Frequency and amplitude                             (d.3) Intensity and frequency Q.4    Which one of the following quantities is conserved in the interference of light waves?          (a.4) Phase difference     (b.4) Amplitude     (c.4) Intensity        (d.4) Path difference

Young’s Experiments to Demonstrate the

Phenomenon of Interference

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Theory of Interference Band

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(Expression for bandwidth or fringe width of an

interference band.)

Expression for path difference S2P – S1P :-

In ∆PNS2 and ∆PMS. By Pythagoras theorem.

S2P2 = S2N2 + PN2

= D2 + (x + d/2)2 ..................... (1)

S1P2 = S1M2 + PM2

= D2 + (x - d/2)2 ...................... (2)

S2P2 – S1P2 = (x + d/2)2 – (x – d/2)2

= x2 + 2x d/2 + d2/4 – x2 + 2x d/2

– d2/4

= 2xd ..................... (3)

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If x, d < < D, then S1P ≈ S2P ≈ D

Equation (3) becomes.

2D (S2P – S1P) = 2xd

S2P – S1P = xd/D ..................... (4)

For a point P to be bright.

S2P – S1P = (2n) λ/2 .................... (5)

From (3) and (4), xd/D = n λ

x = n λ D/d

Let xn, xn – 1 = distances of nth and n + 1th bright

bands from O.

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Using these in (6),

xn = nλ D/d and xn - 1 = (n + 1)λ D/d

X = Band width of bright band

= xn+1 – xn = (nλ D/d) + (λ D/d) – (n λ D/d)

X = λD/d ..................... (7)

e. For a point P to be dark

S2P – S1P = (2n – 1) λ /2 ..................... (8)

From (3) and (8), xd/D = (2n – 1) λ/2

x = (2n – 1) λD/2d ...................... (9)

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Let xn, xn -1 = distances of nth and n - 1th dark

bands from O. Using these in (9).

Xn = (2n – 1) λ D/2d

Xn – 1 = [2(n + 1) – 1] λD/2d = (2n + 1) λD/2d

X = Band width of dark band

= Xn-1 – Xn

= (2nλD/2d) + (λ D/2d) λ – (2n D/2d) + (λ D/2d)

X = λ D/d ...................... (10)

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Assuming the Expression for Path Difference

Obtain an Expression for Band Width of An

Interference Band

a. From above diagram, bright and dark bands are

placed alternately. Let, n = 0, 1, 2, 3....................

λ = Wavelength of the light used.

D = Distance between the sources and the screen.

d = Distance between the two sources.

For a bright band.

xn = n λ D/d and xn+1 = (n + 1) λ D/d

X = λ D/d

b. For a dark band.

Xn = (2n -1) λ D/2d

Xn-1 = [2 (n + 1) - 1] λ D/d = (2n + 1) λ D/2d

X = λ D/d

Conditions for Obtaining Well Defined Steady

Interference Pattern 12

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a. Statement :-

Two sources must be

i. equally bright

ii. monochromatic

iii. coherent

iv. as narrow as possible

v. as closed as possible

b. Explanation :-

i. This will not give a well-defined interference

pattern.

ii. This will not give a well defined interference

pattern.

iii. This produces unstable interference pattern.

iv. These will not a well defined interference pattern.

v. Since X = λ D/d. For λ, D = constant

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X α 1/d. Therefore smaller the distance between

the sources higher the bond width. This gives a well

defined interference pattern.

Biprism

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Two thin prisms of extremely small refracting angle

are connected bases to base. Such a prism formed

is called as “Biprism”. Biprism is used for producing

two coherent sources from a single source.

Biprism Experiment

This experiment is performed to find unknown

wavelength of monochromatic source of light.

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a. Measurement of D :- Distance between slit and the

eyepiece is measured directly from the scale on the

optical bench.

d. Measurement of X :- For this purpose vertical wire

of the cross wire is made coincide with one edge of

the white band and corresponding reading (xn) on

the micrometer scale is recorded using slow motion

screw. The cross wire is moved through known

number of bands (n). Vertical wire is again made

coincide with one edge of the bright band and

corresponding heading (xn) on the micrometer is

recorded.

Mean band width X is obtained by the formula.

X = Xn – X0/n16

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e. Measurement of d (method of conjugate foci) :-

S1, S2 are virtual coherent sources. Distance

between these sources (d) cannot be measured

directly.

1. A convex lens is interposed in between biprism

is so adjusted that real, bright and enlarged

images S1’ and S2’ of S1, S2 are observed in the

focal plane of the eyepiece. Using slow motion

screw distance (d1) between S1’ and S2’ is

measured.

u = object distance v = Image distance

object size / Image size

= object distance / Image distance

d/d1 = u/v ......................... (1)17

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2. Convex lens is displaced towards eyepiece and

its position is so adjusted that real, diminished

and reduced images S1’’ and S2’’ of S1, S2 are

observed in the focal plane of the eyepiece.

Using slow motion screw distance (d2) between

S1” and S2” is measured.

d/d2 = v/u .........................(1)

Multiplying (1) by (2),

d/d2 = u/v × v/u

d2/d1d2 = 1

d2 = d1d2 d =

f. Unknown wavelength l is calculated by :-

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λ = X d/D = X / D

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