CBSE Class 12 · 2016 · All India · Wave OpticsLong Answer

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(a) Derive an expression for the path difference in Young's double slit experiment and obtain the condition for constructive and destructive interference at a point on the screen.
(b) The intensity at the central maxima in Young's double slit experiment is

I_{0}

. Find out the intensity at a point where the path difference is

\frac{λ}{6}

\frac{λ}{4}

and

\frac{λ}{3}

Solution

(a) For slits

S_{1}, S_{2}

separated by

d

and a screen at distance

D

, a point

P

at distance

x

from the centre gives

S_{2} P^{2} - S_{1} P^{2} = 2 x d \Rightarrow (S_{2} P + S_{1} P) (S_{2} P - S_{1} P) = 2 x d

With

S_{2} P + S_{1} P \approx 2 D

, the path difference is

Δ = S_{2} P - S_{1} P = \frac{x d}{D}

Constructive:

Δ = nλ

;

Destructive:

Δ = (2 n - 1) \frac{λ}{2}

.
(b) The intensity is

I = I_{0} cos^{2} \frac{ϕ}{2}

with

ϕ = \frac{2 π}{λ} Δ

(where

I_{0}

is the central maximum).
For

Δ = \frac{λ}{6} : ϕ = \frac{π}{3} \Rightarrow I = I_{0} cos^{2} \frac{π}{6} = I_{0} \cdot \frac{3}{4} = \frac{3 I _{0}}{4}

.
For

Δ = \frac{λ}{4} : ϕ = \frac{π}{2} \Rightarrow I = I_{0} cos^{2} \frac{π}{4} = \frac{I _{0}}{2}

.
For

Δ = \frac{λ}{3} : ϕ = \frac{2 π}{3} \Rightarrow I = I_{0} cos^{2} \frac{π}{3} = \frac{I _{0}}{4}

I_{λ /6} = \frac{3 I _{0}}{4}, I_{λ /4} = \frac{I _{0}}{2}, I_{λ /3} = \frac{I _{0}}{4}

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CBSE Class 12 · 2016 · All India · Wave OpticsLong Answer

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