In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If $\mathrm{I}_{\mathrm{m}}$ be the maximum intensity, the resultant intensity I when they interfere at phase difference $\phi$ is given by

(D) $\frac{I_{m}}{9}\left(1+8 \cos ^{2} \frac{\phi}{2}\right)$

(A) $\frac{\mathrm{I}_{\mathrm{m}}}{9}(4+5 \cos \phi)$

(B) $\frac{I_{m}}{3}\left(1+2 \cos ^{2} \frac{\phi}{2}\right)$

(C) $\frac{I_{m}}{5}\left(1+4 \cos ^{2} \frac{\phi}{2}\right)$

JEE Main · 2012 · 29 Apr 2012 · Offline · Wave Optics

In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If

I_{m}

be the maximum intensity, the resultant intensity I when they interfere at phase difference

ϕ

is given by

(A) $\frac{I _{m}}{9} (4 + 5 cos ϕ)$
(B) $\frac{I _{m}}{3} (1 + 2 cos^{2} \frac{ϕ}{2})$
(C) $\frac{I _{m}}{5} (1 + 4 cos^{2} \frac{ϕ}{2})$
(D) $\frac{I _{m}}{9} (1 + 8 cos^{2} \frac{ϕ}{2})$ ✓ Correct

Answer: (D)

Solution

Let the two slits send amplitudes

a

and

2 a

to the screen. Where they meet with phase difference

ϕ

, the intensity goes as

a^{2} + (2 a)^{2} + 2 (a) (2 a) cos ϕ = a^{2} (5 + 4 cos ϕ)

. The brightest point, where

cos ϕ = 1

, gives the maximum

I_{m} \propto 9 a^{2}

. Dividing,

I = \frac{I _{m}}{9} (5 + 4 cos ϕ)

, and writing

cos ϕ = 2 cos^{2} \frac{ϕ}{2} - 1

turns the bracket into

1 + 8 cos^{2} \frac{ϕ}{2}

. Hence the resultant intensity is

I = \frac{I _{m}}{9} (1 + 8 cos^{2} \frac{ϕ}{2})

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option

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JEE Main · 2012 · 29 Apr 2012 · Offline · Wave Optics

Solution

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