In Young’s double slit experiment, one of the slits is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $I_{m}$ is the maximum intensity, the resultant intensity I when they interfere at phase difference $ϕ$ , is given by

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$\frac{I_{m}}{9} (4 + 5 \cos ϕ)$

$\frac{I_{m}}{3} (1 + 2 \cos^{2} \frac{ϕ}{2})$

$\frac{I_{m}}{5} (1 + 4 \cos^{2} \frac{ϕ}{2})$

$\frac{I_{m}}{9} (1 + 8 \cos^{2} \frac{ϕ}{2})$

answer is D.

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Detailed Solution

$W e k n o w t h a t, I n t e n s i t y \propto A m p l i t u d e^{2} G i v e n : A_{1} = 2 A_{2} ∴ I_{1} = 4 I_{2} I_{\max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(3 \sqrt{I_{2}})}^{2} = 9 I_{2} = I_{m} I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ = 4 I_{2} + I_{2} + 2 \sqrt{4 I_{2} . I_{2}} \cos ϕ = 5 I_{2} + 4 I_{2} \cos ϕ = \frac{I_{m}}{9} (5 + 4 \cos ϕ) [∵ I_{0} = \frac{I_{m}}{9}] = \frac{I_{m}}{9} [1 + 4 (1 + \cos ϕ)] = \frac{I_{m}}{9} (1 + 8 \cos^{2} \frac{ϕ}{2}) [∵ \cos ϕ = 2 \cos^{2} \frac{ϕ}{2} - 1]$