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

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

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

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

$\frac{I_{m}}{9} (4 + 5 \cos ϕ)$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{I_{1}}{I_{2}} = \frac{1}{4} = {(\frac{A_{1}}{A_{2}})}^{2}; I_{max} = {(\sqrt{I_{0}} + \sqrt{4 I_{0}})}^{2} = 9 I_{0} and \\ I = I_{0} + 4 I_{0} + 2 \sqrt{I_{0}} \sqrt{4} I_{0} \cos θ \\ = 5 I_{0} + 4 I_{0} \cos θ = I_{0} + 4 I_{0} [1 + \cos θ] \\ = I_{0} [1 + 4 (2 \cos^{2} \frac{ϕ}{2})] = \frac{I_{max}}{9} [1 + 8 \cos^{2} \frac{ϕ}{2}] \end{array}$