$(a)\;\frac{1}{\sqrt 2} \\ (b)\;\frac{\sqrt 3}{2} \\ (c)\;\frac{1}{2} \\ (d)\;\frac{3}{4} $

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$I= I_0 \cos^2 (\large\frac{\phi}{2})$

Where $\phi$ is phase difference .

Also phase difference $\phi =\large\frac{2 \pi}{\lambda} $$\times path\; differences$

$\qquad= \large\frac{2 \pi}{\lambda} \times \large\frac{\lambda}{6}=\frac{\pi}{3}$

$\therefore \large\frac{I}{I_0} =\cos^2 \bigg(\large\frac{ \phi}{2}\bigg)$

$\qquad= \cos^2 \bigg( \large\frac{\pi}{6} \bigg)$

$\qquad= \bigg( \large\frac{\sqrt 3}{2} \bigg)^2$

$\qquad= \large\frac{3}{4}$

Hence d is the correct answer.

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