$(a)\;2 \;mm \\ (b)\;0.5\;mm \\ (c)\;1\;mm \\ (d)\;4\;mm $

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$I= 4I_0 \cos ^{2} \bigg( \large\frac{\phi}{2} \bigg)$

$I_0 =4 I_0 \cos ^2 \bigg( \large\frac{\phi}{2} \bigg)$

$\therefore \cos \bigg(\large\frac{\phi}{2} \bigg)=\large\frac{1}{2}$

or $\large\frac{\phi}{2}=\frac{\pi}{3}$

or $\phi =\large\frac{2 \pi}{3}= \bigg( \large\frac{2 \pi}{\lambda}\bigg).$$ \Delta x$

or $\large\frac{1}{3} =\bigg( \large\frac{1}{\lambda}\bigg) .y. \large\frac{d}{D}$

$( \Delta x =y. \large\frac{d}{D}\bigg)$

$\therefore y= \large\frac{\lambda}{3 \times \Large\frac{d}{D} } $

$\qquad=\large\frac{ 6 \times 10^{-7}}{3 \times 10^{-4}}$

$\qquad= 2 \times 10^{-3} m$

$\qquad= 2 \;mm$

Hence a is the correct answer.

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