$(a)\;\frac{\lambda D}{3d} \\ (b)\;\frac{\lambda D}{d} \\ (c)\;\frac{3 \lambda p}{d} \\ (d)\;\frac{2 \lambda D}{3d} $

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$\large\frac{I_{\Large max}}{I_{\Large min}}- \bigg( \large\frac{\sqrt {I_1/I_2}+1}{\sqrt {I_1 /I_2}-1}\bigg)^2=\large\frac{9}{1}$

or $\large\frac{x+1}{x-1} $$=3\quad (x =\sqrt {\large\frac{I_1}{I_2}})$

$\therefore x=2$

$\therefore \large\frac{I_1}{I_2}$$=4$

$I_1=4 I_2$

ie if $I_2=I_0$

then $I_1=4 I_0$

$I_0 =4 f_0 \cos ^2 \large\frac{\phi}{2}$

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

$\therefore \bigg(\large\frac{2 \pi}{\lambda} \bigg) \bigg( y \large\frac{d}{D} \bigg) =\large\frac{2 \pi}{3}$

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

Hence a is the correct answer.

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