$(a)\;I_0 \\ (b)\;\frac{I_0}{2} \\ (c)\;0 \\ (d)\;None $

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For microwave as $C= f \lambda$

$\lambda=\large\frac{c}{f} =\frac{3 \times 10^8}{10^6}$$=300\;m$

and as $\Delta x =d \sin \theta$

$\phi =\large\frac{2 \pi}{\lambda} $$(d \sin \theta)$

$\qquad= \large\frac{2 \pi}{300} $$(150 \sin \theta)$

$\qquad= \pi \sin \theta$

So, $I=I_1+I_2+2 (\sqrt {I_1I_2})( \cos \phi)$

with $I_1=I_2 $ and $\phi =\pi \sin \theta$

reduces t $I_R=2 I_1 [1+ \cos (\pi \sin \theta)]$

$\qquad= 4I_1 \cos ^1 (\pi \sin \theta/2)$

and as $I_R$ will be max when

$\cos ^2 [(\pi (\sin \theta) /2]=max =1$

So that $(I_R)_{max}=4 I_1=I_0 (given)$

hence $I=I_0 \cos ^2 [ (\pi \sin \theta)/2]$

If $\theta =90 \qquad I= I_0 \cos ^2 \bigg( \large\frac{\pi}{2} \bigg)$$=0$

Hence c is the correct answer.

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