# In an interference arrangement similar to YDSE , Slits $S_1\;and \;S_2$ are illuminated with coherent microwave sources and are synchronized to have zero phase difference The slits are separates by distance $d= 150 \;m$. The intensity $I_0$ is measured as a function of $\theta$ where $\theta$ is defined as shown. If $I _0$ is max intensity, calculate $I_0$ for $\theta=90^{\circ}$

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

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.