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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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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 $

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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.
answered Feb 14, 2014 by meena.p

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