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Q. In an interference arrangement similar to Young's
double-slit experiment, the slits $S_1$, and $S_2$ are illuminated
with coherent microwave sources, each of frequency $10^6$ Hz.
The sources are synchronized to have zero phase
difference. The slits are separated by a distance d = 150.0 m.
The intensity I$(\theta)$ is measured as a function of $\theta$, where $\theta$ is
defined as shown. If $I_0$ is the maximum intensity, then I$(\theta)$
for $0 \le \theta \le$90$^{\circ}$ is given byPhysics Question Image

IIT JEEIIT JEE 1995

Solution:

The intensity of light is $I(\theta)=I_0 cos^2 \big(\frac{\delta}{2}\big)$
where, $ \delta=\frac{2 \pi}{\lambda}(\Delta x)$
$ =\big(\frac{2 \pi}{\lambda}\big)(d \, sin \, \theta)$
(a) For $\theta$ = 30$^{\circ}$
$ \, \, \, \, \lambda=\frac{c}{v}=\frac{3 \times 10^8}{10^6}=300 m$ and d = 150 m
$ \delta=\big(\frac{2 \pi}{300}\big)(150)\big(\frac{1}{2}\big)=\frac{\pi}{2}$
$\therefore \, \, \, \, \, \, \, \, \frac{\delta}{2}=\frac{\pi}{4}$
$\therefore \, \, \, \, \, \, \, I(\theta)=I_0 cos^2 \big(\frac{\pi}{4}\big)=\frac{I_0}{2} $ [option (a)]
(b) For $\theta$=90$^{\circ}$
$\delta=\big(\frac{2 \pi}{300}\big)(150)(1)=\pi \, or \, \, \frac{\delta}{2}=\frac{\pi}{2} \, \, and \, \, I(\theta)=0$
(c) For $\theta=0^\circ,\delta=0 \, \, or \, \, \frac{\delta}{2}=0$
$\therefore I(\theta)=I_0 $ [option (c)]