Question

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 by

• A. $I(\theta )=I_0/2$ for $\theta$=30${}^{\circ }$
• B. $I(\theta )=I_0/4$ for $\theta$=90${}^{\circ }$
• C. $I(\theta )=I_0$ for $\theta$=0${}^{\circ }$
• D. $I(\theta )$is constant for all values of $\theta$

Answer 1

Answer: (A), (C)

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