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: (C)

$I (\theta)=I_0$ for $\theta$=0$^{\circ}$

Answer 2

The intensity of light is $I(\theta)=I_0 cos^2 \big(\frac{\delta}{2}\big)$
where, \hspace20mm \delta=\frac{2 \pi}{\lambda}(\Delta x)
\hspace25mm =\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
\hspace10mm \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} \hspace10mm [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 \hspace35mm [option (c)]