Optics Ques 154
- 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 $0 \leq \theta \leq 90^{\circ}$ is given by
$(1995,2 M)$
(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 $\theta$
Show Answer
Answer:
Correct Answer: 154.(a,c)
Solution:
Formula:
- The intensity of light is $I(\theta)=I _0 \cos ^{2} (\frac{\delta}{2})$
where, $\quad \delta=\frac{2 \pi}{\lambda}(\Delta x)=(\frac{2 \pi}{\lambda})(d \sin \theta)$
(a) $\operatorname{For} \theta=30^{\circ}$
$ \begin{aligned} & \lambda=\frac{c}{\nu}=\frac{3 \times 10^{8}}{10^{6}}=300 m \text { and } d=150 m \\ \therefore \quad & \frac{\delta}{2}=(\frac{2 \pi}{300})(150) (\frac{\pi}{1})=\frac{\pi}{2} \\ \therefore \quad & \frac{\delta}{2}=\frac{\pi}{4}\\ \therefore \quad & I(\theta)=I _0 \cos ^{2} (\frac{\pi}{4})=\frac{I _0}{2} \end{aligned} $
[option (a)]
(b) For $\quad \theta=90^{\circ}$
$ \delta=(\frac{2 \pi}{300})(150)(1)=\pi $
or
$ \frac{\delta}{2}=\frac{\pi}{2} \text { and } I(\theta)=0 $
(c) For $\theta=0^{\circ}, \delta=0$ or $\frac{\delta}{2}=0$
$ \therefore \quad I(\theta)=I _0 $
[option (c)]
BETA
