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. The slits are separated by a distance $d=150.0m$. The intensity $I\left(\theta \right)$ is measured as a function of $\theta$, where $\theta$ is defined as shown in figure. If $I_0$ is the maximum intensity, then $I\left(\theta \right)$ for $0\le \theta \le {90}^o$ is given by :

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

Answer 1

The correct option is C $I\left(\theta \right)=I_0$ for $\theta ={30}^o$

Angle is small therefore, $sin\theta =tan\theta =\theta$

$sin\theta =\frac{\delta x}{d}$

$\delta x=dsin\theta ,\delta x=\frac{d\times y}{D}$

$phasedifference=\varphi =\frac{2\pi \delta x}{\lambda }=\frac{2\pi dy}{D\lambda }=\frac{2\pi dtan\theta }{\lambda }$

Where, $I=Ioco{s}^2\left(\frac{\varphi }{2}\right)$

$=Ioco{s}^2\left[\frac{\pi d}{\lambda }tan\theta \right]$

$\lambda =\frac{v}{freq.}=\frac{3\times {10}^8}{{10}^6}=300m$

$I=Ioco{s}^2\left[\frac{\pi 150tan\theta }{300}\right]$

$=Ioco{s}^2\left[\frac{\pi tan\theta }{2}\right]$

at $\theta ={30}^0$

$I=Io\times \mathrm{1.99.}$

at $\theta =0^0$

$I=Io$.