Question

Two identical sources each of intensity $I_0$ have a separation $d=\frac{\lambda }{8}$, where $\lambda$ is the wavelength of the waves emitted by either source. The phase difference of the sources is $\frac{\pi }{4}$. The intensity distribution $I\left(\theta \right)$ in the radiation field as a function of $\theta$, which specifies the direction from the sources to the distant observation point $P$ is given by

• A. $I\left(\theta \right)=I_0{\cos }^2\theta$
• B. $I\left(\theta \right)=\frac{I_0}{4}{\cos }^2\left(\frac{\pi \sin \theta }{8}\right)$
• C. $I\left(\theta \right)=4I_0{\cos }^2\left[\frac{\pi }{8}\right(\sin \theta +1\left)\right]$
• D. $I\left(\theta \right)=I_0{\sin }^2\theta$

Answer 1

The correct option is C $I\left(\theta \right)=4I_0{\cos }^2\left[\frac{\pi }{8}\right(\sin \theta +1\left)\right]$

The expression for intensity at a point on the screen is given by

$I\left(\theta \right)=4I_0{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$


The path difference $\Delta x$ is,

$\Delta x=S_{2}P-S_{1}P=d\sin \theta =\frac{\lambda }{8}\sin \theta$

Here, total phase difference will be due to path difference and initial phase difference at source,

$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)+\Delta {\varphi }_{intial}=\frac{2\pi }{\lambda }\left(\frac{\lambda }{8}\sin \theta \right)+\frac{\pi }{4}=\frac{\pi }{4}(1+\sin \theta )$

As, $I\left(\theta \right)=4I_0{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$

$\therefore I\left(\theta \right)=4I_0{\cos }^2\left[\frac{\pi }{8}\right(1+\sin \theta \left)\right]$

Hence, option $\left(C\right)$ is correct.