Question

Two identical sources each of intensity $I_0$ have a separation $d=\lambda /8$, where $\lambda$ is the wavelength of the waves of the waves emitted by either source. The phase difference of the sources os $\pi /4$. 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. $I\left(\theta \right)=\frac{I_0}{4}{\cos }^2\left(\frac{\pi \theta }{8}\right)$
• C. $I\left(\theta \right)=4I_0{\cos }^2\left[\frac{\pi }{8}(\sin \theta +1)\right]$
• D. $I\left(\theta \right)=I_0{\sin }^2\theta$

Answer 1

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

Let the source $S_2$ is ahead of $S_1$ by a phase of $\frac{\pi }{4}$ and further also the phase difference arises because of the path difference between the lights reaching the point P.

$\Delta x=dsin\theta =\frac{\lambda }{8}sin\theta$

Using, $\delta =\frac{2\pi }{\lambda }\Delta x$

$⟹{\delta }_1=\frac{\pi }{4}sin\theta$

So total phase difference will be $\delta =\frac{\pi }{4}+\frac{\pi }{4}sin\theta =\frac{\pi }{4}[1+sin\theta ]$

Now using, $I=4I_{o}co{s}^2\frac{\delta }{2}$

$⟹I\left(\theta \right)=4I_{o}co{s}^2[\frac{\pi }{8}(sin\theta +1)]$