Question

Two identical radiators have a separation of $d=\frac{\lambda }{4}$ where $\lambda$ is the wavelength of the waves emitted by either source. The initial phase difference between the source is $\frac{\lambda }{4}$. Then the intensity on the screen at a distant point situated at an angle, $\theta ={30}^{\circ }$ from the radiators is (here, $I_0$ is intensity at that point due to one radiator alone)

• A. $I_0$
• B. $2I_0$
• C. $3I_0$
• D. $4I_0$

Answer 1

Answer: (B)

$2I_0$

The intensity at a point on screen is given by,

$I=4I_{0}co{s}^2\left(\frac{\varphi }{2}\right)$

where, $\varphi$ is the phase difference

In this problem, $\varphi$ arises,

• due to initial phase difference of $\frac{\pi }{4}$

• due to path difference for the observation point situated at $\theta ={30}^0$

Thus,

$\varphi =\frac{\pi }{4}+\frac{2\pi }{\lambda }\left(dsin\theta \right)=\frac{\pi }{4}+\frac{2\pi }{\lambda }\frac{\lambda }{4}\left(sin{30}^0\right)=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$

Thus,

$\frac{\varphi }{2}=\frac{\pi }{4}$

and

$I=4I_{0}co{s}^2\left(\frac{\pi }{4}\right)=2I_0$