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

Question

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

Dr. Abubacker · Accepted Answer

(b) The intensity at a point on screen is given by
$I = 4I_{0}\cos^{2}{(\phi/2)}$
Where $\phi$ is the phase difference. In this problem $\phi$ arises (i) due to initial phase difference of $\pi/4$ and (ii) due to path difference for the observation point situated at $	heta = 30{^\circ}$. Thus
$\phi = \frac{\pi}{4} + \frac{2\pi}{\lambda}\left( d\sin	heta ight) = \frac{\pi}{4} + \frac{2\pi}{\lambda}.\frac{\lambda}{4}\left( \sin{30{^\circ}} ight) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
Thus $\frac{\phi}{2} = \frac{\pi}{4}$ and $I = 4I_{0}\cos^{2}{(\pi/4}) = 2I_{0}$

Explanation

Reviewed by Dr. Abubacker

Want more practice?