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 source is $\pi / 4$. 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. $2 I_{0}$
• C. $3 I_{0}$
• D. $4 I_{0}$

Answer 1

Answer: (B)

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