Question

In the Young's double slit experiment, the resultant intensity at a point on the screen is $75 \%$ of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (C)

Given, $I_{R}=75 \%$ of $I_{\max }$
$=\frac{3}{4} I_{\max }$
$=\frac{3}{4}\left(4 a^{2}\right)=3 a^{2}$
$\Rightarrow 4 a^{2} \cos ^{2} \frac{\phi}{2}=3 a^{2}$
$\cos ^{2} \frac{\phi}{2}=\frac{3}{4}$ or $\cos \frac{\phi}{2}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{\phi}{2}=\frac{\pi}{6} $ or $\phi=\frac{\pi}{3}$