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 $I_{\max }$
$=\frac{3}{4}{I}_{\max }$
$=\frac{3}{4}\left(4a^2\right)=3a^2$
$\Rightarrow 4a^2{\cos }^2\frac{\varphi }{2}=3a^2$
${\cos }^2\frac{\varphi }{2}=\frac{3}{4}$ or $\cos \frac{\varphi }{2}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{\varphi }{2}=\frac{\pi }{6}$ or $\varphi =\frac{\pi }{3}$