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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } I_R=75 \% 	ext { of } I_{\max } \ &=\frac{3}{4} I_{\max } \ &=\frac{3}{4}\left(4 a^2ight)=3 a^2 \ & \Rightarrow \quad 4 a^2 \cos ^2 \frac{\phi}{2}=3 a^2 \ & \cos ^2 \frac{\phi}{2}=\frac{3}{4} 	ext { or } \cos \frac{\phi}{2}=\frac{\sqrt{3}}{2} \ & \Rightarrow \quad \frac{\phi}{2}=\frac{\pi}{6} 	ext { or } \phi=\frac{\pi}{3}\end{aligned}$

In the Young's double slit experiment, the resultant intensity at a point on the screen is $75 \%$ of the…

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