JEE Main & Advanced Sample Paper JEE Main - Mock Test - 4

In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \[\lambda \] is K, (\[\lambda \] being the wave length of light used). The intensity at a point where the path difference is \[\lambda /4,\] will be:

A) K

B) \[K/4\]

C) \[K/2\]

D) Zero

Correct Answer: C

Solution :

For path difference \[\lambda \], phase difference \[=2\pi \] rad.

For path difference \[\frac{\lambda }{4},\] phase difference \[=\frac{\pi }{2}\] rad.

As \[K=4{{I}_{0}}\] so intensity at given point where path difference is \[\frac{\lambda }{4}\]

\[K'=4{{I}_{0}}{{\cos }^{2}}\left( \frac{\pi }{4} \right)\left( \cos \frac{\pi }{4}=\cos 45{}^\circ \right)=2{{I}_{0}}=\frac{K}{2}\]