KVPY Sample Paper KVPY Stream-SX Model Paper-1

The path difference between two interfering waves at a point on the screen is\[\lambda /6.\] The ratio of intensity at this point and that at the central bright fringe will be: (Assume that intensity due to each slit in same)

A) 0.853

B) 8.53

C) 0.75

D) 7.5

Correct Answer: C

Solution :

At path difference \[\frac{\lambda }{6},\] phase difference is \[\frac{\pi }{3}\]

\[I={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos \frac{\pi }{3}=3{{I}_{0}}\]

\[{{I}_{\max }}=4{{l}_{0}}\]

So the required ratio is \[\frac{3{{I}_{0}}}{4{{I}_{0}}}=0.75\]