2020 Wave OpticsYoung's Double Slit Experiment Easy Question 4 of 8 in Young's Double Slit Experiment

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

A half
B four times
C one-fourth
D double

The answer is B.

Solution

Fringe width $\beta = \frac{\lambda D}{d}$ $d' = \frac{d}{2}$ and $D' = 2D$ ⇒ $\beta' = \frac{\lambda(2D)}{d/2} = \frac{4\lambda D}{d} = 4\beta$ So the fringe width becomes four times.

Concept fringe width dependence on d and D

Young's Double Slit Experiment in past papers

8 questions from this chapter have appeared across 8 exam years.