Q120 / 180 · NEET 2020 (Phase 1) · Physics

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:

Solution

$β = \frac{λ D}{d}$

$β^{'} = \frac{λ D^{'}}{d^{'}}$

$D^{'} = 2 D, d^{'} = \frac{d}{2}$

$β^{'} = \frac{λ \times 2 D}{\frac{d}{2}} = \frac{4 λ D}{d}$

$β^{'} = 4 β$

Fringe width becomes $4$ times.

Hence, the correct answer is option: 'B'.