Question

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. Four times
• B. One-fourth
• C. Double
• D. Half

Answer 1

The correct option is A

Four times

Fringe width in YDSE is given by,
$\beta =\frac{\lambda D}{d}$

If the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width,
${\beta }^{{}^{\prime }}=\frac{\lambda 2D}{d/2}=\frac{4\lambda D}{d}=4\beta$