Question

In Young's double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

• A. unchanged
• B. halved
• C. doubled
• D. quadrupled

Answer 1

The correct option is D quadrupled

Fringe width $\beta =\lambda D/d$

Fringe with on making separation d/2, distance between screen and slit 2D = $\lambda \ast 2D/\left(d/2\right)=4\lambda D/d$

So fringe with is quadrupled.

Answer. D) quadrupled