Question

In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduced to half. As a result the fringe width

• A. is halved
• B. becomes four times
• C. remains unchanged
• D. is doubled

Answer 1

Answer: (B)

Fringe width, $\beta =\frac{\lambda D}{d}$
where $D$ is the distance between slits and screen and $d$ is the distance between the slits.
When $D$ is doubled and $d$ is reduced to half, then fringe width becomes
${\beta }^{\prime }=\frac{\lambda (2D)}{(d/2)}=\frac{4\lambda D}{d}=4\beta$