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Q. 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

NEETNEET 2013Wave Optics

Solution:

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
$\beta^{\prime}=\frac{\lambda(2 D)}{(d / 2)}=\frac{4 \lambda D}{d}=4 \beta$