Question

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

• A. same.
• B. quadrupled
• C. halved.
• D. one$-$third

Answer 1

The correct option is B quadrupled

We have fringe width,

$\beta =\frac{D\lambda }{d}$

where $D$ is the distance between the screen and the slits; $d$ is the distance between the slits and $\lambda$ is the wavelength.

Given that the separation between the slits is halved, i.e., $d^{\prime }=\frac{d}{2}$ and distance between slits and screen is doubled, i.e., $D^{\prime }=2D$.

Hence now the modified fringe width (${\beta }^{\prime }$) is given by

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