Physics - Wave Optics Question with Solution | TestHub

The fringe width in Young's double slit experiment is given by $\beta =\frac{\lambda D}{d}$

Given that the new distance between the slits and the screen is $D^{\prime }=2D$

The new separation between the slits is $d^{\prime }=\frac{d}{3}$

The new fringe width is ${\beta }^{\prime }=\frac{\lambda D^{\prime }}{d^{\prime }}$

Substituting the values, we get ${\beta }^{\prime }=\frac{\lambda (2D)}{\frac{d}{3}}=6\frac{\lambda D}{d}=6\beta$

Therefore, the fringe width becomes $6$ times the original.

Answer: becomes $6$ times the original.