Physics - Wave Optics Question with Solution | TestHub

The fringe width is given as $\beta =0.4$ mm.

The position of the $n$-th bright fringe from the central maximum is $y_n=n\beta$.
For the 5th bright fringe, $y_5=5\beta$.

The position of the $m$-th dark fringe from the central maximum is $y_m^{\prime }=(m-0.5)\beta$.
For the 3rd dark fringe, $y_3^{\prime }=(3-0.5)\beta =2.5\beta$.

The distance between the fifth bright and third dark band on the same side is:
$\Delta y=y_5-y_3^{\prime }=5\beta -2.5\beta =2.5\beta$

Substituting the value of $\beta$:
$\Delta y=2.5\times 0.4=1$ mm

Answer: $1$