Physics - Wave Optics Question with Solution | TestHub

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

The position of the $n^{\text{th}}$ bright fringe from the central maximum is given by $y_n=n\beta$.
For the $6^{\text{th}}$ bright fringe, $y_6=6\beta$.

The position of the $n^{\text{th}}$ dark fringe from the central maximum is given by $y_n^{\prime }=(n-0.5)\beta$.
For the $4^{\text{th}}$ dark fringe, $y_4^{\prime }=(4-0.5)\beta =3.5\beta$.

The distance between the $4^{\text{th}}$ dark band and the $6^{\text{th}}$ bright band on the same side is:
$\Delta y=y_6-y_4^{\prime }$
$\Delta y=6\beta -3.5\beta =2.5\beta$

Substituting the value of $\beta$:
$\Delta y=2.5\times 0.4$ mm $=1.0$ mm

Answer: $1.0$ mm