Question

If fringe width is $0.4$mm, the distance between fifth bright and third dark band on same side is.

• A. $1$mm
• B. $2$mm
• C. $3$mm
• D. $4$mm

Answer 1

The correct option is A $1$mm

Position of $n$th bright fringe from central maxima $x_{n_1}=\frac{n_1\lambda D}{d}$ here $n_1=5$
$\therefore x_{n_1}=\frac{5\lambda D}{d}$
Position of $n$th dark fringe from central maxima.
$x_n=\frac{(2n-1)\lambda D}{2d},$ Here $n=3$
$x_n=\frac{5}{2}\frac{\lambda D}{d}$
$x_{n_1}-x_n=\frac{2.5\lambda D}{d}=2.5\beta$
Given $\beta =0.4$mm
$\Rightarrow x_{n_1}-x_n=1$mm