Question

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

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

Answer 1

Answer: (A)

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 $\beta =0.4mm$
$\Rightarrow x_{n_1}-x_n=1mm$