Question

The distance between the first dark and bright band formed in Young's double slit experiment with band width $B$ is

• A. $\frac{B}{4}$
• B. $B$
• C. $\frac{B}{2}$
• D. $\frac{3B}{2}$

Answer 1

Answer: (C)

Position of nth bright fringe $x_1=\frac{n\lambda D}{d}$
For first bright fringe $n=1$
$\therefore x_1=\frac{\lambda D}{d}$
Position of nth dark fringe
$x_2=\frac{(2n-1)\lambda D}{2d}$
For first dark fringe $n=1$
$\therefore x_2=\frac{\lambda D}{2d}$
Now, $x_1-x_2=\frac{\lambda D}{2d}$
If $B$ is the band width, then
$x_1-x_2=\frac{B}{2}$