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

J & K CETJ & K CET 2007Wave Optics

Solution:

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