Question

Derive the conditions for bright and dark
fringes produced due to diffraction by a
single slit.

Answer 1

To derive:

Conditions for bright and dark

fringes produced due to diffraction by a

single slit.

Solution:

For bright fringes, the interference should be constructive.

So, path difference between the two waves should be whole number multiples of wavelength.

$\therefore \: \dfrac{xd}{D} = n \lambda$

$\implies \: x = n \times \bigg( \dfrac{ \lambda D}{d} \bigg)$

$\boxed{here \: n = \pm1, \pm2, \pm3 \: and \: so \: on}$

For dark fringe , the path difference should be:

$\therefore \: \dfrac{xd}{D} = \bigg(n + \dfrac{1}{2} \bigg)\lambda$

$\implies \: x = \bigg(n + \dfrac{1}{2} \bigg) \dfrac{ \lambda D}{d}$

So, the dark fringes and bright fringes are equally placed, which can illustrated by calculating the fringe width.