Question

A double slit interference experiment is conducted with light waves of wavelength of $650\text{nm}$ passing through two $3.0\mu m$ slits that are $18\text{m}$ apart

Find out the minimum order of the bright fringe that would not be visible

Answer 1

First order diffraction minima is formed when
$a\sin \theta =\lambda$
and maxima at that location is
$d\sin \theta =n\lambda$
Now, taking division we obtain the order of bright fringe at first minima (minimum order)
$\frac{a\sin \theta }{d\sin \theta }=\frac{\lambda }{n\lambda }$
$n=\frac{d}{a}$
$n=\frac{18}{3}=6$