A parallel beam of light ($\lambda =5000 \AA$) is incident at an angle $\alpha =30°$ with the normal to the slit plane in a young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is I₀. Point O is equidistant from S₁ & S₂.The distance between slits is 1mm.   

Path difference at O is $d\sin \alpha =0.5 mm$ corresponding phase difference, $∆\varphi =\frac{2\pi }{\lambda }\times ∆p$

$=\frac{2\pi \left(0.5\times {10}^{-3}\right)}{5000\times {10}^{-10}}=2000\pi =2\pi \times 1000$

O is a point corresponding-1000th maxima.

The point at 1 m below O corresponds to central maxima.

So 4 m from O will be maxima position.

Fringe width =$\frac{\lambda D}{d}=\frac{5000\times {10}^{-7}\times 2}{{10}^{-3}}mm = 1 mm$