A parallel beam of light $\left(\mathrm{\lambda } = 5000 \stackrel{0}{\mathrm{A}}\right)$ is incident at an angle $\mathrm{\alpha } = {30}^0$ 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 ${\mathrm{I}}_0$ . Point O is equidistant from ${\mathrm{S}}_1 \mathrm{and} {\mathrm{S}}_2$  (see the following figure). The distance between slits is 1 mm

The path difference due to inclination of the incident ray is

$$\begin{gathered} \mathrm{dsin\alpha } = 1\mathrm{mm} \mathrm{x} \sin {30}^0 = \frac{1}{2}\mathrm{mm} = 0.5\mathrm{mm} \\ \Rightarrow \mathrm{\lambda } =5000\stackrel{0}{\mathrm{A}} \end{gathered}$$

At point O, the path difference is 0.5 mm, that is,

$5000 \mathrm{x} {10}^3\stackrel{0}{ \mathrm{A}} = 1000\mathrm{\lambda }$

Therefore, O has maxima and 4${\mathrm{I}}_0$ is the intensity at O. At a point located at 1m below point O, the path difference is zero. Therefore, the intensity is 4${\mathrm{I}}_0$.