Question

A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance $d$ and the screen is placed parallel to the plane of the slits. Show that if the incident bean makes an angle $\theta ={\sin }^{-1}\left(\frac{\lambda }{2d}\right)$ with the normal to the plane of the slits, there will be a dark fringe at the centre $P_0$ of the pattern.

Answer 1

It can be seen from the figure that the wavefronts reaching $O$ from $S_1$and $S_2$ will have a path difference of $S_{2}X$
In the $\Delta S_1{S}_{2}X$,
$\sin \theta =\frac{S_{2}X}{S_1{S}_2}$
So path difference $=S_{2}X=S_1{S}_2\sin \theta =d\sin \theta =d\times \lambda /2d=\lambda /2$
As the path difference is an odd multiple of $\lambda /2$ there will be dark fringe at point $P_0$.