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 beam makes an angle $\theta=\sin^{-1}\Big(\frac{\lambda}{2\text{D}}\Big)$ with the normal to the plane of the slits, there will be a dark fringe at the centre $P_0$ of the pattern.
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It can be seen from the figure that the wavefronts reaching O from $S_1$ and $S_2$ will have a path difference of $S_2X.$
In the $\Delta\text{S}_1\text{S}_2\text{X},$
$\sin\theta=\frac{\text{S}_2\text{X}}{\text{S}_1\text{S}_2}$
So, path difference $=\text{S}_2\text{X}=\text{S}_1\text{S}_2\sin\theta=\text{d}\sin\theta=\text{d}\times\frac{\lambda}{2\text{d}}=\frac{\lambda}{2}$
As the path difference is an odd multiple of $\frac{\lambda}{2},$ there will be a dark fringe at point $P_0.$
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