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. Slow that if the incident beam makes an angle $\mathrm{\theta }={\sin }^{-1}\left(\frac{\lambda }{2d}\right)$ with the normal to the plane of the slits, there will be a dark fringe at the centre P₀ of the pattern.

Answer 1

Let the two slits are S₁ and S₂ with separation d as shown in figure.



The wave fronts reaching P₀ from S₁ and S₂ will have a path difference of S₁X = ∆x.
In ∆S₁S₂X, $\sin \theta =\frac{{\mathrm{S}}_1\mathrm{X}}{{\mathrm{S}}_1{\mathrm{S}}_2}=\frac{∆x}{d}$
$\Rightarrow ∆x=d\sin \theta$
Using $\mathrm{\theta }={\sin }^{-1}\left(\frac{\lambda }{2d}\right)$, we get,
$\Rightarrow ∆x=d\times \frac{\lambda }{2d}=\frac{\lambda }{2}$
Hence, there will be dark fringe at point P₀ as the path difference is an odd multiple of $\frac{\lambda }{2}$.