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 \[\theta = \sin^{- 1} \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.
Solution
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,
\[\sin\theta = \frac{S_1 X}{S_1 S_2} = \frac{∆ x}{d}\]
\[\Rightarrow ∆ x = d\sin\theta\]
Using \[\theta = \sin^{- 1} \left( \frac{\lambda}{2d} \right)\] 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 \[\frac{\lambda}{2}.\]
