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 P_{0} of the pattern.

#### Solution

Let the two slits are S_{1} and S_{2} with separation d as shown in figure.

The wave fronts reaching P_{0}_{ }from S_{1} and S_{2} will have a path difference of S_{1}X = ∆x.

In ∆S_{1}S_{2}X,

\[\sin\theta = \frac{S_1 X}{S_1 S_2} = \frac{∆ x}{d}\]

\[\Rightarrow ∆ x = d\sin\theta\]

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 P_{0} as the path difference is an odd multiple of \[\frac{\lambda}{2}.\]