Answer to Question #194468 in Optics for Mohd Arsalan

Question

in the fraunhofer diffraction pattern due to a single slit, the intensity of the central spot is maximum. explain on the basis of geometrical considerations.

Answer 1

Suppose a parallel beam of light is incident normally on a slit of width b. According to Huygens’ principle, each and every point of the exposed part of the plane wavefront (i.e., every point of the slit) acts as a source of secondary wavelets spreading in all directions.

Let us consider a point P which collects the waves originating from different points of the slit at an angle θ. Figure shows the perpendicular from the point A to the parallel rays. This perpendicular also represents the wavefront of the parallel beam diffracted at an angle θ. The optical paths from any point on this wavefront to the point P are equal. The optical path difference between the waves sent by the upper edge A of the slit and the wave sent by the centre of the slit is $\dfrac{b}{ 2} sinθ$ . Consider the angle for which $\dfrac{b}{ 2} sinθ =\dfrac{ λ}2.$ The above mentioned two waves will have a phase difference

$\delta=\dfrac{2\pi}{\lambda}.\dfrac{\lambda}{2}=\pi$

The two waves will cancel each other. The wave from any point in the upper half of the slit is exactly cancelled by the wave from the point b/2 distance below it. The whole slit can be divided into such pairs and hence, the intensity at P will be zero. This is the condition of the first minimum, i.e., the first dark fringe.

So,

$\dfrac{b}{2}\sin\theta=\dfrac{\lambda}{2}$

$b\sin\theta=\lambda$ (first minimum)

Similar arguments show that other minima (zero intensity) are located at points corresponding to b sinθ = 2λ, 3λ, …

or, $b\sin\theta=n\lambda$ (dark fringe)