Question

Fraunhofer diffraction circular aperture derivation

Answer 1

 

Engineering Physics by Dr. Amita Maurya, Peoples University, Bhopal.‎ > ‎Unit-II‎ > ‎

Fraunhofer diffraction at a circular aperture

Fraunhofer diffraction at a circular aperture

The problem of diffraction at a circular aperture was first solved by Airy in 1835. A circular aperture of diameter ‘d ’ is shown as AB in Fig. 14.13. A plane wave front WW’ is incident normally on this aperture. Every point on the plane wave front in the aperture acts as a source of secondary wavelets. The secondary wavelets spread out in all directions as diffracted rays in the aperture. These diffracted secondary wavelets are coverged on the screen SS′ by keeping a covex lends (L) between the aperture and the screen. The screen is at the focal plane of the convex lens. Those diffracted rays traveling normal to the plane of aperture [i.e., along CPo] are get coverged at Po.

Figure 14.13 Fraunhofer diffraction at a circular aperture



All these waves travel some distance to reach Po and there is no path difference between these rays. Hence a bright spot is formed at Po known Airy’s disc. Po corresponds to the central maximum.

Next consider the secondary waves traveling at an angle θ with respect to the direction of CPo. All these secondary waves travel in the form of a cone and hence, they form a diffracted ring on the screen. The radius of that ring is x and its centre is at Po. Now consider a point P1 on the ring, the intensity of light at P1 depends on the path difference between the waves at A and B to reach P1. The path difference is BD = AB sin θ = dsin θ. The diffraction due to a circular aperture is similar to the diffraction due to a single slit. Hence, the intensity at P1 depends on the path difference d sin θ. If the path difference is an integral multiple of λ then intensity at P1 is minimum. On the other hand, if the path difference is in odd multiples of , then the intensity is maximum.

i.e., d sin θ = nλ for minima _________ (14.39)

and d sin θ = (2n−1)  for maxima _________ (14.40)

where n = 1, 2, 3, …. etc. n = 0 corresponds to central maximum.

The Airy disc is surrounded by alternate bright and dark concentric rings, called the Airy’s rings. The intensity of the dark ring is zero and the intensity of the bright ring decreases as we go radially from P0 on the screen. If the collecting lens(L) is very near to the circular aperture or the screen is at a large distance from the lens, then



Where f is the focal length of the lens.

Also from the condition for first secondary minimum [using equation (14.39)]



Equations 14.41 and 14.42 are equal



But according to Airy, the exact value of x is



using equation (14.44) the radius of Airy’s disc can be obtained. Also from equation (14.44) we know that the radius of Airy’s disc is inversely proportional to the diameter of the aperture. Hence by decreasing the diameter of aperture, the size of Airy’s disc increases.

Comments

You do not have permission to add comments.

View as Desktop My Sites

Powered By Google Sites