Circular Aperture Diffraction

Let AB be a small aperture (say a pin hole) and S is a point source of monochromatic light. XY is a screen perpendicular to the plane of the paper and P is a point on the screen. SP is perpendicular to the screen. O is the centre of the aperture and r is the radius of the aperture. Let the distance of the source from the aperture be a (SO = a) and the distance of the screen from the aperture be b (OP = b). P₁OQ₁ is the incident spherical wavefront and with reference to the point P, O is the pole of the wavefront. To consider the intensity at P, half period zones can be constructed with P as centre and radii b + λ/2, b + 2λ/2 etc. on the exposed wavefront AOB. Depending on the distance of P from the aperture (i.e. the distance b) the number of half period zones that can be constructed may be odd or even, if the distance a is such that only one half period zone can be constructed, then the intensity at P will be proportional to m₁² (where m₁ is the amplitude due to the first zone at P). On the other hand, if the whole of the wavefront is exposed to the point P, the resultant amplitude is m₁/2, or the intensity is proportional to m₁²/4. The position of the screen can be altered so as to construct 2, 3 or more half period zones for the same area of the aperture. If only 2 zones are exposed, the amplitude = m₁ – m₂ (minimum) and if 3 zones are exposed, the amplitude m₁ – m₂ + m₃ (maximum) and so on. Thus, by continuously altering the value of b, the point P becomes alternately bright and dark depending on whether odd or even number of zones are exposed by the aperture.

Now consider a point P’ on the screen XY. Join S to P’. The line SP’ meets the wavefront at O’. O’ is the pole of the wavefront with reference to the point P’. Construct half period zones with the point O’ as the pole of the wavefront. The upper half of the wavefront is cut off by the obstacle. If the first two zones are cut off by the obstacle between the point O’ and A and if only the 3rd, 4th and 5th zones are exposed by the aperture AOB them the intensity at P’ will be maximum. Thus, if odd number or half period zones are exposed, point P’ will be of maximum intensity and if even number of zones are exposed the point P’ will be minimum intensity. As the distance of P’ from P increases, the intensity of maxima and minima gradually decreases, because, with the point P’ far removed from P, the most effective central half period zones are cut off by the obstacle between the points O’ and A. With the outer zones, the obliquity increases with reference to the point P’ and hence the intensity of maxima and minima also will be less. If the point P’ happens to be of maximum intensity, then all the points lying on a circle of radius PP’ on the screen will also be of maximum intensity. Thus, with a circular aperture, the diffraction pattern will be concentric bright and dark rings with the centre P bright or dark depending on the distance b. The width of the rings continuously decreases.

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