Fresnel''s and fraunhofer''s classes of diffraction, Other Engineering

Q. Distinguish between Fresnel's and Fraunhofer's classes of diffraction. Show that the relative intensities of successive maxima of Fraunhofer diffraction at single slit are

Sol. (ii) When slit width is large compared to wavelength the diffraction is limited to boundary only. Some of the main difference between Fresnel's and Fraunhofer's class of diffraction are listed below -

Fraunhofer's diffraction at a single slit qualitative treatment : Let S be a monochromatic light source of wave length l. A rectangular slit S1 S2 of width 'a' is placed perpendicular to the plane of paper.

As we know that in Fraunhofer class diffraction source S and screen XY should be effectively at infinity distance from diffracting element, for this purpose two convex lenses L1 and L2 are used.

Here L1 is a collimating lens which is placed at a distance equal to focal length of lens L1 from source S. light emerging form converging lens L1 is retained in form of a parallel beam of light. This parallel beam of light (plane wave front) is incident on slit of width 'a'. The light passing through the slit is get diffracted and focused by another lens L2 on the screen XY, which is placed in the focal plane of lens L2.

Now in accordance with the theory of Ray optics, and intense sharp image of slit should by obtained on the screen as a diffraction pattern, but in place of this a diffraction pattern consist of a central bright fringe having alternate dark and bright band of sharply decreasing intensity on both the side of central bright fringe is obtained, known as diffraction pattern due to a single slit.

Let the angle of diffraction for secondary wavelets emerging from slit of width S1 S2 = a is such that

Let the whole slit width is divided into 10 equal parts (Fig. (b) ). Now the secondary waves from part 1, 2, ...... 5 and those from the corresponding parts 6, 7, ............ 10 have a path difference of l/2 i.e. they reach of a point P on screen with a phase difference π since they are emerging with parts of equal area so they have equal amplitude.

In result at point P they cancel each other's effect. See wavelet from part 1 cancel contribution of secondary wavelet form part 6. Similarly part 2, cancel contribution of part 7 and so on. Therefore the resultant disturbance at point P will be minimum i.e. zero. In general we can write a sin ? =l, 2l ............ minima (condition for zero Intensity).

Here the waves from 1st part and 2nd part having a path difference of land cancel cut each other's effect but the wavelets from 3rd part contribute to disturbance at point P and produce some intensity.

Explanation : Let S1 S2 be a slit of width 'a'. The monochromatic light of wavelength l incident normally on this slit width 'a'. Then according to Huygens principle, each point in slit S1 S2 sends out secondary wavelets in all directions. From Fig (c), it is clear that all the secondary wavelets starts from different points in slit S1 S2, are in the same plane. The secondary wavelets, diffracted in the direction of incident ray are focused at point P0 on serene. Since point P0 is equidistant for all the secondary wavelets starting from slit S1 S2. So the wavelets reaching at point P0 will be in phase and a maximum intensity will be found at point P0. Our purpose is now to find the resultant intensity at a give point P on screen due to secondary wavelets those are diffracted by an angle '?'.

From Fig. (d) , it is clear that the path difference between the secondary wavelets focusing at point P and diffraction through an angle '?' by the slit S1 S2 increases continuously from extreme points of slit S1 to S2.

To evaluate this path difference between secondary wavelets emitting from S1 and S2. Let S1K be a perpendicular to S2K. Since the optical path from the plane S1K to point P are equal, the path difference between wavelets from S1 to S2 in a direction ? can be given as

Equation (2) gives the phase difference between extremum secondary wavelets starting from S1 and S2 and it is clear from fig.(c) that phase difference  increases continuously from S1 to S2.

Suppose that width of the slit is divided into p equal parts, and then phase difference between wavelets from two consecutive elements will be. Similarly amplitude of each secondary wavelets is assumed to be a0, since each part has the same width.

Now the resultant disturbance from these entire elements can be calculated by using vector polygon method or phasor diagram method. For this purpose, let us construct a polygon of each of polygon method or phasor diagram method. For this purpose, let us construct a polygon of each of equal length 'a0' and having a successive phase start from point S1 and ends at S2.

This line S1 S2 will represent the resultant disturbance. Since the number of secondary wavelets is very large then this stepped curve can be consider as a continuous are. The phase difference between the wavelets from extremum ends of the slit S1 S2 is the angle between tangents of S1 & S2.

Posted Date: 7/11/2012 2:40:55 AM | Location : United States

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