Intensity in single-slit diffraction.

Intensity in single-slit diffraction.


1 Answers

Sachin Tyagi
31 Points
13 years ago

Figure given below shows a slit of width a divided into N parallel strips, each of width δx. The strips are very narrow, so each strip can be regarded as a radiator of Huygens wavelets, and all the light from a given strip arrives at point p with the same phase. The waves arriving a P from any pair of adjacent strips have the same phase difference Δ?, which can be found from

Phase difference/2? = path difference/λ


Δ? = 2?/λ δx sinθ

Where δx sin θ, as the detail of figure below shows, is the path difference for rays originating from corresponding points of adjacent strips. If the angle θ is not too large, each strip produces a wave of the same amplitude δE0 at P. the net effect at P is due to the superposition of N vectors of the same amplitude, each differing in phase from the next by Δ?. To find the intensity at P, we must frist find the net electron field of the N vectors.

Let us consider some examples of the addition of phasors in single slip diffraction. We first consider the resultant electric field at point P0. In this case θ=0, and Δ?=0 as  the phase difference between adjacent strips. According to the method of some section, we then lay N vector of length δE0 head to tail and parallel to one another. The resultant Eθ. This is a clearly maximum value that the resultant of these N vectors can take, so we label it Em. as we move away from θ=0, the phase difference Δ? assumes a definite nonzero value. Again laying the N vectors head to tail, each differing in direction from the previous one by Δ?, we obtain the resultant 10b.

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