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Question:
intensity in single slit pattern
Answer:

If we consider the Fraunhofer conditions,we see that the wave arrives at the single slit as a plane wave. Divided into segments, each of which can be regarded as a point source, the amplitudes of the segments will have a constant phase displacement from each other, and will form segments of a circular arc when added as vectors. The resulting relative intensity will depend upon the total phase displacement  according to the relationship.

I = I0 sin2[δ/2] /[δ/2]2

The total phase angle can be related to the deviation angle θ.

δ = 2πa sinθ / λ

The intensity as a function of angle θ -

I = I0 sin2[πa sinθ/λ ] /[πa sinθ/λ]2

5

 

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