Question

Discuss the special cases on first minimum in Fraunhofer diffraction.

Answer 1

1. Divide the slit AB into two halts AC and CB.
2. Two different points on the slit which are separated by the same width is corresponding points.
3. The path difference of light waves from corresponding points meet at p and interfere destructive to make the first minimum.
   δ = `"a"/2`  sin θ
   The condition for p to first minimum
   `"a"/2 sin θ = lambda/2`
   a sin θ = λ (first minimum)
   a sin θ = 2 λ (second minimum)
   a sin θ = 3 λ (third minimum)
   `("a" sin theta)/(2"n") = lambda/2`(nth minimum)