A plane monochromatic light wave with intensity `I_(0)` falls normally on an opaque screen with a round aperture. What is the intensity of light `I` behind the screen at the point for which the aperture
(a) is equal to the first Fresnel zone, to the internal half of the first zone,
(b) was made equal to the first Fresnel zone and then half of it was closed (along the diameter) ?

Explanation

When the aperture is equal to the first Fresnel Zone:-
The amplitude is `A_(1)` and should be compared with the amplitude `(A)/(2)` when the aperture is very wide. If `I_(0)` is the intensity in the second case the intensity in the first case will be `4I_(0)`.
When the aperture is equal to the internal half of the first zone:- Suppose `A_("in")` and `A_("out")` are the amplitudes due to the two halves of the first Fresnel zone. Clearly `A_("in")` and `A_("out")` differ in phase by `(pi)/(2)` because only half the Fresnel zone in involed. Also in magnitude `|A_("in")| = |A_("out")|`. Then
`A_(1)^(2) = 2|A_("in")|^(2)` so `|A_("in")|^(2) = (A_(1)^(2))/(2)`
Hence following the argument of the first case. `I_("in") = 2I_(0)`
(b) The aperture was made equal to teh first Fresnel zone and then half of it was closed along a diameter. In this case the amplitude of vibration is `(A_(1))/(2)`. Thus