A plane monochromatic light wave with intensity `I_(0)` falls normally on an opaque disc closing the first Fresnel zone for the observation point `P`. What did the intensity of light `I` at the point `P` become equal to after
(a) half of the disc (along the diameter) was removed,
(b) half of the external half of the first Fresnal zone removed (along the diameter)?

Explanation

Suppose the disc does not abstruct light at all. Then
`A_(disc) + A_("remainder") = (1)/(2)A_(disc)`
(because the disc covers the first Fresnel zone only).
So `A_("remainder") =- (1)/(2)A_(disc)`
Hence the amplitude when half of the disc is removed along a diameter
`= (1)/(2)A_(disc) + A_("remainder") = (1)/(2)A_(disc) - (1)/(2)A_(disc) ~~0`
Hence `I = 0`.
(b) In this case
`A = (1)/(2)A_("external") +A_("remainder")`
`= (1)/(2)A_("external") - (1)/(2)A_(disc)`
We write `A_(disc) = A_(in) + iA_(ou)`
where `A_(in) (A_(out))` stands for `A_("internal") (A_("external"))`. The factor `i` takes account of the `(pi)/(2)` pahse difference between two halves of the first Fresnel zone. Thus
`A =-(1)/(2)A_("in")` and `I = (1)/(4)A_("in")^(2)`
On the other hand `I_(0) = (1)/(4) (A_("in")^(2) + A_(out)^(2)) = (1)/(2) A_("in")^(2)`
so `I = (1)/(2)I_(0)`.