Question

A plane monochromatic light wave with intensity I_o 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 Fresnel zone was removed (along the diameter)?

Answer 1

(a) Suppose the disc does not obstruct 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

where  A_in (A_out) stands for A_internal (A_external). The factor i takes account of the π/2 phase difference between two halves of the first Fresnel zone. Thus