A plane light wave with wavelength `lambda` and intensity `I_(0)` falls normally on a large glass plate whose opposite side serves as an opaque screen with a round aperture equal to the first Fresnel zone for the observation point `P`. In the middle of the aperture there is a round recess equal to half the Fresnel zone. What must the depth `h` of that recess be for the intensity of light at the point `P` to be the highest? What is this intensity equal to?

Explanation

The contribution to the wave amplitude of the inner half-zero is
`(2pia_(0)e^(-ikb))/(b)int_(0)^(sqrt(blambda//2)) e^(-ikrho^(2)//2b) rho d rho`
`=(2pia_(0)e^(-ikb))/(b) int_(0)^(blambda//2) e^(-ikx//bdx)`
`= (2pia_(0)e^(-ikb))/(b) (e^(-iklambda//4) - 1) xx (1)/((-ik)/(b))`
`= (2pi ia_(0)e^(-ikb))/(k) (-i -1) = + (A_(1))/(2) (1-i)`
With phase factor this becomes `(A_(1))/(2)(1 + i)e^(-i delta)` where `delta = (2pi)/(lambda) (n - 1)h`. The contribution of the remaining aperture is `(A_(1))/(2) (1 - i)`
(so thet the sun of the two parts when `delta = 0` is `A_(1)`)
Thus the complate amplitude is
`(A_(1))/(2)(1 + i) e^(i delta) + (A_(1))/(2)(1 - i)`
and the intensity is
`I = I_(0)[(1-i)e^(i delta) + (1-i)][(1 -i)e^(-i delta) + (1+i)]`
`= I_(0) [2 + 2+(1 - i)^(2) e^(-i delta) + (1+i)^(2) e^(i delta)]`
`= I_(0)[4 -2ie^(-i delta) + 2ie^(i delta)] = I_(0) (4 - 4 sin delta)`
Here `I_(0) = (A_(1)^(2))/(4)` is the intensity of the incident light which is the same as the intensity due to an aperture of infinite extent (and no recess). Now
`I` is maximum when `sin delta =-1`
or `delta =2k pi+(3pi)/(2)`
so `h = (lambda)/(n - 1) (k+(3)/(4))` and `(b)I_(max) = 8I_(0)`.