InterviewSolution
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InterviewSolution
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A plane light wave with wavelength `lambda = 0.57mu m` falls normally on a suface on a surface of a glass `(n = 1.60)` disc which shuts one and a half Fresnel zones for the observation point `P`. What must the minimum thickness of that disc be for the intensity of light at the point `P` to be the highest ? Take into account the interference of light on its passing through the disc. |
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Answer» We follow the argument of `5.103`. We find that the contribution of the first Fresnel zone is `A_(1)=-(4pi i)/(k)a_(0)e^(-ikb)` For the next half zone it is `-(A_(2))/(2)(1 + i)` (The contribution of the remaining part of the `2^(nd)` Fresnel zone will be `-(A_(2))/(2)(1 - i))` If the disc has a thickness `h`, the extra phase difference suffered by the light wave in passing through the disc will be `delta = (2pi)/(lambda) (n - 1)h`. THus the amplitude at `P` will be `E_(P) = (A_(1)-(A_(2))/(2)(1+i))e^(-i delta) -(A_(2))/(2)(1 - i) + A_(3) - A_(4) - A_(5)+...........` `~~((A_(1)(1 -i))/(2))e^(-i delta) + (iA_(1))/(2) = (A_(1))/(2)[(1 - i)e^(-i delta) + i]` The corresponding intensity will be `I = I_(0) (3-2 cos delta - 2 sin delta) = I_(0) (3-2 sqrt(2) sin(delta + (pi)/(4)))` The intensity will be a maximum when `sin(delta + (pi)/(4)) =-1` or `delta + (pi)/(4) = 2kpi + (3pi)/(2)` i.e., `delta = (k + (5)/(8)).2pi` so `h = (lambda)/(n - 1) (k + (5)/(8)), k = 0, 1, 2,.........` Note:- It is not clear why `k = 2` for `h_(min)`. The normal choice will be `k = 0`. If we take `k = 0`, we get `h_(min) = 0.59mum`. |
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