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A plane light wave with wavelength lambda = 0.60 mu m and intensity I_(0) falls normally on a large glass plate whose side view is shown in Fig. At what height h of the ledge will be intensity of light at points located directly below be (a) minimum, (b) twice as low as I_(0) (the losses due to reflection are to be neglected). |
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Answer» Solution :Just below the edge the amplitude of the wave is given by `A = (1)/(2)(A_(1) - A_(2) + A_(3) - A_(4)+………)e^(-I delta)` `+(1)/(2)(A_(1) - A_(2) + A_(3) - A_(4)+………)` Here the quantity in the brackets is the contribution of various Fresnel zones, the factor `(1)/(2)` is to TAKE account of the division of the plate into two parts by the ledge, the phase factor `delta` is given by `delta = (pi)/(lambda)H (n -1)` and takes into account the extra length traversed by the waves on the left. Using `A_(1) - A_(2) + A_(3) - A_(4) +......~~(A_(1))/(2)` we get `A = (A_(1))/(4)(1 + e^(i delta))` and the corresponding intensity is `I = I_(0) (1 + cos delta)/(2)`. where `I_(0) prop ((A_(1))/(2))^(2)` (a) This is minimum when `cos delta =-1` So `delta =(2k +1)pi` and `h =(2k +1) (lambda)/(2(n - 1)), k = 0,1, 2,.......` using `n = 1.5, lambda = 0.60mu m` `h = 0.60 (2k+1)MU m`. (B) `I = I_(0)//2` when `cos delta = 0` or `delta = kpi + (pi)/(2) = (2k + 1)(pi)/(2)` This in this case `h = 0.30 (2k + 1)mum`. 
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