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A parallel beam of white light falls on a thin film whose refractive index is equal to n = 1.33. The angle of indices is theta_(1) = 52^(@). What must be the film thickness be equal to for the reflected light to be coloured yellow (lambda = 0.60 mu m) most intensity ? |
Answer» Solution : Path difference between `(1)&(2)` is `2n d SEC theta_(2) - 2D tan theta_(2) sin theta_(1)` `=2d(N-(sin^(2)theta_(1))/(n))/sqrt(1-(sin^(2)theta_(1))/(n^(2)))=2d sqrt(n^(2) - sin^(2) theta_(1))` For bright fringes this must equal `(k + (1)/(2)) lambda` where `(1)/(2)` comes from the phase CHANGE of `pi` for `(1)`. Here `k = 0, 1, 2,....` Thus `4d sqrt(n^(2) - sin^(2) theta_(1)) = (2k + 1) lambda` or `d = (lambda(1 + 2k))/(4sqrt(n^(2) - sin^(2) theta_(1))) = 0.14 (1 + 2k) mu m...` |
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