InterviewSolution
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InterviewSolution
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Refractive index of a thin soap film of a uniform thickness is 1.34. Find the smallest thickness of the film that gives in interference maximum in the reflected light when light of wavelength `5360 Å` fall at normal incidence. |
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Answer» Condition for maximum interference in the reflected light, in case of thin film interference, can be expressed as `2 mu t cos r = (2 n - 1) lamda//2, n = 1, 2,…` [for plane parallel films] where `mu` is the refractive index of film relative ot the surrounding: `t` is the thickness of film , and r is the angle of refraction. For normal incidence, `r = 0` `:. 2mu t = (2n - 1) lambda//2, n = 1, 2, 3 ,...` For minimum thickness, `n = 1` `:. 2 mu t_(min) = lambda//2` or `t_(min) = (lambda)/(4 mu)` Given: `lambda = 5360 Å` and `mu = 1.34` `:. t_(min) = (5360)/(4 xx 1.34)` or `t_(min) = 1000`. |
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