Find the thinnest oil film of refractive index 1.52 on the surface of water for

Question

Find the thinnest oil film of refractive index 1.52 on the surface of water for which constructive interference for a red light of vacuum wavelength 632.8nm takes place by reflection. What is the thinnest (non-zero) thickness when the film appears dark? (Assume the light beam incident to the film perpendicularly and the index of refraction for soap water is n=1.33)

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Explanation / Answer

Given that refractive index of oil is n = 1.52 wavelength of light = 632.8 nm -------------------------------------------------------- The condition for the destructive interference is           2nt =m For thinnest film m =1           2nt = So the thickness of the film is             t = /(2n)               = (632.8 nm)/(2(1.52))                = 208.15789 nm

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