A uniform layer of water (n=1.33) lies on a class plane (n=1.52). Light shines p

Question

A uniform layer of water (n=1.33) lies on a class plane (n=1.52).  Light shines perpendicularly on the layer.  Because of constructive interference, the layer looks maximally bright when the wavelength of the lights is 432 nm in vacuum and also when it is 648 nm in vacuum. (a) Obtain the minimum thickness of the film.  (b) Assuming that the film has the minimum thickness and that the visible spectrum extends from 380-750 nm, determine the visible wavelength(s) (in vacuum) for which the film appears completely dark.

Explanation / Answer

Here,nw= 1.33 and ng= 1.52 by Snell's law nw* sin = ng* sin1 or sin1= (nw* sin/ng) Here, = 90o or sin1= (nw/ng) for constructive interference d * sin1= m * or d = (m * /sin1) Here,m = 1 and = 432 nm = 432 * 10-9 m substituting values we get minimum thickness of film. (b)for destructive interference d * sin1 = (m + (1/2)) * or = [d * sin1/(m + (1/2))] Here,m = 1,2,3,............... for above wavelengths the film appears completely dark. or sin1= (nw/ng) for constructive interference d * sin1= m * or d = (m * /sin1) Here,m = 1 and = 432 nm = 432 * 10-9 m substituting values we get minimum thickness of film. (b)for destructive interference d * sin1 = (m + (1/2)) * or = [d * sin1/(m + (1/2))] Here,m = 1,2,3,............... for above wavelengths the film appears completely dark.

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