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

Question

A uniform layer of water (n=1.33) lies on a glass plate (n=1.52). Light shines perpendicularly on the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the light 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 to 750 nm, determine the visible wavelength(s) (in vacuum) for which the film appears completely dark.

Explanation / Answer

= 432 nm, ' = 648 nm path difference = 2nt where n = 1.33 2nt = k = k'' k/k' = 648/432 = 3/2 a) for minimum t, k = 3, k' = 2 t = 3/(2n) = 487 nm b) 2nt = (k" + 1/2)" 1296 = (k" + 0.5)" " = 1296/(k" + 0.5) nm k" = 2, " = 518 nm

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