Question

A 659-nm-thick soap film (nfilm=1.33) rests on a glass plate (nglass=1.52). White light strikes the film at normal incidence.

What visible wavelengths will be constructively reflected from the film? If there is more than one answer, separate them by a comma. (Units in nm)

Answer 1

Both of these reflections suffer a ½-wave (180°) phase change as they meet a boundary of greater ref. index.
When these refletions re-combine constructively their path difference in the film has been Nλ'

where N = 1, 2, 3, 4 ... and λ' = wavelength in the film (λ' = λair / nfilm)

For a film thickness t the wave path diff. = 2t (down and back up) = Nλ'

2t = N(λair/nf)
λair = 2t.nf / N = 2 x 659*10^-9m x 1.33 / N
λair = 1.76*10^-6 / N

Assuming that visible light has a range 400 - 700nm .. {4.0 - 7.0}^-7 m
For ..
N = 1 .. λair = 17.6*10^-7 m
N = 2 .. λair = 8.8*10^-7 m
N = 3 .. λair = 5.86*10^-7 m
N = 4 .. λair = 4.4*10^-7 m
N = 5 .. λair = 3.52*10^-7 m

The wavelengths for constructive interference in the visible range are ..
►5.86*10^-7 m and 4.4*10^-7 m