# A soap film(n=1.33) in air is 320nm thick.if it is illuminated with white light at normal incidence,what colour will it appear to be in reflected light???

1
by Deleted account
which topic is dis

2015-04-07T02:21:32+05:30

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λ = light ray's wavelength :  380 nm  to  760 nm  for visible light (white rays).

μ_film = 1.33
thickness of film = d = 320 nm
θ1 = 0°  = angle of incidence at the top layer of film
μ_air = 1
θ2 = angle of refraction into the soap film = angle of reflection at the film-air interface.
μ_air Sin θ1  = μ_film  Sin θ2          --- (1)

There is a phase shift of 180 deg. when light rays are reflected at the top surface of the film, as μ_air < μ_film.  There is no phase shift when rays are reflected at the  film-to-air interface.  The OPD (optical path difference) is:

2 μ_film  d  Cos θ2  = (m - 1/2) λ   for constructive interference.    --(2)
2 μ_film  d  Cos θ2  = m λ    for destructive interference.    -- (3)
here m = an integer.

As θ1 = 0°,  θ2 = 0°.
2 * 1.33 * 320 nm  = Optical path difference OPD
= 661.12  nm

In equations  (2) and (3),  the RHS is λ/2, λ, 3λ/2 ...  We have the range of white light wavelengths mentioned above.
661.12 = λ for red color  satisfies equation  (3).  So red color is destroyed.
661.12 = 1.5 λ for  λ = 440.7 nm satisfies equation (4),   ie., violet color is constructively added.