|Apr18-07, 09:30 PM||#1|
A question on thin film
Consider a soap film where both sides of the film is air.
So reflected light has a π phase change.
Consider constructive interference,
2 d "nfilm" + 1/2λ = nλ , where "nfilm" is the refractive index of the film and d is the thickness the film.
Due to the gravity, the film at the bottom is thicker.
My teacher said that if the film is too thick, the above equation can be satisfied by more than one wavelength in the visible spectrum. So different colour combine to give a white colour.
I don't understand what is the meaning of this statement.
Is it mean that for different colour (λ), there will be a different order (n) to satisfy 2 d "nfilm" + 1/2λ ??
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|Apr19-07, 08:06 AM||#2|
The interference is between the reflected ray at the first interface and the reflected ray at the second one.
As in any case there is a phase inversion (a phase change of pi radians) at one of the two reflections. Then if the path of light inside the film equals an integer number of wavelengths (in the film), the interference will be destructive. Is it is equal to an integer plus one half, the interference will be constructive.
Imagine that the film thickness is [tex]1000.5\lambda_1[/tex] (wavelengths in the film). What is the next value of [tex]\lambda[/tex] for which the interference will be constructive? It will be when the (same) thickness of the film is equal to [tex]999.5\lambda_2[/tex] and the next for [tex]998.5\lambda_3[/tex].
Compute the ratio between consecutive lambdas. You will realize that they are really close.
And in-between you have lambdas with destructive interference.
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