A soap film is viewed in white light. If the film is much thinner than the wavelength of blue light, what is the appearance of the film?

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The effect involved in here is called thin film interference. It occurs when light reflected from the upper boundary (or first surface) of a material interferes with the light reflected from the lower boundary (or second surface). This interference can be destructive (when the phase difference is equal to `pi` ) or constructive (no phase difference).

We need to use a fact to understand what we will see in our case. For a soap film (or water), when light hits the first surface (from air to "water"), the reflected light has a phase change of `pi`Some of the light rays get transmitted (with no phase change) and reflected on the second surface (with no phase change as well). At the end, the first reflected light ray has a phase change of `pi` while the second reflected light ray has no phase change (when compared to the original incident light ray).

Now, when we assume that the thickness of our film is much smaller than the wavelength of the blue light, we see that the difference in the path lengths (difference in path lengths give rise to phase differences too) of the two reflected light rays is negligible, thus, no new phase change is created and the reflected rays have a phase difference of `pi` ( because `pi` - 0 = `pi` ). So for the blue light, all interference is destructive for the reflection. And since the blue light is the one with the smallest wavelength that we can see, all the other colors get destructive interference as well (for the same reason). Thus, the soap film becomes black (reflectionless would be a better term because we can still see through the soap film), because no light is being reflected from it!

If we increase the thickness of our soap film to around one-quarter of the wavelength of the blue light, only red and green light will be reflected, and the soap film will appear yellow.


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