Thin-film Interference

Have you ever noticed the colorful sheen of oil on top of a puddle, just after a rain shower? Or the colors in a soap bubble? Or the iridescence of a butterfly's wings, peacock feathers or beetles?

The surface of butterfly wings, hummingbirds, beetles and peacock feathers have microscopic ridges. Light reflecting off the high points and the low points of the ridges incur a path difference, as shown in the diagrams below. For those colors (wavelengths) where the path difference is λ (or 2λ or 3λ), constructive interfence occurs and those colors are present in the reflected light. For other colors (wavelengths) for which the path difference amounts to a half-wavelength (or 1½ or 2½) shift and therefore results in destructive interference. Those colors are eliminated from the reflected light. Which colors get reflected depends on the height of the ridges (which can vary from place to place on a wing, feather or shell, and also on the angle from which you are viewing. This phenomenon is called iridescence.

In the case of a sheen of oil or a soap film, the colors you see are the result of so-called thin-film interference. Light rays reflect off both the front/top surface and the back/bottom surface of the film. The ray that went through the film and reflected off the back/bottom surface had to travel a longer distance -- twice the film's thickness for normal incidence. If that extra distance amounts to an integer number of wavelengths, the waves will constructively interfere. If that extra distance is an odd number of half-wavelengths, then destructive interference occurs. This is very similar to iridesence, with a few twists. First, the wavelength in the film is what matters, and that is λ/n, where λ is the wavelength in the air, and n is the index of refraction of the film (also called n₁ in the diagrams). The second complication is that light expereiences a phase shift when it reflects off a transparent surface with a higher index of refraction. In other words, crests become troughs and vice versa. You saw this if you did Activity 1 in the righthand margin. Such a surface is called a hard boundary. In the first diagram below it doesn't really matter, because the phase shift happens at both the air-to-film surface and at the film-to-whatever-the-film-is-sitting-on interface.

However, in the case where the back of the film (the second boundary) is a soft boundary (n2<n1) then no phase change occurs at that reflection. This effectively reverses the conditions for constructive and destructive interference. That is, in this situation, destructive interference now happens when the physical path difference is an integer number of whole wavelengths.

CREDIT: the diagrams above came from the blog of Michael Mow, unknown if he is the creator.

Activities & Practice
to do as you read

1. Play again with the PhET simulation of a transverse wave on a string. Specifically, set it up with very little friction, in Pulse mode and with a fixed end. Send a pulse down the string. What happens to it when it reflects off the far (fixed) end? Now hit 'Reset', change the end to be loose and send another pulse. How is the behavior of this pulse's reflection different than the first?

Additional Activities & Practice