Homework Help: Need help with thin film interference

1. Dec 9, 2004

pnazari

A student diving in a swimming pool (filled with water with index of refraction
1.33) creates thin films of air. Viewed underwater, what are the first two non-zero thicknesses of an air film for which there will be constructive interference for reflection off the surfaces of the bubble for light of wavelength 655 nm in water
a. 123 nm, 369 nm
b. 218 nm, 653 nm
c. 218 nm, 873 nm
d. 246 nm, 492 nm
e. 436 nm, 871 nm
f. 436 nm, 1091 nm

The answer is B, but I keep getting D. Here is what I have done:

Lambda=2nd/m (m=1,2,3..,n=index,d=thickness)

2. Dec 10, 2004

Andrew Mason

Tricky problem.

It has been a while since I have looked at this area but I do recall from optics that quarter wave coatings minimize reflected glare by making the reflected wave from the air/film surface out of phase by $\pi$ with the reflected wave from the film/glass surface, resulting in these two reflections destructively interfering. Since the light from a lower index of refraction medium meeting a higher one (air to film and film to glass) will reflect with a 180 degree phase shift at both surfaces, we don't have to worry about the phase change on reflections. So the distance of travel through the film (i.e from air/film surface to the glass surface and back to the air/film surface) just has to be a half wavelength longer. So the thickness of the film has to be $\lambda/4$.

When light (or any wave) goes from a higher index to a lower index medium it
does not have a 180 degree phase shift. The reflected wave is in phase with the incident wave. But the reflected wave from the other surface (air to water) does have a 180 degree phase shift because it is going from a low to high index medium. That is the key to this problem.

To provide CONSTRUCTIVE interference (maximum reflection), a phase shift of $2\pi$ is required. A $\pi$ shift is provided by the reflection from the air to water surface. So the two way travel through air has to produce a $\pi$ phase shift. This means that the thickness of air has to produce a $\pi/2, 3\pi/2, 5\pi/2 ...$ phase shift, so the minimum would be a quarter of the wavelength of the light in air and the next 3/4 wavelength. Since $\lambda_{air} = 1.33 \times 655 nm = 871 \text{nm}$ you have your answer.

AM

Last edited: Dec 10, 2004