What is the thinnest soap film (excluding the case of zero thickness) that appears black when viewed by reflected light with a wavelength of 510 nm ? the index of refraction of the film is 1.33, and there is air on both sides of the film?

There are two ways to approach this problem. Chose the one based on what your prof says.

We are told that the soap film appears black when viewed. This means that it is destructive reflection.

Now we have to determine whether there is half-cycle relative phase shift or no relative phase shift. Since air has an index of refraction of 1 and the film has index of refraction 1.33, there is a half-cycle relative phase shift. So we can use 2t=mλ (m=1 for smallest thickness). t being the thickness of the film and λ being wavelength in the material. Thus you have to do (510*10^-9)/1.33

Solve for t and we get:



Another way, unfortunately harder, is use this equation:



Here's the hard part. We have to determine phi. The phi with subscripts can be either π or 0. Phi can be 2π or π. For destructive interference phi is π. For the first reflection, air's index of refraction is less than the film, therefore phi sub r2 has a phase difference of 0. For phi sub r1, the index of refraction is higher than the air, so there is phase difference, so it equals π. So we get:



Finally compute the equation and we get:



Hope this helps!!