Question

A thin film suspended in air is 0.400 µm thick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.58. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?

Answer 1

Here we will use the thin film conditions for the case,

Incident light is white light (that means it contains all the wavelengths from the visible region), and the incidence is normal to the surface of the film. So we don't have to bother with pesky angles. The first ray, let's say at a wavelength is reflected back into the air. Since the reflected ray will be phase shifted by 180 degrees. This phase difference can be translated into path difference, which for 180 degrees is .

Now, consider some part of the same wavelength is transmitted into the thin film. Inside, at the second interface, the ray is reflected but at this 'wall' so the reflected ray will have 0 phase difference. It will head back again through the first interface and meet the first ray that was reflected at the surface. The second ray will then have traveled the extra distance of 2 times the thickness of the film.

So, when these two rays meet, one phase shifted by 180 and the other having a path difference of , constructive interference can only occur if the total path difference is the integral multiple of the incident wavelength .

Hence,

The wavelength we use here is for the thin film and to get the corresponding wavelength in vacuum, we divide by the refractive index of the thin film.

So, we have our condition. Let's see now what wavelength will give constructive interference,

Which is and is way beyond the visible region. So no wavelength for the visible region will reflect and give fully constructive interference.