You are given a film with n = 1.60 that is deposited on glass. You are also given the wavelength of light that is 510nm. You are asked to find the thinnest film that will produce constructive interference in the reflection of light. To solve this, find the wavelength of the film and that is 510nm/1.62 = 314.815nm. Because there is constructive interference, there is a phase reversal on the reflection of the above surface and no reversal on the second surface reflection. Therefore, the light will traverse into the film thickness two times. 314.815nm/4 = 78.704nm.
