Question

An oil film of thickness 0.12 micrometer rests on a pool of water. If a ray strikes the film at an angle of 60°,what is the wavelength reflected in the first order?

Answer 1

The condition of the constructive interference in the oil-water film (see https://en.wikipedia.org/wiki/Thin-film_interference#Oil_film):

$$2d\sqrt{n_{oil}^2 - \sin^2i} + \dfrac{\lambda}{2} = m\lambda$$

where $n_{oil} = 1.5$ is the index of refraction of oil, $i = 60\degree$ is the incident angle, $d = 0.12\times10^{-6}m$ is the thickness of the film and $m = 1,2,3,...$. The first wavelenght will be at $m = 1$, thus:

$$\lambda = 4d\sqrt{n_{oil}^2 - \sin^2i} = 4\cdot 0.12\times 10^{-6}\sqrt{1.5^2 - \dfrac{3}{4}} \approx 623.5\space nm$$

Answer. 623.5 nm.