Light is incident on a lens with a thin coating. The light first travels through the thin coating, and then the lens. The thin coating has an index of refraction of n = 1.3, and the lens has an index of refraction of n=1.6. What is the minimum thickness of this coating that will destructively reflect light of wavelength 500 nm?

The minimum thickness of the coat is 96.15nm

Explanation:

firstly we will have to understand the geometry of the problem, that we have a lens coated with a thin film. Light approaches the film and approximately 50% of the light that gets reflected, gets reflected from the outer surface of the the film coating. The rest is reflected from the filmcoat-lens interface. We need these two reflected rays to meet destructively. For that to happen they must be completely out of phase that is they must differ by λf/2 where λf is the wavelength of the light in the coating medium given by λf = λo/nf, where nf is the refractive index of the film nf = 1.3 and λo is the wavelength in vacuum given in the question that is λf = 500nm. But since the second wave that gets reflected through the film-lens interface travels through the film coating twice, that means that the film coating must only be half of λf/2 that is h the film coating thickness will be given by h = (1/2) (λf/2) = (1/2) (λo / (2*nf)) = 96.15nm.

3λf/2, 5λf/2, 7λf/2 ... phase differences between the two reflected waves would also result in destructive interference but the thickness they would yield for the film would not be minimum.