Answer:
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.
