Answer
(a) $102nm $
(b) thinner
Work Step by Step
(a) The required thickness can be determined as
$ t=\frac{\lambda_{vacuum}}{4n_{mf}}$
We plug in the known values to obtain:
$ t=\frac{565\times 10^{-9}m}{4(1.38)}$
$ t=102nm $
(b) We know that wavelength is inversely proportional to the thickness of the film and frequency is inversely proportional to the wavelength. Thus, we conclude that the reflection is suppressed with light of higher frequency and hence the coating of magnesium fluoride should be thinner.
