 Solution

Reflections from the two film surfaces interfere with propagation phase $4\pi nd\cos {\theta }_t/{\lambda }_0$ plus the relative reflection phase. Reflection from a lower to a higher refractive index contributes a $\pi$ phase reversal. For a higher-index film surrounded by air there is one relative reversal, henceThickness or angle variation therefore produces fringes or colours under broadband illumination. An antireflection coating between indices $n_0,n_s$ instead uses, at normal incidence, $n_f=\sqrt{n_0{n}_s}$ and $n_{f}d={\lambda }_0/4$: the two reflected amplitudes then cancel at the design wavelength. Absorption, polarization, unequal amplitudes and multiple reflections determine the full contrast.