 Solution

For a lossless antireflection coating, cancellation requires both equal reflected wave amplitudes and an odd half-cycle difference in their round-trip wave phase. At normal incidence the interface amplitude coefficients are $r_{01}=(n_0-n)/(n_0+n)$ and $r_{12}=(n-n_s)/(n+n_s)$. Choosing $n_0<n<n_s$ makes both negative: both reflections acquire the same $\pi$ phase reversal. A quarter-wave optical thickness supplies their remaining relative phase $\pi$.

Equal amplitudes require $r_{01}=r_{12}$, which reduces to $n^2=n_0{n}_s$. ThusThe exact multilayer reflection amplitude is zero for these choices, not merely the sum of the first two reflected rays: its numerator is $r_{01}+r_{12}{e}^{2i\beta }=0$, with $\beta =2\pi nd/{\lambda }_0=\pi /2$. Any positive odd multiple of this thickness has the same cancellation at the design wavelength; the smallest is the thin-layer choice. absorption, index dispersion, angular spread and deviations from the selected wavelength limit the practical cancellation.