= 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 $n^2=n_0n_s$. Thus
$$
\boxed{n=\sqrt{1.4(1.6)}=1.49666,\qquad
d_{\min}=\frac{500\ {\rm nm}}{4n}=83.52\ {\rm nm}.}
$$
The 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.
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