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Deposit hydrogenated silicon nitride by plasma-enhanced chemical vapor deposition. A plasma activates a silicon-containing precursor and a nitrogen source, enabling a conformal thin film at a lower substrate temperature than a purely thermal process. Tune gas composition, power and time to control thickness, refractive index, stress and hydrogen content.

The film serves as an antireflection coating and can also provide chemical and field-effect surface passivation. As a simple optical check, a quarter-wave thickness is $d\simeq {\lambda }_{\mathrm{design}}/(4n)$; $d\simeq 75\mathrm{nm}$ and $n\simeq 2$ correspond to a design wavelength near $600\mathrm{nm}$. The textured surface and broadband illumination require a fuller optical optimization, so this check is an order-of-magnitude explanation rather than an exact reflectance calculation.

A thin antireflection coating makes reflections from its two interfaces interfere destructively near a chosen wavelength, often in the green part of the visible spectrum. For a single lossless layer at normal incidence, $n_{f}d={\lambda }_0/4$ and ideally $n_f=\sqrt{n_0{n}_s}$ minimize reflection at ${\lambda }_0$. reflection is less completely cancelled in the red and blue, so the residual reflected light can appear purple. This is primarily a reflection colour, not a claim that the transmitted beam is strongly purple. Multilayer coatings can broaden the low-reflectance band.

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.

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.