What's the reason behind metals' high optical reflectivity?
[Physics] Why do metals have high optical reflectivity
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In the optics community we usually refer to $\text{Al}_2\text{O}_3$ as sapphire instead of corundum which is the geologists name for it. However, when we talk about sapphire in optics we are usually referring to the crystalline form, and I'm not sure if your $20\ \mu\text{m}$ coating will be crystalline or not. Another difficulty lies in the fact that the optical properties of sapphire can change dramatically with small amounts of doping which is why the gemstones can be found in all sorts of different colors
Unfortunately, I was unable to find a lot of data on Sapphire at $10.6\ \mu\text{m}$. All of the plots I could find (such as the one below) show that the tranmissivity at that wavelength is very low. Unfortunately, the losses of many optical material at such long wavelengths can be very high. I know that in fused silica the absorption coefficient is around 90%. So, the low transmittance doesn't necessarily translate to high reflectance.
*Image From: http://www.crystran.co.uk/optical-materials/sapphire-al2o3
The resonances are due mainly to plasma oscillations. But these metals are not simple plasmas; there are interband transitions as well as collective plasma oscillations, and the frequencies of these various excitations can overlap causing interactions. In silver, there is no interband transitions at the plasm frequency, so the plasma resonance is not damped, whereas in other metal, that is not the case. Even in silver interband transitions play a role in shifting and sharpening the resonance.
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I assume you’re referring to the visible range of the spectrum, and so the answer to this question comes down to essentially three things:
Facts (1) and (2) lead to a large, Drude-like conductivity of the metal, which in the visible range is primarily imaginary (meaning the oscillating electrical current in the metal excited by the light is essentially $\pi/2$ out of phase with the light). Thus, the re-emitted light is completely out of phase, and so the light’s electric field basically goes to zero at the metal surface. This condition only applies if there is very little auxiliary absorption (i.e. if fact (3) is true, which it is for silver, say, but not for gold in the blue/green part of the spectrum, which is why gold has its color).
Since the light field goes (close) to zero at the surface, it has very little penetration into metal at all, meaning the vast majority of the light power is reflected. This can be understood as the free electrons in the metal moving at the surface to effectively screen out the light field from getting into the bulk, and, in doing so, re-radiating the power outwards again.