Exercise 6.3.6: Metal mirrors

For aluminium

and in the same way silver $|\Gamma |=0.982$, $R=96.4\%$, gold $R=78.8\%$ at $550$ nm and $97.1\%$ at $700$ nm. Glass: $\Gamma =-0.2$, $R=4\%$.

$N^2=(0.96{)}^2-(6.69{)}^2-\mathrm{j}2\times 0.96\times 6.69=-43.8-\mathrm{j}12.8$. The real part of the permittivity is negative: below its plasma frequency the electron gas of the metal behaves like the plasma of Section 6.2. Then

a small and mostly reactive impedance: the surface is close to a short circuit, so $\Gamma$ is close to $-1$ and little power can be absorbed.

The field falls to $1/\mathrm{e}$ at $x=1/({\beta }_0\kappa )={\lambda }_0/(2\pi \kappa )$ and the power density, $\propto {\mathrm{e}}^{-2{\beta }_0\kappa x}$, at half of that:

A $100$ nm aluminium film already attenuates the power by ${\mathrm{e}}^{-200/13.1}=2\times 10^{-7}$: mirror coatings need only about $100$ nm of metal.

Silver and aluminium reflect all visible colours almost equally. Gold reflects $97\%$ of red light but only $79\%$ of green (and less of blue): the reflected white light is depleted at the short wavelengths and looks yellow.

Absorbed fraction of aluminium from the impedance, Eq. (6.3.17), in units of ${\eta }_0$:

and $1-0.921=0.079$ $✓$.