Light Energy Absorption In Mirror

Question

Let the amount of energy in one pulse of (laser) light be $E$, and the wavelength be $\lambda$.

This pulse goes straight to the mirror, and it is reflected by the mirror.

Let the reflectivity of this mirror be $r_{\lambda}$.

Question. How can I get the formula of $r_{\lambda}$?

Any answers (or hints) will be appreciated, thank you.

Answer 1

You can obtain the reflection coefficient using the Fresnel equations: http://en.wikipedia.org/wiki/Fresnel_equations

For a metal mirror and perpendicular incidence the (intensity) reflectivity is given by:

$r_\lambda=|R_\perp|^2=\frac{(1-n(\lambda))^2+\kappa(\lambda)^2}{(1+n(\lambda))^2+\kappa(\lambda)^2}$

Here $n$ is the real part of the refractive index (e.g. n(530 nm)=0.95 for Aluminium at green (530 nm) light) and $\kappa$ is the imaginary part (e.g. $\kappa(530 nm)=6.4$ for Aluminium). So for Aluminium you obtain $r_\textrm{green}=91\%$.

You can find values for other metals and wavelengths here: http://refractiveindex.info/?group=METALS&material=Aluminium .

In general $r$ increases with the frequency of light. The electrons can't follow the field for high frequencies. One can measure $n$ and $\kappa$ of a material by Ellipsometry http://en.wikipedia.org/wiki/Ellipsometry . $\kappa$ is related to the electrical conductivity $\sigma(\omega)$ of the metal.