I have a very basic doubt: If we shine light on a metal surface, it is well known that frequency components exceeding the threshold frequency of the metal get absorbed to contribute to photo-electrons. What happens to the frequencies lower than the threshold? Are they completely reflected? or a part of it is absorbed till the skin depth which leads to heating up of the metal? In the latter case, is there any way to calculate how much fraction of radiation is reflected and how much is absorbed (excluding PE effect)?
To complete what Alan just wrote; once you get the dispersion curve (n,k vs wavelength), the reflectance is calculated with (assuming an opaque metal, and normal incidence):
R = [n2(1+k2)+1-2n]/[n2(1+k2)+1+2n].
All the light that is not reflected is absorbed in the metal (since it is opaque).
There are published measurements of the real and imaginary components of the complex dielectric constant (relative permittivity). Use these to calculate the fraction of power reflected using the Fresnel equations with complex refractive index.
http://refractiveindex.info has a materials database and on-line reflectance calculator.
http://refractiveindex.info/?shelf=3d&book=metals&page=copper
http://refractiveindex.info/?shelf=3d&book=metals&page=iron
https://en.wikipedia.org/wiki/Refractive_index#Complex_refractive_index
To complete what Alan just wrote; once you get the dispersion curve (n,k vs wavelength), the reflectance is calculated with (assuming an opaque metal, and normal incidence):
R = [n2(1+k2)+1-2n]/[n2(1+k2)+1+2n].
All the light that is not reflected is absorbed in the metal (since it is opaque).
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