We have a question that asks us what thickness of copper is necessary to achieve optical reflectivity. I know that the answer is just basically several times the skin depth (says 10 times). the only problem is that this contradicts a previous result that I've seen. We can prove that when a wave is incident on a metallic surface that the power dissipated per unit surface area is equal to (H^2)/2sigma*delta where sigma is the conductivity and delta is the skin depth. We get this by integrating the poynting vector at z=0. This seems to suggest that the energy reflected from the surface of the conductor is independent of the depth of the copper, so surely it shouldn't matter thow thick the actual layer of copper is? thanks for your help.
It seems that the surface finish will have a large effect on the amount of reflected optical energy. But assuming that the surface is highly polished, then maybe the two effects that you mention are not completely at odds. Only some of the optical energy will be reflected, regardless of how many skin depths deep the copper is. If the copper is only 1 skin depth thick, then some of the optical energy gets transmitted, some gets absorbed, and some gets reflected. If the copper is many skin depths thick, then none gets transmitted, some gets absorbed and some gets reflected. Can you reconcile the amount absorbed with the amount reflected?