- #1
sachi
- 75
- 1
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.
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.