How Does the Refractive Index Influence Light Interaction with Metals?

[sudu.ghonge]

When a light wave of the form E=E₀e^iω hits the surface of a metal, the refractive index if given(considering restoring force of metal atoms and damping equal to 0), by n²=1-p²/ω² where
p=plasma frequency and
ω= frequency of the wave.
In the case where n²<0,
n is complex. and hence the light wave is completely damped.
What my question is, does that meant that the wave is reflected back or, does it mean that the energy is lost as heat?

Also, to explain the case of an n² > 0, an example of the ionosphere is often quoted where n keeps varying with altitude and hence radio waves that we send from the Earth's surface turn and get reflected back to the surface. Does that mean that more the substance is metallic, more it acts like a reflector? At this point I'm pretty much confused about what I even want to ask. Please enlighten. Thank you.

 

[DrDu]

Yes, and ideal metal described by the Drude formula you gave is completely reflecting up to the plasma frequency. As n^2 is the dielectric function and this expression is real, there is no dissipation (generation of heat). In real metals the formula for n^2 contains some complex terms.
This has some puzzling consequences: The metal will absorb some of the incident light and reflectivity will decrease but also transmittance will increase. Probably you would expect transmittance to decrease with rising absorptivity, e.g. when looking at a dilute solution of some dye. An example of this behaviour is gold: Gold does not reflect very well in the green-blue part of the spectrum due to transitions from the d band to the conduction band. However very thin sheets of gold (as in some light bulbs) will transmit just this greenish-blue part of the spectrum. The reason is the following: A metal can only absorb some of the light if it can enter the medium (i.e. if not all of the light gets reflected). But then this light has some chance to escape the metal on the other side unabsorbed.