If a non-absorbing film is put on a perfect electric conductor the reflectivity of the system is still unity. In case of real metals, interference effects become obvious through interference fringes in the reflection spectra. Since transmittance is zero in these systems, this means that absorption takes place.
In fact, this is known and these systems have been used for a long time e.g. as as high-temperature solar absorbers.
What I could not find in the corresponding literature is any hint on the microscopic nature of the absorption. It could take place in the metal due to a non-zero skin depth, but normally light cannot couple to bulk plasmons at the surface of a metal as the corresponding dispersion relation does not cross the light line. The excitation of a surface plasmon could be possible, but still light has to be coupled in. Since you also find the effect above for normal incidence it can not be a kind of prism coupling etc. Probably I miss something very basic, but what is it that is excited by this absorption and how is the energy converted to heat?
Single electron absorption, though forbidden in metals, is made possible by electron-phonon and electron-electron interections. Collective excitations like plasmons are not important here. Though the single electron transitions might be weak but the interference in the dielectric layer increases their occurrence.
First of all thank you both for your answers! Indeed, coherent interaction of the incoming wave with its reflected part is assumed, otherwise nothing would be absorbed.
Actually it is somewhat hard for me to grasp the concept that there can be absorption without an underlying mechanism except that obviously Maxwell's equations demand that there is absorption (certainly the equations just describe the underlying physics, which is what actually demands absorption). I did a couple of FDTD simulations to see where the energy is actually absorbed, and the surprising result is that most of the energy is absorbed in the layer (overall the absorptance is proportional to the electric field intensity which is confirmed by my results, and the electric field intensity is naturally much higher in the layer than in the surface of the metal). I did not think of that beforehand, because it is even more counterintuitive, since the index of absorption of the layer is by definition zero, but this is were the fields interfere...
Dear Jörn, thanks again! I checked again and have to revise my earlier posting. I have checked the energy loss inside the metal, and it is about 80 % of the overall loss. However, when I choose a perfect electric conductor and a weakly absorbing layer on top I also see the interference fringes and in this case (of course) there is no absorption within the metal. Concerning stored energy, it does not seem in the simulation as if electric field intensity is staying in the system. I'm still puzzled...
Definitely, but for a paper just about that I would need a better idea of what is exactly going on... ;-)
I was also using FDTD Solutions from Lumerical (even if the problem above can, of course, be solved fully analytically).
You mean that there is an excitation of bulk plasmons? Yes, this could explain most of the absorption. Still, a part also seems to take place in the transparent layer...
Yes, this is also what I thought in the beginning.
Fact is, there is absorption even without a corresponding peak in the imaginary part. This is a very simple system which you can solve analytically and I certainly investigated the system first analytically and used the numerical calculations just for visualization always checking if numerical und analytical solutions agree.
Sorry, I am somewhat busy at the moment. But you have the Lumerical file and you also can do the calculation by yourself... ;-)
Then what you see are the numerical errors.
By the way, I found a new system where the same phenomenom takes place, but without a metal. It involves attenuated total reflection and the latter occurs also in spectral regions where k = 0...
Your question is rather vague. What is meant here: "interference effects become obvious through interference fringes in the reflection spectra"?
In a REAL metal absorption is always the case. There are mainly two mechanisms: intraband (Drude) and interband. The first dominates at low frequencies (from 0 up to near IR), the second usually becomes essential at visible. Intraband mechanism: electrons move under the action of the field, get an additional energy from the field, are scattered by phonons and other lattice imperfections giving their energy to the lattice, which is finally heated. The interband mechanism is similar but now the energy is enough for transitions between different energy bands.
If the metal surface is flat plasmons cannot be excited by the incident radiation (neither at a normal nor at an oblique incidence).
Well, thank you very much for the undergraduate level introduction into metal optics! I try not to do the same to you concerning wave optics, but a perfect electric conductor always has perfect reflectance, which means even if you have an electric field standing wave effect (or interference effect) in an overlayer, you don't see any indication of it in a reflectance spectrum. This is not the case for real metals. There you see a reduction in reflectance in the spectrum at the spectral position of the standing wave. Accordingly you get comparably strong absorption there, much stronger than for the case without non-absorbing overlayer. I guess I already mentioned in my question, that for flat metal surfaces plasmons cannot be excited. So what causes the absorption if you have a non-absorbing overlayer above the metal surface?
What system do you consider? A semiinfinite metal (ideal or real) and a finite-width dielectric layer on top of it? And you are talking about the interference in the dielectric layer? This is your system?
It seems that I've understood the question. But the answer is simple. If the metal is ideal, the field does not penetrate to it, the boundary condition at the surface is E_tangential=0. The wave is reflected with R=1, as you write. If the metal is real the wave penetrates into the metal (the non-zero skin depth as you write), i.e. the electric field has a sinusoudal x-dependence in the overlayer and exp(-kappa*x) in the metal (if the frequency is smaller than the plasma frequency). This field causes the electron motion and intraband absorption in the metal. Neither bulk nor surface plasmons are excited in this case but this is not needed to get the finite absorption and the interference picture in the reflection spectrum.
Many thanks, this was exactly the answer I was looking for! It would be very kind if you could provide me additionally with a reference where I could do some reading and which I could cite should that be necessary.
It is much simpler to solve the problem than to search for the solution in the literature. Sorry, I do not know any reference about this.
It depends on what does the word "transparent" mean. It "transparent" means that the layer does not absorb radiation then the absorption is evidently zero. In general, the absorption is determined by the imaginary part of the dielectric function of the layer (=by the real part of the conductivity).
Ok, yes, this is still an open point. According to the simulation, only about 80% is absorbed in the metal. The rest is absorbed in the transparent layer. And transparent in the context of this question means that the imaginary part of the dielectric function is set to zero. After having given the issue some additional thought I think that it must be a numerical error of some kind, even if I can't think of one which could have such a drastic effect...
Remind me, how many orders of magnitude are the errors that you found smaller than 20 %?... ;-)
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