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Recently I was asked to explain the difference between reflection and total internal reflection from a purely conceptual standpoint (no math).

Let me explain what I already know. Reflection and refraction at the quantum level are the same thing. Light is a photon. A photon is a discrete particle that has wave characteristics (a finite wave traveling like a bullet). As the photon travels it collides with electrons in the matter of the medium it is traveling in. Depending on the energy of the photon and the allowed energy bands of the medium the photon cause the electron to jump up a level. If the photon is absorbed then the medium will increase its motion (at the macro scale increasing its temperature). If the photon is not absorbed it will be re-emitted (really as a new photon). I have read and watched Feynman's QED lectures and book and have a pretty good understanding of his process for determining how all these paths come together to give the net path of the photon. The general rule of thumb is that the photon wants to take the path which requires the least amount of time. I understand how this principle goes to explain refraction and reflection.

What I don't seem to understand is why does one material seem to cause a higher percentage of refraction compared to another (metal vs. glass). What about the electron configuration of a the medium changes the net effect of the absorption and re-emissions of the photons? Is there a change in the probability of the photon being re-emitted in a reverse direction? Is there a farther distance the photon can travel before being incident onto an electron? This is the part where my understand breaks down.

When you have hit the critical angle in a medium that refracts and the light completely reflects, are the photons moving is the same manner as they would be in a material that always reflects? How does this connect to the question in the previous paragraph?

I know I have a bunch of mini questions embedded in answering this one larger question. Any help on any of the parts would be greatly appreciated?

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1 Answer 1

First, there is a qualitative difference between metal and glass: metal is a conductor, while glass is a dielectric. Under so called "plasma frequency" EM waves do not travel in conductors except that near the surface (see skin effect). For higher frequencies (usually far above the visible light) metal is transparent and refraction does occur. Specifically, answering to your question

What about the electron configuration of a the medium changes the net effect of the absorption and re-emissions of the photons?

Metal has a substantial amount of free electrons, glass doesn't. From the EM point of view, the difference originates from the relative effect of two terms in the Maxwell equation (roughly speaking, the corresponding dimensionless parameter depends on conductivity and wave frequency). From the QM point of view, the interaction with electron gas is substantially different from the interaction with an atom.

When you have hit the critical angle in a medium that refracts and the light completely reflects, are the photons moving is the same manner as they would be in a material that always reflects? How does this connect to the question in the previous paragraph?

Again, from the EM point of view both of these processes result in an evanescent wave and are quite similar. I am not sure what exactly happens with a specific photon (as a particle) here, as the evanescent wave is a general wave effect. Perhaps someone more knowledgeable in QED can provide a QM perspective on this.

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From the ensemble of photons the classical EM wave emerges naturally, though not simply :) . Lubos has an article in his blog on this motls.blogspot.gr/2011/11/… –  anna v Jul 10 '13 at 14:43
    
I pretty much understand how light viewed as a wave explains the situations, but I guess what I am looking for is how the QED approach explains it. –  gmccabe Jul 12 '13 at 12:04

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