Reflection of light

[chilehed]

Back when I was in college physics, when we were talking about optics, my professor told us in passing that reflection isn't actually like balls bouncing off of a wall, that what really happens is that the incident light is absorbed by and then immediately re-emitted from the atoms of the surface.

 

After class I told him of my understanding that exited atoms emit photons at random times and in random directions, and asked him how it could then be that in the case of reflected light the direction corresponds to the angle of incidence. If reflection actually involves an absorption/emission process, I'd expect reflections to be a lot more diffuse than they are.

 

He began nodding his head before I'd half asked the question (he knew what was coming), and answered "we really don't know".

 

This evening my 12-year old son was asking questions about reflections and polarization, and it reminded me of all of this. Can anyone tell me if my prof was correct, and if he was explain how it works (without too much technical detail because I'm just a Mechanical Engineer and my background in theoretical physics is very limited)?

 

Thanks.

[Norman Albers]

I am just really getting into quantum wave theory, but I'll hazard a guess about the structure of the problem. Look first at the classical field analysis, where by equating boundary conditions at the surface of two materials, you derive angle bending and also strength of transmission and reflection. I suspect that a quantum theorization derives a similar sum of possible paths, but interprets the intensity (squared) as probability that one photon will go somewhere. I am curious here to be sure: in a thread on 'wave packets' scalbers mentioned a thickness of dielectric material and quantum 'choice' of reflection/transmission. On the other hand I just read Feynman Lectures on his really classical treatment of reflections of a metallic surface. The first point is to distinguish between this, where the bulk of the material does not pass light, but the surface shows an imaginary index of refraction (to me this speaks of skin current); and a transmitting medium (a glass). In conversation a few days ago with solidspin he said, yes and we treat the Hamiltonian of the whole surface. Mathematically I am starting to wrap my head around the idea of quantum results being seen by accepting the reality of normal modes of the system. ]I have worked with these as a product of Fourier-transform on the system, and it's what you come up with in k-space is the complete set of 'analytic' modes, a basis set of physical possibilities. In either case, the condition of the surface is important. A good mirror has no variations on a scale comparable to the light you are considering, and this pertains to metal or glass. If you think of a specular source, say a pointed wall, the surface will have some characterization of "smoothness" and some bumpiness which will scatter light. Think now about the highway mirage in desert driving.

Edited September 11, 2008 by Norman Albers

[elas]

After class I told him of my understanding that exited atoms emit photons at random times and in random directions, and asked him how it could then be that in the case of reflected light the direction corresponds to the angle of incidence. If reflection actually involves an absorption/emission process, I'd expect reflections to be a lot more diffuse than they are.Thanks.

 

I'm no expert, but surely the answer is in the question. An excited atom is bouncing all over the place emitting photons in directions determined by the direction of the atom at any given moment. But, a single photon striking an atom is either retained or ejected; if ejected then the ejection angle is determined by its arrival angle regardless of whether or not it is momentarily absorbed.

Consider large drops of rain hitting a pool, the 'reflected' water is not the original raindrop but it does contain the original energy. Density determines what matter is 'reflected'; energy carries the matter away.

The speed at which the reflected matter proceeds is partly determined by charge in that charged particles are subject to resistance, zero charged particles are not subject to resistance but they are subject to density so the speed of zero charged particles in water is slower than zero charged particles in air; but in both cases the speed is constant. Whereas the speed of a free electron in water is steadily decreasing its final speed being determined by the excited state of the water molecules (temperature).

[swansont]

Back when I was in college physics, when we were talking about optics, my professor told us in passing that reflection isn't actually like balls bouncing off of a wall, that what really happens is that the incident light is absorbed by and then immediately re-emitted from the atoms of the surface.

 

After class I told him of my understanding that exited atoms emit photons at random times and in random directions, and asked him how it could then be that in the case of reflected light the direction corresponds to the angle of incidence. If reflection actually involves an absorption/emission process, I'd expect reflections to be a lot more diffuse than they are.

 

He began nodding his head before I'd half asked the question (he knew what was coming), and answered "we really don't know".

 

This evening my 12-year old son was asking questions about reflections and polarization, and it reminded me of all of this. Can anyone tell me if my prof was correct, and if he was explain how it works (without too much technical detail because I'm just a Mechanical Engineer and my background in theoretical physics is very limited)?

 

Thanks.

 

 

Yes he was, and sort of, but no, not really. i.e. yes, he was correct, but no, any complete answer is going to involve quantum physics.

 

Your instinct about absorption is correct — but only for real states of an atom or molecule. But there are ways of looking at the system where the photon is absorbed in what is known as a virtual state; the photon is almost immediately re-emitted, and it follows Snell's law or the law of reflection, depending on which way it goes, from conservation of energy and momentum. This is why, in the QM picture, light slows down in a medium — the absorption and re-emission take time.

 

It also explains the existence of Brewster's angle for (non)reflection of polarized light — a dipole does not radiate along its axis, so light polarized perpendicular to the plane cannot reflect, because the light cannot be re-emitted in that direction. It must be transmitted.

[solidspin]

I agree -

 

so, to flesh out your concept, as an exercise, think of all the means by which light is absorbed/reflected - gorgeous colors from transition metals are absorbing "white" light, but their d electrons (think cadmium yellow, cobalt or prussian (Fe) blue, etc.) are transmitting only the wavelengths of the color we see so vividly w/ our horribly wavelength-restricted eyes. Then, think about radio antennae absorbing all of those photons at a particular, favorite radio station, right? then a more QM picture comes into view, of the plasmons on the surface of the metal being polarized and resonating according to the initial polarization and resonant frequency (90.7Mhz out here in NYC!!) - though this is largely only absorption. Coincidentally, it's the same frequency on my magnet for 79bromine, to which I'm tuning right now. Of course, then the classic mirror example comes into maybe a more mature view, right? The mirror is just silver-backed glass, but the same plasmon phenomena are occurring, as noted above, w/ Brewster's angle. Do a simple experiment - take some polarizing mylar, shine regular flashlight and look at how effective the polarized light 1) comes through and 2) the efficiency w/ which the polarized light is subsequently reflected off the mirror, onto a surface where you can observe the results...