I am currently reading about the interactions between light and matter, but I keep coming across conflicting explanations.
My initial understanding (using classical electrodynamics) was that light (EM) waves do not actually travel through glass. Instead, they are absorbed by the atoms and/or electrons within the glass, which then emit new EM waves. The new EM waves emitted by the energised particles make up the forward-radiation, which is the light observed from the opposite side of the glass. So it appears as if light is travelling through the glass, but it's actually being absorbed and then re-emitted with (mostly) the same properties.
The article How Does Light Travel Through Glass? by Chad Orzel, based on a related SE question, confirms the above:
To understand the propagation of a wave through a medium, you can
think of each component of the medium– atoms, in the case of a glass
block– as being set into motion by the incoming wave, and then acting
as a point source of its own waves.
But an answer to a different but related question, If light is made of particles, how does it pass through glass? conflicts with the above:
the electrons in glass are tightly bound to atoms so they are not free
to move like the electrons in a metal and therefore they do not absorb
the photons.
I know the above is referring to the quantum electrodynamics understanding, but I think it still conflicts as it claims that EM waves are not absorbed by the electrons in glass because they are too tightly bound to their atoms.
Also, the Wikipedia page on reflection provides a small explanation of the mechanics behind refraction within glass, which kind of conflicts with both of the above:
In the case of dielectrics such as glass, the electric field of the
light acts on the electrons in the material, and the moving electrons
generate fields and become new radiators. The refracted light in the
glass is the combination of the forward radiation of the electrons and
the incident light. The reflected light is the combination of the
backward radiation of all of the electrons.
From my understanding, the above claims that some of the light is absorbed and re-emitted by the electrons (which conflicts with the linked answer), and that the rest of the incident light travels through the glass (makes up the rest of the refracted light) (which I think conflicts with the linked article, claiming the light is absorbed (although the article doesn't technically state whether all of the light is absorbed and re-emitted, so maybe that's where I'm going wrong(?))
I'm sure this is my misunderstanding, so I would be very grateful if you could clear it up for me:
When a light wave is incident on a sheet of glass, do the particles that make up the glass absorb the waves and re-emit them in the forward direction? Or do the light waves manage to travel through the glass, without being absorbed and then exit the glass?
Best Answer
The short answer to the core of your question,
is that both processes are at play, though there are important subtleties involved in the term 'absorb'. The transmitted light is produced by the interference of the original beam and the radiation produced by the excitations it produces inside the glass, and it is this interference that produces the additional phase encoded in the glass's refractive index, which then trickles out to the phase velocity and the change in the wavelength.
To see why this is the case, then the best place to start is with the macroscopic-electrodynamics description of the glass slab, in which the essential concept is that of the glass's polarization density $\mathbf P$. Electrodynamics is a linear theory, and it sees the problem of light propagating through glass as a superposition of two different sources:
The field after the glass slab is produced by the interference of the initial beam and the radiation produced by the oscillating charges inside the glass.
However, here is the important thing: the glass can do this induced-oscillations-then-emission game without absorbing any energy at all. In the steady state, the charge oscillations inside the glass are exactly 90° out of phase with the electric field that drives them, which means that the net power that they emit is being put in, in equal amounts, by the driver.
If you want to go further than that and talk about photons, though, then you need to be a good bit more careful. If you're thinking about the light as a quantized object, then it is a quantized object that is coupled to the matter it is travelling through: in other words, it is a joint excitation of the EM field and the charge oscillations it produces in the glass.
In essence, then, the photon briefly becomes a polariton while it is in the glass ─ though the term is generally reserved for situations with much stronger coupling to resonant energy levels where there is a substantial population in those excited energy levels (or, more precisely, a substantial probability amplitude of excitation of each individual emitter to those energy levels) than what happens in an everyday dispersive dielectric. In glass, the probability of excitation (i.e. the "probability that the atom has absorbed a photon") is negligible as far as its contributions to the internal energy go (so it is often called a 'virtual transition'), but that probability amplitude does contribute to charge oscillations which re-shape the phase of the beam without altering its amplitude.
That said, of course, you still do need to provide the initial bit small of energy to take that small population to the excited states, and this is normally taken from the leading edge of the pulse while the monochromatic situation is being set up ─ and generally returned back to the field when the pulse leaves, as long as the medium is transparent. (Up to a point, of course - there's some strict constraints on how dispersive a medium can be without absorbing energy. However, there's plenty of transparent media where the absorption is extremely small and this can be disregarded.)