The particles associated with the electromagnetic waves, described by Maxwell's equations, are the photons. Photons are massless gauge bosons, the so called "force-particles" of QED (quantum electrodynamics).
While sound or the waves in water are just fluctuations (or differences) in the densities of the medium (air, solid material, water, ...), the photons are actual particles, i.e. excitations of a quantum field. So the "medium" where photons propagate is just space-time which is still there, even in most abandoned places in the universe.
The analogies you mentioned are still not that bad. Since we cannot visualize the propagation of electromagnetic waves, we have to come up with something we can, which is unsurprisingly another form of a wave, e.g. water or strings.
As PotonicBoom already mentioned, the photon field exists everywhere in space-time. However, only the excitation of the ground state (the vacuum state) is what we mean by the particle called photon.
The short answer to the core of your question,
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?
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 initial light beam, which is present at the glass slab as well as beyond it, and
- the charge oscillations that initial light beam excites in the glass, which primarily emit (because they are essentially a phased array) in the initial beam's direction.
- (There are also surface terms produced by those charge oscillations, which produce reflections at the boundaries, but I'll ignore those here.)
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.)
Best Answer
I'm firmly of the opinion that the best way to think of the 'wave-particle duality' of quantum mechanics is to simply accept that photons, electrons, etc are neither particles, nor waves; what is 'real' is that they are described by a wave-function governed by the laws of quantum mechanics. This wave-function has properties which remind us, in the macroscopic world, as waves in some ways and particles in other ways.
This is false. Maxwell's equations from which the wave equation for EMR is derived need not make reference to a medium.
You're right! But the vibrations of spacetime itself actually yield another type of wave: gravitational waves.