Mean free path of a photon in a fiber Is there a way to calculate, or a reference table I can to look up which provides the average distance a photon travels before it encounters an electron and is absorbed or re-emitted in a fiber optics cable? I asked a fiber manufacturer but all they could come up with is the index of refraction. I'm just looking for an approximation.
 A: This is not a straightforward calculation: you need to know the detailed quantum optics of the interaction between the photon and the fibre's matter. What you are after is the absoption cross sections of the optically active "atoms" (I include molecules in this term, which stands for "interactors with light), as well as the lifetime of the excited states, i.e. how long an absorber hangs on to a photon before re-emitting it. 
If you want to think of the photon as a tiny bullet being cyclically absorbed and re-emitted, you can relate the "mean free path" to the refractive index by the following  simplistic idea, with reference to the drawing below:

If the probability of absorption per unit length is $p$, and the lifetime of the excited state that delays the photon is $\tau$, then the mean transit time of the photon through a short length $\delta z$ is:
$$(1-p\,\delta z)\frac{\delta z}{c} + p\,\delta z\,\frac{\delta z}{c} + p\,\delta z\,\tau = \frac{\delta z}{c} + p\,\,\delta z\,\tau$$
but, in terms of the refractive index $n$, this is simply $n\,\frac{\delta z}{c}$, so, on equating this to the expression above, we get:
$$p\,\tau = \frac{n-1}{c}$$
This expression is almost what you want, for $p^{-1}$ is the mean free length, i.e.:
$$L = p^{-1} = \frac{\tau\,c}{n-1}$$
But, as you can see, you also need to know the lifetime of the excited states, so you need to dig deep into the light-matter interaction details. A femtosecond absorption in glass of $n=1.5$ yields $0.6{\rm\mu m}$, i.e. an optical wavelength or so.
This is the "simplistic", bullet picture. I explain in detail in the references at the end, the way to think about a lone photon propagating through the fibre is as a quantum superposition of free space plane waves (pure, free photon) and excited atoms at all the positions in the fibre at once. With the atoms coupled to the electromagnetic field, such a quantum superposition is how we describe the fibre's eigenmodes, and they correspond to the eigenmodes as found by solving Maxwell's equations. In this viewpoint, note that there is no dynamic absorption and re-emission cycle: basis states with atoms in ground state with the photon free and atoms in excited state exist as quantum superpositions. In this viewpoint, the absorption cross sections and raised state lifetimes determine the magnitude of the complex weights in the quantum superposition.
A colleague and I give the quantum description of single photons propagating through optical fibres in a set of three JOSA-A papers:
JOSA B, Vol. 24 Issue 4, pp.928-941 (2007)
JOSA B, Vol. 24, Issue 4, pp. 942-958 (2007)
JOSA B, Vol. 24 Issue 6, pp.1369-1382 (2007)
a good summary of these is (probably more readable):
R. Vance, F. Ladouceur "One-Photon Electrodynamics in Fibre-Fluorophore Systems"
A: I'm not an expert, but is this not simply the single-mode attenuation in the fibre? As far as I can see, the absorbed photon will either
a) be re-emitted into a different mode (contributes to attenuation.)
b) turn into phonons (contributes to attenuation.)
c) be re-emitted into the same mode by stimulated emission (probably very rare, and would you care about this anyway if the emission was coherent?)
d) be re-emitted into the same mode by spontaneous emission (contributes to attenuation as incoherent.)
Fibre attenuations are of the order 1 dB/km (I read here), suggesting an mfp of around 3km.
I look forward to an answer from someone who knows what they're talking about.
