The principle behind a fibre optic cable is that light is reflected along the cable until it reaches the other side, like in this diagram:

Fibre optic cable diagram

Although I know that the light is slowed down somewhat because it's not going through air, I've always wondered about another factor: what about the fact that the light path is zig-zagged rather than straight? Doesn't that significantly increase the distance that the light has to travel? If so, by how much does it slow down the time the light takes to travel through the cable?

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    $\begingroup$ actually, it's more of a waveguide optic effect rather than classical multiple reflection. Good luck doing the maths ;-) $\endgroup$ Commented Oct 8, 2013 at 15:04
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    $\begingroup$ I assume this is for a multi-mode cable? In a single-mode cable the light will pretty much follow the cable, without bouncing around much. Although there are probably some other complexities that arise because of this. $\endgroup$
    – JSQuareD
    Commented Oct 8, 2013 at 15:38
  • $\begingroup$ @JSQuareD Clearly a MMF $\endgroup$
    – Brethlosze
    Commented Sep 24, 2017 at 19:30
  • $\begingroup$ Some real-world data: The IEX stock exchange routes their traffic through 61km of wound up fiber as a speed bump for traders, which introduces 350 µs delay. See exchange.iex.io/about/speed-bump and youtu.be/d8BcCLLX4N4?t=159. The minimal delay from that should be 61km/c = 204 µs.They do not specify where the additional 146 µs delay comes from (e.g. reduced speed of light inside the fiber, longer travel distance, maybe even delay from the electric components that send and receive the signal). $\endgroup$
    – Socowi
    Commented Oct 27, 2022 at 20:04

3 Answers 3


We can make a 1st-order approximation by assuming the following:

  • $L=3$ m is the length of the fibre optic cable
  • $d=3\cdot10^{-6}$ m is the diameter of the cable
  • the cable is perfectly straight
  • $\theta=0.785$ rad (~45$^\circ$) is the angle of reflection inside the cable
  • photons are classical balls
  • reflection is perfectly elastic
  • photons still travel at $c$

Simple geometry shows that the particle travels $h=\frac{d}{\sin\theta}=4.24\times10^{-6}$ m over a linear distance of $x=3\cdot10^{-6}$ m. Do this a million times, you find that the photon traveled 4.24 meters instead of 3 meters!

Given speed of light in vacuum, it would take 14.1 nanoseconds for the photon to travel the reflected path, whereas it would take 10.0 nanoseconds to travel 3 meters linearly. Both the distance & duration are about 40% increases!

Since $L=3$ m and $t=14.1\cdot10^{-9}$ s, then the "linear" photon speed in the fibre optic cable is $v_{\gamma,fo}=2.13\cdot10^8$ m/s, a reduction of about 30%.


As per the request in the comment, using $2c/3$, the reflecting photon would take 21.2 nanoseconds to travel the cable while the linear photon travels the distance in 15 nanoseconds. This would then lead to $v_{\gamma,fo}=1.42\cdot10^8$ m/s (instead of $\sim2\cdot10^8$ m/s) which is still a reduction of about 30%.

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    $\begingroup$ The refractive index of glass is about 1.5. So instead of $c$, why not use $\frac{2}{3}c$? This is definitely a very significant difference. $\endgroup$
    – JSQuareD
    Commented Oct 8, 2013 at 15:34
  • $\begingroup$ @JSQuareD: not as significant as you might think. The reflected photon still takes about a 30% reduction in its speed compared to its linear counterpart. But I did add in the requested information as an edit. $\endgroup$
    – Kyle Kanos
    Commented Oct 8, 2013 at 15:41
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    $\begingroup$ So that's for light bouncing at 45 degree angles, and it causes a large slowdown. Does light get beamed down fibre optic cables at much smaller angles, then, to minimize this bouncing? I found this article talking about operating at 99.7% the speed of light; wouldn't that require almost a completely straight path for the light beam? $\endgroup$
    – Jez
    Commented Oct 8, 2013 at 15:42
  • $\begingroup$ Of course the relative difference is going to be the same. But there is a very significant absolute difference. The OP seems to ask two questions: (1)'How fast does light travel trough a fiber optic cable?', (2)'How much does the reflection matter'. You already answered (2), now you also answered (1) :) Plus, it's always nice if your answer is right in numbers as well as concept :) $\endgroup$
    – JSQuareD
    Commented Oct 8, 2013 at 15:44
  • $\begingroup$ @Jez: Total internal reflection for glass-air is $\sim42^\circ$, so anything larger than this would work, which would then lead to faster travel times. $\endgroup$
    – Kyle Kanos
    Commented Oct 8, 2013 at 15:49

The full picture is attained by solving for the modes of the whole waveguide (in theory for the refractive index profile right out to and beyond the jacket) by the methods described, for example, in Chapter 12 through 15 of:

A. W. Snyder and J. D. Love, "Optical Waveguide Theory", Chapman and Hall, 1983.

and the spectrum of effective indices ($c$ divided by the axial propagation speed of mode in question) for the bound modes lies between the maximum and minimum index of the refractive index profiles. So this translates to agree with George Smith's answer, for example, which gives a good intuitive description of the range in terms of a ray picture.

For many fibres, though, the range of bound mode effective indices is much narrower than the other answers would imply. This is because the modes in question are confined very tightly near the core, so only this region is relevant in setting the effective index and, especially for single mode fibres, the difference between core and cladding indices is miniscule. For a step index profile fibre, the fibre $V$ parameter:

$$V = \frac{2\,\pi\,\rho}{\lambda}\sqrt{n_{core}^2-n_{clad}^2}$$

where $\rho$ is the core radius, $\lambda$ the freespace light wavelength and $n_{core},\,n_{clad}$ the core and cladding refractive indices, must be less than the first zero of the first-kind, zeroth order Bessel function $J_0$, or about 2.405 if the fibre is to be single moded. For core radiuses that are readily manufactured (to wit, more than 1 micron), this means that $n_{core}$ and $n_{clad}$ are typically less than one percent difference. Let's plug in $\lambda = 1550\mathrm{nm}$ and $\rho=1\mu\mathrm{m}$ with $n_{clad} = 1.48$ (pure silica), then we find that the maximum $n_{core}$ we can have for $V\leq2.405$ is 1.59. This is an extreme example. More typically, for this wavelength, we would have $\rho = 5\mu\mathrm{m}$, when $n_{core}\leq1.4847$, a difference between $n_{core}$ and $n_{clad}$ of $0.3\%$.

So, for a single mode optical fibre, you can almost always say that the propagation speed is $c$ divided by the cladding index, and your error in assuming so will typically be less than half a percent.


Well the OP doesn't say anything much about the nature of the "fiber optic" cable. There's a whole range of things describable as fiber optic cables, from cut up fishing lines, glued as a bundle onto an LED to make an ornament, up to transoceanic single mode high bandwidth cables. Let's assume (for different reasons) that we are not talking about either of those; the latter because it's quite complex, and the former is just a toy. So let's consider just cables of up to a few meters, intended for practical signal transmission, from audio signals in hi-fi equipment of common computer data signal (digital). The shorter multimode types could be made from plastic fibers, but these are fairly lossy. More serious multi-mode cables would be glass fibers. For a glass refractive index of 1.5, the critical angle for Total Internal Reflection (TIR) at the glass-air interface would be 41.8 degrees (from the surface normal) or 48.2 degrees from the fiber axis. Light at this angle would have the longest possible path of 1/cos48.2.

Well this comes out to 1.5 times the fiber length; exactly the refractive index (no surprise). On top of that you would have the light speed slowdown, also a factor of 1.5, so the longest travel time would be the square of the index times the cable length divided by c of course. So most of the light would take between a factor of n and a factor of n^2 times the air length time.

Well there's a complication I haven't mentioned. Serious fiber optic cables ; even multimode fibers made of glass, are almost certain to be "clad" fibers; consisting of a core glass with a refractive index of perhaps 1.6, coated with an outer layer of lower index, perhaps 1.5. The OP may want to research what actual values are found in available FO cables. This cladding separates the environment and contamination from the reflecting interface that confines the radiation. So the light is travelling in the core glass, of index 1.6, but the effective critical angle is now arcsin (1.5/1.6) = 69.6 degrees, so the maximum path length is now 1.6/1.5 (1.0667) times the fiber length, not 1.5 times.

So the angles shown in OP's diagram have greatly exaggerated off axis paths, and the path extension is only a few percent. But the higher core index, also means a slower light speed in the fiber.

You can actually determine much of this without knowing the fiber index, if the manufacturer specifies the "numerical aperture" of the fiber, which is a measure of the maximum acceptance angle of the cone of light which can propagate along the cable.

I don't want to belabor this too much, because there are so many variables. Unclad fibers are too sensitive to contamination to be of much use. Even touching them with your fingers and getting sweat on the surface, will cause serious loss of light out the side of the fiber.


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