Take the 2-minute tour ×

Why aren't other electromagnetic waves used in optical fibres instead of visible light? Is it because the wavelength of light fits the internal reflection/refractive index of the material used for the fibre? e.g. Is the material that refracts light cheaper compared to material that refracts EM waves with other wavelengths?

share|improve this question

3 Answers 3

up vote 10 down vote accepted

Almost all fibers (and certainly all telecomms fibers) do not use visible light but infrared light. A typical wavelength used is 1550 nm. At this wavelength the combined effects of absorption and (Rayleigh) scattering in the fiber attains a minimum value, thereby allowing long transmission distances.

share|improve this answer
There's a more elaborate explanation here thefoa.org/tech/wavelength.htm –  raindrop Feb 28 '13 at 20:46

The total internal reflection picture of an optical fiber is a geometric optics construct that only applies for high frequencies (i.e. visible light, wavelength much smaller than fiber cross section). For lower frequencies it becomes a boundary value waveguide problem. You attempt to solve for the transverse EM field in the fiber respecting appropriate boundary conditions, and you will find below some frequency there are no propagating modes. That's why for RF waves you use coaxial cable or some other waveguide which has propagating modes at low frequency.

share|improve this answer
I love the reference to coaxial cable! –  raindrop Feb 28 '13 at 0:53
It isn't always visible light, either. Frequently, infrared wavelengths are used, especially because they have desirable properties in nonlinear media that allow very fast switching. –  KDN Feb 28 '13 at 1:24
Single mode fiber indeed act as a waveguide. however, that doesn't mean the photons allowed thru are visible photons. In fact, standard single mode fiber is designed to transmit infrared photons at 1550 nm wavelength. –  Johannes Feb 28 '13 at 13:20

There are more reasons than propagation, absorption and scattering. (Near)-optical frequencies also have the advantage of effective sources and receivers. Terahertz receivers are very hard to build. Small UV-frequency sources are non-trivial either.

Next there's a bandwidth issue. If the frequency is too low, a high bandwidth can only be achieved by a very high Signal to Noise Ratio (SNR). Yet at "low" frequencies (<1Thz), simple thermal noise becomes a limitation. To overcome this would require a high signal level, with associated power consumption.

share|improve this answer

protected by Qmechanic Mar 15 '13 at 11:09


Would you like to answer one of these unanswered questions instead?
