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Why do optical fibers usually have a cladding? Ok, if you put make a bundle of optical fibers this prevents that light leaks from one fiber to another fiber in contact. However, are there other reasons to use claddings? Are there applications of optical fibers without claddings?

(I am mainly interested in optical fibres which can be described by geometrical optics, i.e. multimode fibers.)

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For a fiber to guide light, even considering the situation using only geometrical optics, there must be total internal reflection (TIR) at the boundary of the fiber core. For TIR to occur, the angle of incidence of the light must be greater than the critical angle $\theta_c$. If the material before the boundary has index of refraction $n_1$ and the material after the boundary has index $n_2$, then the critical angle is:

$$ \theta_c = \arcsin(\frac{n_2}{n_1})$$

Of course, if $\frac{n_2}{n_1} > 1$ then the critical angle is undefined -- TIR cannot occur. This is why TIR is only possible when light encounters an interface with a material of lower index. This should be familiar to you if you think about it: you can see the bottom of a pool from above the water, but the water surface looks silvery and reflective from underneath; the reflective face of a roof prism looks reflective when you try to look through it from the glass-to-air side, but if you look from the air side, you can see clearly.

On its own, this doesn't require a fiber to have cladding. Air has an index very near 1, so any other material will have a higher index, and therefore guide light without a cladding. However, this situation would be very unstable. Any material contacting the fiber could produce an area where TIR does not occur, and suddenly the fiber would begin leaking. Consider how easy it would be for this to happen! Any sort of dirt or oil on the fiber would either eliminate TIR completely, or raise the critical angle such that some of the light could escape. The same would be true of most materials you might use to mount the fiber, or even any type of coating you might apply as a mechanical barrier to dirt.

Thus, in practice, fibers have a cladding to ensure that the optical guiding characteristics continue to work in real world conditions. They also provide engineers with another way of adjusting the properties of the fiber, so its definitely not something we are unhappy about.

Do keep in mind though, that there are examples of unclad "fibers" although they are usually just acrylic rods used in teaching demonstrations, so I wouldn't really call them a fiber in the common sense of the word.

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  • $\begingroup$ Actually this answer is not entirely right. Yes, cladding allows for total internal refraction, but in most cases it's there to lower the critical angle. This prevents modal dispersion (i.e light entering the fibre steeply goes away whereas light entering shallowly is internally reflected). Hence, this means that the difference between the 'slowest' light and 'fastest' light for a single bit is as small as possible, increasing the rate of transmission $\endgroup$
    – yolo
    Commented Jun 15, 2021 at 19:46
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Do you think a fiber that worked until is got some water on it would be acceptable?

In order for the fiber to have a reliable, known behavior (transmission speed, attenuation, wavelength acceptance) in all environments, it is necessary to the total internal reflection to happen against a well defined external material.

That's the cladding.

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an optical fibre without cladding exists but is hard to make profit of it:

The contrail of jets is an optical fiber under special circumstances: Ice crystals in the form of hexagonal plates are created in wet and very cold temperatures at high altitudes, typically above 10 km, and form a tube, nearly cylindrical, that can extend for hundreds of kilometres. If the atmosphere is quiet the ice crystals will descend slowly, parallel to the ground oscilating up to a maximum of 8 degrees from the vertical, and if It is lit by the Sun or the Moon, not visible at sea level, perpendicular to the cylindrical tube at the opposite site in relation to the observer he will see the light.

The light will bounce from crystal to crystal under total reflection conditions with very low absortion, and subjected to polarization.

The observational evidence:

The perceived usual shapes are like 'discs, cigars, triangles' and correspond to some sort of intersection of a plane with a cylinder.

A slight change of angle between the sun/moon, the contrail, and observers will make a significant change observed in the lights, which are a decomposition of white light. When observing a change in velocity inconsistent with the motion of any known object remember that the light moves at c speed ;-) and the Earth's rotation around its axis contributes to the changing angles. The distribution of the observations of UFOs in space and time historically accompanies the evolution of aviation.

In colder areas like Alaska and northern Norway, with more candidates to observers and with polar commercial air routes means that there are more reports of observations. Besides the light reflected by the ice cap may also inject more light into the contrail. In areas with little commercial air routes as the South Pacific there are hardly any observations of UFOs.

They are what we call UFOs by Night.
This is not the physics involved in the day light observations.

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  • $\begingroup$ These are interesting claims, but do not appear to be very relevant to the question that was asked. Also, claims this interesting should perhaps be backed up with a few citations? $\endgroup$
    – Andrew
    Commented Mar 16, 2011 at 18:42
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    $\begingroup$ It is an original discovery of mine. I read hundreds of reports from pilots and consulted databases with the distribution of the observations. I put the pertinent questions: Where is the source of the light? What exist in the atmosphere here that nobody sees by night? Slept upon the subject and the unconscious brought me the answers: Sun and contrails. Then went looking for software such as the Hplanets.exe and other by USA's Navy to put myself in time and places of the past. Of course I'm not expecting much of this issue. $\endgroup$ Commented Mar 16, 2011 at 19:32

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