Why not regular reflection?

Question

Fiber optics make use of total internal reflection for transmitting signals through them. Associated with fiber optics are many kinds of losses such as loss of signals due to light rays entering at angles more than the acceptance angle.

If in theory, we could make fibers that are "mirrored" on the inside which would reflect all rays of light regardless of their angle of incidence, would that improve the capacity of these fibers?

One possible reason why I think mirrored fibers should not be used is that rays entering at very high angles (approaching 90 degrees) will undergo a large number of reflections due to which they will arrive at the other end very late. This could cause interference with other data that has been sent.

Answer 1

The thing is that there are no really good mirror coatings. From this plot from Edmund Optics, a reseller for technical optics, you can see that for 1-1.5 µm (telecommunication wavelength) the reflectance doesn't go higher than 98% (that's normal incidence, it will be a little higher for grazing incidence).

In contrast, the inner surface of a glass fiber with a clean surface has 100% reflectance. It's also important to note that the transmission within fused quartz fibers of 1-1.5 µm is particularly high, so there are very little transmission losses either, as can be seen in this plot (theoretical model) from RP Photonics:

Answer 2

I think the main reason is practicality. In the normal process of manufacturing of optical fiber, the material is simply pulled out of the preform, and the core and the cladding result directly from this (having been formed in the preform before pulling). If a mirror surface were used instead of low-index cladding, you'd have to find a material that would:

1. have near 100% reflectivity,
2. be pullable at the temperatures of pulling of the glass from which the fiber is made,
3. retain its reflectivity and smoothness after pulling,
4. retain enough thickness to avoid becoming translucent.

The answer by zonksoft also explains that even for normal mirrors the metallic coatings are still far from being 100% reflective, unlike the total internal reflection at the core-cladding interface of the fibers. Add this to the above points, and you'll see that regular reflection is not really practical for optical signal transmission over long distances.

Answer 3

TLDR:

TIR is Total reflection

Whereas optical signals guided by our best metallic mirrors (98% reflective) would lose half their signal strength after every 34 reflections.

TIR:

Total Internal Reflection gives you Total reflection - but wait there's more: these reflections also have zero phase shift and zero polarization concerns! That is a much better mirror than any mirror we can construct to reflect light at higher angles of incidence. Even Dielectric mirrors at 99.999% reflectivity are not even close to Total reflection.

Some reflection happens at every material boundary. For example, at the boundary between air and glass, about 4% of light is reflected. But except for the case of TIR, some of the wave energy is reflected and the rest is refracted, additionally more complicated phenomena arise with polarization, phase shifts, etc.

The amount of reflectance of a given material boundary is calculated with Fresnel's Equations. At incident angles near Brewster's Angle particular polarizations are completely diffracted while polarizations orthogonal to those are completely reflected. On top of that, reflections (other than TIR) generally have a 180° phase shift.

The reason metallic surfaces can reflect a high percentage of high angle of incidence EM waves is because the damping constant of metals is so high, so light can't penetrate very far. The deeper explanation involves complex numbers. The Hagen-Rubens relation shows how better conducting materials are also better reflectors, but no metals are perfect conductors. Suffice it to say that our best metallic mirrors are only 98% reflective.

Dialectic mirrors can be more reflective than metallic surfaces but must trade-off between reflectivity and the range of wavelengths for which they will reflect at all, and they are very challenging to manufacture.

And even if we did (very expensively) build dielectric mirrors around optical fibers, as you suggested, the longer path lengths taken by the higher angle reflections would just smear out the signal.

That TIR is so "perfect" of a reflection phenomenon, and selective of photons on the faster track, is one of the things that makes fiber optics work so well.