Phase shift in back-reflection generated at the mating point of two fibers

If you have light traveling from glass to air at normal incidence, e.g., in an optical fiber with the endface open, the Fresnel reflection coefficient would roughly be $R = |(1.5-1)/(1.5+1)|^2 = 0.04$ which translates to $-14\,$dB loss. Also, the reflected light that travels back in the opposite direction would not have any phase shift w.r.t. to the input light beam.

But when you have another optical fiber mated together with the first fiber, the reflection is typically much lower, e.g., $-35$ to $-40\,$dB with FC/PC type connectors. As far as I understand, the physical construction of two fibers mating inside a connector typically involves at least a part of the fibers (or ferrules) being in physical contact. And it is due to the manufacturing precision and polishing etc. that the reflection losses are reduced tremendously. Is that correct?

And here are my main questions:

1) Is the interface that describes the Fresnel reflection correctly now glass-air-glass or glass-glass or something in between?

2) If it is glass-air-glass, does a component of the reflection also come due to the air-glass part? If so, the phase-shift from that part should be $\pi$. In that case, what would be the net phase shift in the back-reflected light?

Treat this as glass-air-glass where the thickness of the air is much smaller than the wavelength.

Here is an article on how to calculate the transmitted and reflected wave. http://www.jedsoft.org/physics/notes/multilayer.pdf The article assumes incident light is in vacuum. But the principal is the same if it is in glass. Likewise if a layer is thinner than a wavelength.

If the thickness was 0, you would ideally get two reflected waves that perfectly canceled. (Or perhaps multiple reflections that perfectly cancel.) In that can, the phase would be irrelevant.

But cancellation is not perfect. One reason might be that the air thickness is not 0. In that case, you would calculate the phase shift as for any multilayer. You add all the reflected waves together. The phase shift at each reflection is either 0 or $\pi$. The waves are not in phase because they travel the air gap distance and back.

Keep in mind the reason for imperfect cancellation might be that the surfaces are not perfectly flat. A better polish gives better results. This limits the accuracy of calculation for the ideal plane case (or as the Wikipedia article below says, the case where the surface is convex.)

Another reason might be that the cores of the fibers are not perfectly aligned. You might get more reasons from the engineering department of a connector company. Here is a list: http://en.wikipedia.org/wiki/Optical_fiber_connector

• This is a good answer. My intuition would be that the surface roughness has more of an effect than the air gap. In principle, fibers that refract from glass to air, would refract the opposite direction back from air to glass anyway, so there's not a ton of loss, right? FYI, they sell index-matching gel that can fill in these pockets of air and effectively let this connection reach about 100 % – Adam Hughes Jan 26 '15 at 7:19

The "phase of the back-reflected light" is only a meaningful quantity to the extent that the back-reflection happens at a known location---known much more accurately than a wavelength.

I don't think that in the context of FC/PC fiber connections, you can pinpoint (with sub-micron accuracy) a single location where the (slight) reflection occurs.

If this surmise is correct, then you are asking a meaningless question which has no answer. But I could be wrong. :-D