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I was reading the chapter about optical fibers of "Fiber - optic communication systems" and I have read this statement:

The fraction of the power contained in the core is given by the confinement factor. Although nearly 75% of the mode power resides in the core for V = 2, this percentage drops down to 20% for V = 1.

where V is the normalized frequency of an optical fiber.

So, the confinement factor represents how much energy is contained in the core (which is the useful energy) with respect to the total energy, that is higher since part of power is contained inside the cladding:

enter image description here

I have a simple question about this fact. An optical fiber is based on the total internal reflection phenomenon, and so I'd say that all losses through the cladding are parasitic effects due to incidence angles lower than critical angle (for instance determined by fiber bending etc). So, since they are small parasitic effects, it seems too strange to me that the confinement factor may be 20% (if the fiber is rightly coupled to the source, I'd say that only few few rays may be transmitted inside the cladding)! Maybe there is something I'm not considering. Can you help me on understanding that?

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  1. Even in the case of total internal reflection there is evanescent leakage of the optical field into the less rare medium. The further above the critical angle you are the smaller this effect. The light penetrating into the cladding in this case is due to this “normal” evanescent leakage and not due to something like over bending the fiber. This leakage would occur for an ideal straight fiber.

  2. The energy which is in the cladding is not useless. It contributes to the total transmitted power. The main issue with the cladding is that it is optically lossier than the core. This means that energy which leaks into the cladding is absorbed in some percentage contributing to transmission losses. For this reason one prefers for the energy to stay confined to the core.

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  • $\begingroup$ Thank you for your answer. About 1, I have always been told that total reflection means no transmitted power (transmisttance equal to 0). I may understand that there may be a little leakage because of the evanescent wave, but 80% of power leakage seems too high $\endgroup$
    – Kinka-Byo
    Jul 9, 2020 at 15:29
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    $\begingroup$ @Kinka-Byo, an evenascent field contains some energy. We say there's no energy transmitted because this energy doesn't propagate indefinitely away from the boundary. $\endgroup$
    – The Photon
    Jul 9, 2020 at 18:05
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    $\begingroup$ Yes, for the case of total internal reflection it is true that there is no power transfer away from the interface in the rare medium. However, there is 1) electric and magnetic fields present in the rare medium which means there is optical energy in the rare medium and 2) The electric and magnetic fields in fact carry energy (power) along the original propagation direction of the beam along the surface of the interface. It is this energy/power which is absorbed by the lossy cladding which we try to avoid in optical fibers. $\endgroup$
    – Jagerber48
    Jul 9, 2020 at 18:32
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    $\begingroup$ See this visualization: olympus-lifescience.com/en/microscope-resource/primer/java/tirf/…. There are some issues with the simulation, I don't think it gets all of the details correct for the polarization but it is interesting to see the "cartwheeling" polarization that you don't see for far field propagating modes. $\endgroup$
    – Jagerber48
    Jul 9, 2020 at 18:33
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    $\begingroup$ Finally, regarding surprising amount of power which leaks into the cladding for the case of fibers with small $V$, single mode fibers. The point with these fibers is that size of the core is close to the far-field diffraction limit for light. That is, the core is only a few optical wavelengths across. Given that the evanescent penetration is also only a few optical wavelengths it is not surprising that the fraction of power contained in the evanescent field becomes large as the core itself become comparable in size to an optical wavelength. $\endgroup$
    – Jagerber48
    Jul 9, 2020 at 18:37

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