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I am trying to understand the concept of waveguiding, scattering, and total internal reflection. I have a question regarding the optical waveguides.

Let's consider an optical waveguide transmitting light via total internal reflection. If one takes its photo, and if the color of the guided light can be easily seen on the waveguide, does this observed color necessarily imply that some light is escaping the waveguide due to scattering?

In a hypothetical ideal waveguide with perfect confinement, would the guided light remain invisible within the structure?

For example, in the photo below, if we waveguide was perfect, we would not be able to see the light within the fiber, correct? (considering we are only dealing at the visible spectrum)

enter image description here

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Yes, you are right in that a sane waveguide will not allow much of the light out.

What you have pictured is probably an optical fiber intentionally made to scatter substantial amount of the light for illumination and decorative purposes.

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    $\begingroup$ All optical fibers have some scattering and absorption looses. $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 11 at 7:35
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    $\begingroup$ @AgniusVasiliauskas but usually it's some noticeable loss per hundreds of kilometers rather than a meter. $\endgroup$
    – Ruslan
    Commented Jan 11 at 7:45
  • $\begingroup$ The problem with this post is that, it's too strict in expression,- "sane waveguide will not allow the light out". This is not true, because all waveguides allow SOME light out, which like you say may be tiny in standard fibers, but still it's greater than zero, while post states that it lets out NONE, which is false. Second thing which is questionable is part "intentionally". Higher attenuation coefficient does not mean that it was made intentionally by itself, it may be unintentionally as well (not knowing what materials to use, mis-information, insufficient on resources, etc). $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 11 at 8:41
  • $\begingroup$ Fixed, as per popular demand, thanks for your comments $\endgroup$
    – fraxinus
    Commented Jan 11 at 15:21
  • $\begingroup$ Theoretically speaking, does this mean that a perfect waveguide would not let any light out? I almost got a sense that you @fraxinus and Agnius Vasiliauskas made the imperfection of the waveguide letting some light out sound like an artifact of the waveguide rather than defect. I am also confused about you mentioning the decorative and illuminative purposes because I briefly partook projects in optics labs and I saw multimode cables where the light proparating was clearly visible. The terminology was a bit unusual (sanity vs insanity :) for me that's why I am asking for elaboration. $\endgroup$
    – user390982
    Commented Jan 11 at 22:03
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Yes, in optical fiber there's scattering and absorption looses, defined by a attenuation coefficient :

$$\tag{1}\alpha=-\frac{1}{l(km)}10\log\frac{P_{out}}{P_{in}}$$

which is measured in $\text{dB/km}$. In an ideal fiber attenuation coefficient would be zero, and as such we would not see light propagating in a fiber, just as we do not see laser beam propagating in space if there's no scattering particles (that's why in a concert with a laser show,- they put a lot of white smoke around).

However, even in an ideal fiber you may see a fiber end shining (not in propagation direction), due to diffraction and other effects.

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  • $\begingroup$ Life irony,- I have Laser physics master's degree, but somehow in this question I can't "sell it" :-D $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 11 at 17:38

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