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[Not a duplicate! Also, the answer to a similar question was unsatisfactory]

I was digging into why we don't use mirrors in place of fibre optics cables. Majorly, the answers were as follows:

It's not easy &/or cost effective to have a cylindrical metal surface layered on glass for large distances.

But then I came across some piece of information:

Metals refractive index is always complex number (and not only for metals). Imagine part shows the extinction coefficient k - absorption in a material. Real and imagine part isn't connected. P.S. For engineering calculations real part sometimes is less than 1.

And I wondered... ignoring the imaginary part depicting absorption (since it has nothing to do with the real part), since Real(Index of Refraction) for metals is less than 1, light from vacuum (n=1) would show TIR for almost all incident light!

enter image description here

Critical angle for a silvered mirror-vacuum interface would be:

arctan(0.15/1) ≈ 0.15 rad ≈ 8.6°

That means it would reflect almost all light except that with angle smaller than 8.6° with normal, solely on the basis of TIR.

Is that how reflection in metalled mirrors work!? That obviously can't be entirely true since we do have the 8.6° critical angle, so what's the entire (true) picture of reflection in mirrors?

Also, just out of curiosity, does Real(n)<l in metals mean light travels faster than that in vaccum?

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For the last question: In a conductor, the wave equation at the surface has exponential sulutions of the form $e^{i k y - x/d}$, that means wave-like along the surface and exponential decaying amplitude into the material.

For the first question then, the free electron gas in the mirror metals conduction band always react to the tangent field in a tube.

This implies energy dissipation.

Light in optical fiber, with filled electron bands, distance to the conduction band far greater than the photon energies, there is no dissipation.

The total reflection is lossless, too, except for quantum scattering losses, because nothing is zero exactly in quantum theory. Mostly by roughness of surface and inhomogenity of the material.

A metallic tube can be used for some 1000 wavelengths only, e.g. for microwaves in the cm range; an optical fiber makes hundreds of km for visible light of 500 nm.

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