# Do mirrors (with metal surfaces) show TIR (Total Internal Reflection)?

[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!

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?

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.