Total internal reflection for perpendicular incidence

Are there any materials, which give total internal reflection for perpendicular incidence?

Here, by this phenomena I mean:

• no absorption (including for evanescent propagation inside),
• finite evanescence depth (even if very small).

I am asking, as some of my calculations gave this result. Yet, I am not sure whether this result is physical (i.e. can happen) or practical (i.e. there is a system exhibiting such phenomenon).

I does works that way for dielectric mirrors. I am curious if there are any "uniform" materials exhibiting such effect. By "uniform" I mean spatial period (or any kind of other structure) much, much smaller than (vacuum) wavelength.

(Obviously, it goes beyond refraction index, so please don't write it is not possible with a fixed $n$, because I know.)

• Total Internal reflection will only work, if the ray of light is incident greater than Critical Angle. Dec 4 '14 at 10:40
• @Sushant23 Did you read the last line? I am not a high-school student, I am a PhD. Dec 4 '14 at 10:40
• But I am still an undergraduate student. Also I was just trying to answer according to my knowledge. Besides even in the case of dielectric mirrors, the mechanism works the same way for the TIR to occur. Dec 4 '14 at 10:45
• perhaps you could give more details in the question as to why TIR at normal incidence does not imply $n=\infty$
– tom
Dec 4 '14 at 11:19
• I'd suggest you to try computing the photonic band structure for your lattice. My guess would be that you're using a frequency which appears in the band gap. In certain conditions (high enough refraction index contrast (which corresponds to strong coupling), even though lattice constant is small) this would indeed be possible. Dec 4 '14 at 12:36

As you know the standard textbook answer to this question would be, for total internal reflection (TIR) the angle of incidence, $\theta$, must be greater than $\theta_c$, which is of course given by $sin^{-1} ({1 \over n}$) so for TIR at $\theta=0$ we need to have $n= \infty$.

In the case of a dielectric mirror, this is a multilayered device a combination of reflections from the top surface and interfaces below the surfaces lead to interference so that you could get total reflection $\theta=0$, but this is not a case of total internal reflection. Instead it is due to interference and reflection from multiple interfaces.

So if I understand correctly, your question is whether a continuous mateial without structures on scales similar to the wavelength of light might exhibit total reflection from the surface.

The only answer that I can think of is a highly polished metal surface, which acts like a mirror. I am sorry that I don't think this is the answer you want.

• Metals - I was thinking about them, but AFAIK they have high absorption, which also gives raise to high reflection (though, through a different mechanism). Or am I wrong? Dec 4 '14 at 17:59
• @PiotrMigdal - looking at the graph Gold (Au) has 100 % reflectance (and thus no absorption) from 1 to 5 $\mu m$ - the colour of gold will be because it reflects above 500 nm very highly but below 500nm less than 40% reflection - NB 500 nm is ~ green - so yellow red reflect - blue absorbed.... the only problem with metals for you is I don't know how the evanescent wave will penetrate - maybe not very far.
– tom
Dec 4 '14 at 18:41
• The point is in absorption per propagation. Dec 4 '14 at 18:53
• ok - so from your point of view that graph is not correct - it has just been normalized to 100%... look at layertec.de/en/capabilities/coatings/metallic - this company is claiming near 100 % reflectance and therefore very little absorption.
– tom
Dec 4 '14 at 19:24
• Ahhh - but you mean absorption per propagation of evanescent wave into material do you... because if you want TIR you don't want propagation into the material do you except via the evanescent wave... ???
– tom
Dec 4 '14 at 19:32