It is in the context of Total Internal Reflection (TIR) that I was staging my question. If you have a negative refraction indice what happens to TIR? In fibers this interaction is at the core to cladding interface and some of the power is lost each time a reflection happens. If the negative refraction indice is exactly right, the core to cladding interface becomes a perfect reflector and no loss can occur. This could result in a fiber that could carry light over extreme distances without loss.
That's right. Separating the phenomenon of reflection from the conditions of total internal reflection: when the light is simply reflected it has nothing to do with the index of refraction. Usually some light is reflected and some is transmitted, but the reflected portion goes right back into the same index of refraction it came from, so Snell's law would not apply to that portion. When the angle of incidence is such that the angle of refraction puts the rest of the refracted beam back into the material, then all the light stays in the material. Bell labs had an optical device that could be used in optical computing that sent the beam exactly parallel to the surface, the critical angle, so all of it entered the substrate with no loss, no reflection.
A perfect vacuum supposedly does this, but any kind of glass will absorb some light proportionally to path length. Water does this, too, and the absorption depends on frequency. There could be materials that do not absorb light at all, but I don't know of any.
That is a good question. Since we are dealing with a negative refractive index material, that would not be the core material since it will always have a lower n than the cladding. However, in total internal reflection, the light beam never propagates in the rarer medium (i.e. cladding in this case). Propagation is entirely within the more optically dense material. What does penetrate into the cladding from the fiber is an evanescent field whose intensity drops exponentially from the interface. Putting a negative index in for the cladding doesn't significantly change the penetration depth. On that basis, energy loss could still occur in the rarer, negative index material so it would not be a perfect reflector. However, things do get interesting if a polarized beam were to be used in an internal reflection geometry. I just tried the basic equations for internal reflection and, with polarized light, I was getting negative penetration depths, but positive ones for unpolarized light. This suggests that the evanescent field is inverted for S and P polarizations. these are just some off the cuff calculations. Looking at the equations, the polarized expression has a n21 factor (which is negative for a negative refractive index material) but this factor is not included in the equation for the unpolarized case. It is also interesting that the critical angle for a negative refractive index material is also negative. I'm wondering what meaning this has. Total internal reflection should occur if you are greater than the critical angle. In all of these experiments, the light is entering from air (n~1) into a lower refractive index material (n~-1). Under such a case, the critical angle is near -90 degrees but, experimentally, the beam enters the metamaterial. Does anyone know if reflection from these surfaces was measured? Me thinks the equations I'm using need to be re-derived for this type of situation.
One more thought. To make these metamaterials, special structures in the amterial need to be assembled and the size of those structures are of similar magnitude to the wavelength of light used in the experiment. If that is the case, the structures need to be within the penetration depth of the evanescent wave, otherwise, the bulk properties of the media in which the structures are embedded will dominate the internal reflection characteristics. This would also mean that the choice of incidence angle would also be very important. There is some serious thesis material here if someone wants to develop the theory.