opticsrefraction
Is it possible to have an index of refraction, $n < 1$, giving the "illusion" of a speed greater than $c$.
As far as I understand it, is only the phase-velocity (which does not carry any information) that exceeds the speed of light.
But what I would really like to know is, under what conditions it is possible to have $n<1$?
One way to look at that:
The envelope of the red light pulse travels at the group velocity, less than $c$.
The fast oscillations within the main pulse travel faster than $c$.
In the end, the torch "lighting up" is given by the group velocity, because the envelope of the pulse is what you receive.
This animation showing a train of pulses is pretty clear. You can see that each pulse goes slowly, but the fast oscillations go fast.
Well at the most fundamental level, the index of refraction of a material is defined as
$ n = \sqrt {\epsilon \mu} $
where $\epsilon $ is the electric permittivity and $\mu$ is the magnetic permeability of the material.
This arises from the solution of Maxwell's equations in a medium.
Also arising from the wave equation, which can be derived from Maxwell's equations, is that the index of refraction is the speed of light in vacuum, $c$ divided by the speed of light in the material $c_m$.
$ n ={ c \over c_m} $
Worth noting is that since $c_m$ is always less than $c$ , the index of refraction is always greater than 1.
Now the phase velocity for an electromagnetic wave of angular frequency $\omega$ is given by
$v_p = {\omega \over k}$ where $k= {2\pi \over\lambda_m}$ is the magnitude of the wave vector and $\lambda_m$ is the wavelength in the medium.
So after a little algebra, we find that the wave vector and index of refraction are related by
$k ={ n\omega \over c}$
Where does all this come into play in refraction and Snell's Law? Well, it is the wave vector that comes into play in satisfying the boundary conditions on the electric (and magnetic) fields at the interface between two media.
To see this, let's look at the simple case of a plane wave of monochromatic light incident on the interface between two media with indices of refraction $n_1$ and $n_2$:
In this simple case, considering the boundary conditions on the electric field at the interface is sufficient to derive Snell's Law.
The boundary condition is given by equation (1) in the diagram, namely that the incident and reflected electric fields minus the transmitted electric field must be zero, or equivalently that the total electric field at the interface must be continuous.
Since we defined y=0 as the plane of the interface, this boundary condition must hold for any value of x. This leads after a little algebra to equation (2) in the diagram. This equation depends only on the angles of incidence $\phi_{inc}$ and refraction $\phi_{tr}$, the wave vector magnitudes in both media $k_1$ and $k_2$, and the transmission and reflection coefficients $T$ and $R$ ( the fraction of energy that is transmitted into the new medium and reflected into the old medium respectively). By symmetry, $\phi_{inc} = \phi_{refl}$.
Because the left side of Equation 2 is independent of angle, so must the right side be. This leads to the term in the exponential being zero, which leads directly to Snell's Law, using the relation between wave vector and index of refraction shown above.
The group velocity never comes into play in the boundary conditions of refraction. It does come into play in the propagation of energy in the media (as opposed to the fields), but that's another question.
Best Answer
I have been reaserching some more, and what I have found is something about waves in dispersive media. (The wave speed depends on the frequency of the wave). Source: (http://www.acs.psu.edu/drussell/Demos/Dispersion/dispersion.html) Apparently when the wave is traveling with a frequency close to the resonans frequency $n$ gets smaller than $1$. But only the group velocity exceeds $c$.