reflectionspecial-relativity
If you move at $5$ $ms^-$$^1$ towards a plane mirror, your reflection moves $10$ $ms^-$$^1$ towards you.
But what happens if you're moving much faster, say $0.8c$?
Would your reflection move at $1.6c$, since it's not a physical object? Or is it still confined to the speed of light and you have to apply the Lorentz factor? Or, does some strange light-reflecting thing occur since you're moving so fast at a mirror?
This question cannot really be answered because you cannot travel at the speed of light. See Accelerating particles to speeds infinitesimally close to the speed of light?
If you were massless, you would always travel at the speed of light. However, in that case you would not perceive the passing of time. In relativity, the time that passes for an observer depends on the proper time. The proper time for a light-like trajectory is always zero, so photons themselves do not experience the passage of time.
If you travel very near to the speed of light - perhaps 99.9% light speed relative to Earth, you would still be able to view yourself normally in a mirror you carried with you. That is ensured by the principle of relativity, which states that all physical processes work the same way at any constant speed.
You are getting reflections from the front (glass surface) and back (mirrored) surface, including (multiple) internal reflections:
It should be obvious from this diagram that the spots will be further apart as you move to a more glancing angle of incidence. Depending on the polarization of the laser pointer, there is an angle (the Brewster angle) where you can make the front (glass) surface reflection disappear completely. This takes some experimenting.
The exact details of the intensity as a function of angle of incidence are described by the Fresnel Equations. From that Wikipedia article, here is a diagram showing how the intensity of the (front) reflection changes with angle of incidence and polarization:
This effect is independent of wavelength (except inasmuch as the refractive index is a weak function of wavelength... So different colors of light will have a slightly different Brewster angle); the only way in which laser light is different from "ordinary" light in this case is the fact that laser light is typically linearly polarized, so that the reflection coefficient for a particular angle can be changed simply by rotating the laser pointer.
As Rainer P pointed out in a comment, if there is a coefficient of reflection $c$ at the front face, then $(1-c)$ of the intensity makes it to the back; and if the coefficient of reflection at the inside of the glass/air interface is $r$, then the successive reflected beams will have intensities that decrease geometrically:
$$c, (1-c)(1-r), (1-c)(1-r)r, (1-c)(1-r)r^2, (1-c)(1-r)r^3, ...$$
Of course the reciprocity theorem tells us that when we reverse the direction of a beam, we get the same reflectivity, so $r=c$ . This means the above can be simplified; but I left it in this form to show better what interactions the rays undergo. The above also assumes perfect reflection at the silvered (back) face: it should be easy to see how you could add that term...
Best Answer
The mirror is equivalent physically to a legitimate person mimmicking you behind an open gap... so apply the same logic as two trains coming towards each other at relativistic speeds.