Reflection on moving mirrors Say I have an endless mirror, in a x y plane, at y=1.
Situation  1: the mirror is stationary and when we send light vertically from the origin, the light reflects back and returns to the origin.
Situation 2: let the mirror move horizontally at a constant velocity. Would the light reflected return to the origin?
 A: A nice way to answer this question is to do the calculation in the rest frame of the mirror. Light propagating in the $y$ direction in the lab frame will propagate in some other direction in the rest frame of the mirror. It will then reflect off in the ordinary way in that frame (angle of reflection equals angle of incidence). Finally, one transforms back to the lab frame to see what direction the reflected beam of light is travelling in.
Here is a calculation using 4-vectors and Lorentz transformation.
Initial 4-wave-vector of beam propagating in +ve $y$ direction, as expressed in lab frame:
$$
K = \left( \begin{array}{c} k \\ 0 \\ k \\ 0 \end{array} \right)
$$
Lorentz transformation for relative motion of the frames in the $x$ direction:
$$
\Lambda = \left( \begin{array}{cccc}
\gamma & -\beta \gamma & 0 & 0 \\
-\beta \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1  \end{array} \right)
$$
where $\beta$ is the speed of the mirror in the $x$ direction.
Hence the 4-wave-vector of the upward propagating beam, in the rest frame of the mirror, is
$$
K' = \Lambda K = \left( \begin{array}{c}
\gamma k \\ -\beta \gamma k \\ k \\ 0 \end{array} \right)
$$
(And for a check let's confirm that this is null: $K' \cdot K' =
-\gamma^2 k^2 + \beta^2 \gamma^2 k^2 + k^2 = k^2(1 - \gamma^2(1-\beta^2)) = 0$.)
Now the reflection process in the rest frame of the mirror simply reverses the vertical ($y$) component of $K'$ so the final wave vector (in mirror frame) is
$$
K'_f = \left( \begin{array}{c}
\gamma k \\ -\beta \gamma k \\ -k \\ 0 \end{array} \right)
$$
and therefore the final wave vector in the lab frame is
$$
K_f = \Lambda^{-1} K'_f =
\left( \begin{array}{cccc}
\gamma & \beta \gamma & 0 & 0 \\
\beta \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1  \end{array} \right)
\left( \begin{array}{c}
\gamma  \\ -\beta \gamma  \\ -1 \\ 0 \end{array} \right) k
= \left( \begin{array}{c} 1 \\ 0 \\ -1 \\ 0 \end{array} \right) k.
$$
Thus we find that the light comes straight back down again: the horizontal motion of the mirror does not matter.
Note that motion of the mirror in some other direction, not parallel to its surface, would change the reflection angle for non-normally incident light.
A comment
When I started on this answer I began to write reasons why, by symmetry, the motion of
the mirror could not matter. Those arguments are probably ok, but it occurred to me
that mirror reflection off a dielectric stack, for example, involves propagation inside
the stack and it is well known that the motion of the medium affects the motion of light inside the medium. So then the symmetry argument was not quite so self-evident as it may first appear. But I think the above calculation via two changes of frame is water-tight.
At least, I find I am more confident of this answer than I would be trying to argue that a moving medium doesn't affect the reflection at normal incidence in the lab.
