About the rotating speed of a light beam which has been reflected off a rotating mirror A light beam generated from a source reaches a rotating mirror (x m away from the source), and is reflected off to a fixed mirror ( x m from the rotating mirror ), and again back to the rotating mirror. When the beam is leaving the rotating mirror for the second time, why does it have a rotating speed twice that of the rotating mirror?
 A: Let's assume that the rotating mirror was at 45 degrees to the emitted beam when the beam reaches it, and this reflection point is the center of rotation. The beam is then reflected by the stationary mirror back along its path to the rotating mirror, so that the return beam intersects the rotating mirror at the same point as it did the first time through. In other words, there are no translational effects to worry about.
Furthermore, let's assume that mirror has rotated 1 degree between reflections. Then the reflected beam will hit the mirror at a 44 degree angle, and make a 44 degree angle in the other direction (Angle of incidence equals angle of reflection, right?). The included angle will be 90 - 44 - 44, or 2 degrees.
In other words, the apparent change in beam angle (what you call beam rotating speed, I think) will be twice the change in the mirror angle (what you call mirror rotating speed).
It gets a bit more complicated if you have to take position shifts into account, but not much, and the result is the same. 
