phyahmad said:
So one of the mirrors is moving away from him and hence the photon needs more time to reach it than the photon of the other mirror?
Indeed, this scenario is probably the simplest to demonstrate the relativity of simultaneity associated with light having the same measured speed in both frames.
Let's consider the frame of reference in which the mirrors are moving. Let's say the mirrors are travelling with speed ##v## (assume this is to the right). The two photons are moving to the left and right at speed ##c## (as measured in this frame). The source of light is moving in this frame and remains at the midpoint between the mirrors. Let's say that (in this frame), it is a distance ##D## from both mirrors.
This implies (by the definition of velocity/speed), that the photon moving to the left will reach the mirror after ##\Delta t_1 = \frac{D}{c+v}##. In this frame of reference the light and the mirror are moving towards each other, so their combined separation speed is ##c + v##. For example, if ##v = \frac c 2##, then the light will move two-thirds of the distance ##D##, and the mirror will move one-third of the distance ##D## in the given time. That must be the case if the light is measured (in this frame) to have speed ##c##.
Likewise, the photon moving to the right will take time ##\Delta t_2 = \frac{D}{c - v}## to reach the mirror. For example, if ##v = \frac c 2##, then the mirror will move a distance ##D## and the light move a distance ##2D## before the light reaches the mirror. Again, this is consistent with the light moving at ##c##.
The important point, however, is that in the rest frame of the source and mirrors, the measured speed of light is also ##c## - for both photons. And the mirrors are at rest. So, the light reaches both mirrors at the same time (as measured in this frame).
It's this transformation between reference frames that is different in SR from what we are used to in classical, Newtonian mechanics.