# Laser in front of a wall - mathemathical physics

Note, however, that if the times are not taken at the origin (after the round trip), but at the wall (after only the forward leg of the trip), then ##t_2' - t_1'## is roughly half ##t_2 - t_1##, so the speed is twice greater. That is why you need to choose whether the speed is measured locally or remotely.

I am much aware of that fact but for the previous part the observer was standing next to the laser. I can see no reason, nor does the problem state anything about changing the position of the observer therefore I am positive that the speed is measured remotely (both times are taken at the origin).

Then basically you need to combine the round trip time you derived previously for a given angle with the spot position for the given angle.

I don't get it. :/

Look at the formula in #25.

yup, where ##t_2-t_1=t_L=\frac{2a}{c_0}\frac{1}{sin(\varphi (t))}## and ## \varphi = \pi /4##.

Your earlier definition of ##t_L## was the round-trip duration at a particular angle. In this case, however, you need the times (not durations) of the arrivals back at the origin from two different (but close) angles.

Right! Ah, how could I miss that -.-

So, total time should be ##t_1=t+t_L## if ##t_L## is the round trip duration. Similar for ##t_2##.

Since the difference ##t_2-t_1## is small, I guess it is the same as if I calculate ##dt## from ##t_1## or from ##t_2##.