Laser in front of a wall - mathemathical physics

  • Thread starter skrat
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  • #26
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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).
 
  • #27
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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.
 
  • #28
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I don't get it. :/
 
  • #29
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Look at the formula in #25.
 
  • #30
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yup, where ##t_2-t_1=t_L=\frac{2a}{c_0}\frac{1}{sin(\varphi (t))}## and ## \varphi = \pi /4##.
 
  • #31
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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.
 
  • #32
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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##.
 

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