Once again, thank you all. I got it.
For the first part of the light's journey, I calculate that it takes 33 ns to reach the leading mirror. This reconciles with the closing velocity being 0.4c over the 4m. For the second part of the journey, the light takes 8.3 ns to reach the trailing mirror...
Thank you both for your replies.timetraveller123: I agree that the time required in the galaxy frame is not simply 2L/c, and in fact this is recognised in the solution to the problem where it calculates the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then uses...
1. I'm fine with this one, I simply calculated L = L0/γ = 5/1.2 5= 4m.
2. This is the one I'm having problems with.
My approach was to say that both observers would measure the light beam to have velocity of ‘c’. Therefore, if 4m is the distance between the mirrors as observed in the frame of...