Time Dilation in Relativity: Solving a Simple Observer Problem

[unscientific]

Homework Statement

An observer fires a laser pulse at a stationary mirror located at distance d away and measures the time taken. A second observer moves in the same direction as the beam with velocity v and sees these events.

According to each observer, how much time will it take for the pulse to make each leg of the trip and back and what is the total time taken from emission to detection?

The Attempt at a Solution

Event 1: Laser pulse is fired

t₁ = 0, x₁ = 0

t₁' = γ(t1 - vx₁/c²) = 0

Event 2: Laser pulse bounces off mirror

t₂ = d/c

t₂'
= γ(t₂ - vx₂/c²)
= γ(d/c - vd/c²)
= (γd/c)(1 - v/c)

Event 3: Laser reaches back observer

t₃ = 2d/c, x₃ = 0

t₃' = γ(t₃ - vx₃/c²)
= γ(2d/c)
= (2d)/√(c² - v²)


Is there anything wrong with my working?? I can get to the first to the last step by simply applying the time dilation effect by using Δt' = γΔt...

 

[TSny]

Everything looks good. And yes, the first to last step can also be obtained using the time dilation formula. Good.

 

[unscientific]

Everything looks good. And yes, the first to last step can also be obtained using the time dilation formula. Good.

Thanks! That was pretty easy for a 10 mark exam question