= Solution
Use emission, reflection and return-to-source events $A,B,C$. In the source-and-mirror rest frame their $(ct,x)$ coordinates are $A=(0,0)$, $B=(d,d)$ and $C=(2d,0)$. The laser and mirror have vertical <worldlines>; the two light rays have slopes $\pm1$ on a diagram with equal spatial and $ct$ scales.
For a frame moving with <velocity> $+v$ along the outgoing beam, let $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$. A <Lorentz transformation> gives
$$
B'=(\gamma d(1-\beta),\gamma d(1-\beta)),\qquad C'=(2\gamma d,-2\gamma\beta d).
$$
Both laser and mirror now move towards negative $x'$, with worldlines $x'=-vt'$ and $x'=d/\gamma-vt'$. Light still has slopes $\pm1$. The moving observer's own worldline is $x'=0$; it should not be confused with the moving source worldline.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-resit-lorentz-light-round-trip.png]
{title=Emission, mirror reflection and return to the original source in the rest frame and a moving Lorentz frame}
Back to article page
