Past exam of the physics 1 paper of the University of Oxford / 2015 / 11 / b / iii / Solution

The photon starts at $x_E^{\prime }=\gamma L$ and travels left to the runner at the fixed position $x^{\prime }=0$. ThusThis distance is not the simultaneous contracted tunnel length $L/\gamma$, because the runner is not at the entrance when the photon is emitted in $S^{\prime }$. Adding the entrance-to-runner gap from part (ii) to $L/\gamma$ indeed gives $\gamma L$.