 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/11/b/iii/solution

The photon starts at

x_{E}^{'} = γ L

and travels left to the runner at the fixed position

x^{'} = 0

. Thus

d_{γ}^{'} = γ L, Δ t_{γ}^{'} = \frac{γ L}{c} .

This distance is not the simultaneous contracted tunnel length

L / γ

, because the runner is not at the entrance when the photon is emitted in

S^{'}

. Adding the entrance-to-runner gap from part (ii) to

L / γ

indeed gives

γ L

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