Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section B / 4 / e / 1 / Solution

Keep the magnetic field fixed in the laboratory and exchange source and observer. The incident unpolarized light now meets $P_2$ first. In fixed transverse laboratory axes the backward glass map is still $R(\theta )$, not $R(-\theta )$: this is optical nonreciprocity of the Faraday effect. The order of the two polarizers is reversed, soThe reverse light intensity is thereforeFor the optimal forward setting $\theta =\pi /4$, it vanishes: the $45^{\circ }$ input selected by $P_2$ rotates to $90^{\circ }$ and is blocked by $P_1$. At a general field, $I_r+I_f=I_0/2$. The result assumes equal incident light intensities in the two experiments. Treating the glass as a reciprocal natural optical rotator would predict cancellation of its laboratory rotation on reversal and would give the wrong answer here.