= 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, so
$$
J_{O\to S}=P_1R(\theta)P_2
=\frac12\begin{pmatrix}\cos\theta-\sin\theta&\cos\theta-\sin\theta\\0&0\end{pmatrix}.
$$
The reverse <light intensity> is therefore
$$
\boxed{I_r=\frac{I_0}{2}\cos^2\left(\frac\pi4+\Upsilon LB_\parallel\right).}
$$
For 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.
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