= Solution
Let $I_0$ be the incident <light intensity>. The first ideal <polarizer> transmits $I_0/2$ and produces <linear polarization> along $\mathbf e_1$. <Faraday rotation> turns that direction through $\theta=\Upsilon LB_\parallel$, and <Malus's law> at the second <polarizer> gives
$$
\boxed{I_f(B_\parallel)=\frac{I_0}{2}\cos^2\left(\frac\pi4-\Upsilon LB_\parallel\right).}
$$
Equivalently, use the input <coherency matrix> $C_{\rm in}=(I_0/2)I_2$ and $I_f=\operatorname{tr}(JC_{\rm in}J^\dagger)$. This produces the same one-half factor required by <unpolarized light>.
Maxima satisfy $\theta=\pi/4+m\pi$, with $m\in\mathbb Z$. For $L=1\,\mathrm m$ and $\Upsilon=9\,\mathrm{rad\,T^{-1}\,m^{-1}}$, the smallest field magnitude is
$$
\boxed{|B|_{\min}=\frac{\pi}{4\Upsilon L}=0.0872665\,\mathrm T=87.3\,\mathrm{mT},\qquad I_{\max}=\frac{I_0}{2}.}
$$
Its direction is the axial direction that rotates the polarization passed by $P_1$ toward the pass axis of $P_2$ through the shorter $45^\circ$. In the right-handed ray-basis convention where $\rho=+\Upsilon B_\parallel$ and the depicted second axis is $+45^\circ$, this is $B_\parallel>0$, along $S\to O$. Reversing that field at the same magnitude gives $\theta=-\pi/4$ and zero transmission. A signed <Verdet constant> is meaningful only together with its rotation convention, so the polarization-alignment description fixes the physical direction without relying on ambiguous clockwise labels in a perspective diagram.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2022-q4-faraday-transmission.png]
{title=Original forward and reverse transmission curves and laboratory polarization directions at the optimum <Faraday rotation>}
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