= Solution
For propagation along z, write $\mathbf E=\operatorname{Re}\{(E_x\widehat{\mathbf x}+E_y\widehat{\mathbf y})e^{i(kz-\omega t)}\}$. A <Jones vector> contains the complex transverse components. Equal <phase> modulo $\pi$ gives <linear polarization>; equal magnitudes with <phase difference> $\pm\pi/2$ give <circular polarization>. A common overall <phase> has no effect on the <polarization ellipse>. A deterministic optical element acts by a <Jones matrix>, with successive matrices ordered from right to left along the <light> path.
A <circular polarizer> uses a linear <polarizer> transmitting x and a <quarter-wave plate> whose principal axes are at $\pm45^\circ$. Up to an overall <phase>, convenient matrices in fixed physical x/y axes are
$$
P=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad
Q=\frac1{\sqrt2}\begin{pmatrix}1&i\\ i&1\end{pmatrix}.
$$
The plate is lossless, $QQ^\dagger=I_2$, and $Q(1,0)^T=(1,i)^T/\sqrt2$ is circular in the supplied convention. \b[Put P on side A and Q on side B]: forward passage is $J_{A\to B}=QP$.
<Unpolarized light> has no single <Jones vector>. Represent its <intensity> by the <coherency matrix> $C_{\rm in}=(I_0/2)I_2$ and propagate it as $C_{\rm out}=JC_{\rm in}J^\dagger$. Forward transmission gives
$$
C_{A\to B}=\frac{I_0}{2}\,\mathbf e_+\mathbf e_+^\dagger,
\qquad\mathbf e_+=\frac1{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}.
$$
Thus \b[the transmitted <light> is circular with <intensity> $I_0/2$]. The initial <polarizer> removes one incoherent linear component; the plate changes <phase> without restoring the lost <intensity>.
Reverse passage encounters Q before P and has $J_{B\to A}=PQ$ in the same fixed transverse axes. It gives
$$
\boxed{C_{B\to A}=\frac{I_0}{2}P.}
$$
The output is linearly polarized along x and again has <intensity> $I_0/2$. A retarder alone leaves an unpolarized <coherency matrix> unchanged. Consequently an unpolarized scene appears darker from either side, although the transmitted polarization differs. No polarization-dependent response of the observer is needed to explain this reduction.
For the mirror calculations below, use an ideal near-normal isotropic reflector, $M=-I_2$ up to a common <phase>. In fixed physical transverse axes <reflection> leaves the <polarization ellipse> unchanged; it reverses circular <helicity> because the propagation direction reverses. A reciprocal plate has the same relative <phase> delay on the return traversal in this fixed-axis convention. In particular $Q^2=i\begin{pmatrix}0&1\\1&0\end{pmatrix}$: two passes exchange x and y linear states.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2013/paper-1-oe2013-polarizer-mirror.png]
{title=Original outward and return polarization-state diagrams for the two orientations of a polarizer-plus-quarter-wave-plate filter in front of a mirror}
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