 Solution

For propagation along z, write $\mathbf{E}=\mathrm{Re}\{(E_x\hat{\mathbf{x}}+E_y\hat{\mathbf{y}})e^{i(kz-\omega t)}\}$. 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 $\pm 45^{\circ }$. Up to an overall phase, convenient matrices in fixed physical x/y axes areThe plate is lossless, $Q{Q}^{†}=I_2$, and $Q(1,0{)}^T=(1,i{)}^T/\sqrt{2}$ is circular in the supplied convention. 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 $C_{\mathrm{in}}=(I_0/2)I_2$ and propagate it as $C_{\mathrm{out}}=J{C}_{\mathrm{in}}{J}^{†}$. Forward transmission givesThus 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 givesThe 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 $Q^2=i\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: two passes exchange x and y linear states.