= Solution
Choose the first <polarizer>'s pass axis as $\mathbf e_1$ and write the second as $\mathbf a=(\mathbf e_1+\mathbf e_2)/\sqrt2$. The positive signed rotation takes $\mathbf e_1$ toward $\mathbf a$ by the short $\pi/4$ angle. Let $B_\parallel$ be signed according to $\theta=\Upsilon LB_\parallel$. The <Jones matrices> of ideal <polarizers> are their <orthogonal projection> matrices,
$$
P_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad
P_2=\mathbf a\mathbf a^{\mathsf T}=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}.
$$
Removing the unimportant common propagation <phase>, the lossless glass block has the <rotation matrix>
$$
R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}.
$$
The elements act from right to left, so
$$
\boxed{J_{S\to O}=P_2R(\theta)P_1
=\frac12\begin{pmatrix}\cos\theta+\sin\theta&0\\\cos\theta+\sin\theta&0\end{pmatrix},\qquad
\theta=\Upsilon LB_\parallel.}
$$
With the common optical <phase> retained, multiply this by $e^{i\bar\phi}$. Extra scalar attenuation would also multiply the result, but none is supplied. A single <Jones vector> does not represent <unpolarized light>; the device's <Jones matrix> is nevertheless well defined for each incident component and can be applied to their statistical mixture.
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