= Solution
\b[General pure state.] Use time dependence $e^{i(kz-\omega t)}$. The most general nonzero <Jones vector> can be written
$$
\mathbf J=Ae^{i\chi}\begin{pmatrix}\cos\alpha\\e^{i\delta}\sin\alpha\end{pmatrix},\qquad A>0,\quad0\leq\alpha\leq\frac\pi2,\quad0\leq\delta<2\pi.
$$
There are four real parameters in the full complex field. Removing overall <intensity> $A^2$ and the unobservable common <phase> $\chi$ leaves \b[two real degrees of freedom for <pure polarization>], $\alpha$ and $\delta$. Their endpoint degeneracies simply reflect the fact that the <phase> of a vanishing component is irrelevant. A single <Jones vector> cannot represent partially polarized light.
\b[Three linear <polarizers>.] Define the transmission axis $\mathbf a_\theta=(\cos\theta,\sin\theta)^T$. An ideal linear <polarizer> has <Jones matrix> $P_\theta=\mathbf a_\theta\mathbf a_\theta^T$. For definiteness label $(1,-i)^T/\sqrt2$ as right-circular in the chosen convention; reversing the handedness convention changes no <intensity> below. The first <polarizer> produces $(1,0)^T/\sqrt2$. The intermediate <polarizer> and final $y$ analyzer then give
$$
\mathbf J_{\rm out}=P_{\pi/2}P_\theta P_0\mathbf J_R=\frac{\sin\theta\cos\theta}{\sqrt2}\begin{pmatrix}0\\1\end{pmatrix},
$$
so <Malus's law> gives
$$
\boxed{\frac{I_{\rm out}}{I_{\rm in}}=\frac18\sin^2(2\theta).}
$$
The maximum is $1/8$ at $\theta=\pi/4+m\pi/2$, and the transmission vanishes at $\theta=m\pi/2$. The first <polarizer> accounts for an initial factor $1/2$; retaining that factor is essential when quoting transmission relative to the incident circularly polarized beam.
\b[<Waveplate> and quarter-wave matrix.] A lossless <waveplate> has <linear birefringence>: its orthogonal fast and slow axes have indices $n_f<n_s$. A thickness $L$ produces slow-minus-fast <phase retardation>
$$
\delta=\frac{2\pi L}{\lambda_0}(n_s-n_f).
$$
Resolving the incident <Jones vector> into the plate axes, advancing the two phases and transforming back gives, up to a common <phase>,
$$
J_\delta(\theta)=R(-\theta)\begin{pmatrix}1&0\\0&e^{i\delta}\end{pmatrix}R(\theta).
$$
A <quarter-wave plate> has $\delta=\pi/2$ modulo $2\pi$. With $c_\theta=\cos\theta$ and $s_\theta=\sin\theta$, its <Jones matrix> is therefore
$$
\boxed{Q(\theta)=\begin{pmatrix}c_\theta^2+i s_\theta^2&(1-i)c_\theta s_\theta\\(1-i)c_\theta s_\theta&s_\theta^2+i c_\theta^2\end{pmatrix}.}
$$
In particular $Q(0)=\operatorname{diag}(1,i)$. Equal incident components along the two axes acquire a relative quarter-cycle <phase>, converting $45^\circ$ <linear polarization> to <circular polarization>; an incident component along either principal axis stays linearly polarized.
\b[Half-wave replacement.] A <half-wave plate> has $\delta=\pi$ and hence
$$
H(\theta)=\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}.
$$
It converts the $x$-polarized output of the first <polarizer> to a linear state at angle $2\theta$. The final $y$ analyzer gives
$$
\boxed{\frac{I_{\rm out}}{I_{\rm in}}=\frac12\sin^2(2\theta).}
$$
The maximum is $1/2$, because the ideal <half-wave plate> loses no light after the first <polarizer>.
\b[Quarter-wave replacement.] The transmitted $y$ amplitude is $(1-i)c_\theta s_\theta/\sqrt2$, giving
$$
\boxed{\frac{I_{\rm out}}{I_{\rm in}}=\frac14\sin^2(2\theta).}
$$
More generally, the retarder gives $I_{\rm out}/I_{\rm in}=\tfrac12\sin^2(\delta/2)\sin^2(2\theta)$. All three requested curves have the same zeros and angular period, but different heights:
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2017-q3-polarization-transmission.png]
{title=Transmission through crossed outer polarizers with three different middle elements}
\b[Preparing an arbitrary ellipse.] Yes: two rotatable <quarter-wave plates> already suffice. Let the input left-circular state be $\mathbf J_L=(1,i)^T/\sqrt2$. Up to an overall <phase>, any target <pure polarization> can be parameterized by its <polarization ellipse>:
$$
\mathbf J_{\rm target}=R(-\beta)\begin{pmatrix}\cos\eta\\i\sin\eta\end{pmatrix},\qquad-\frac\pi4\leq\eta\leq\frac\pi4.
$$
Here $\beta$ is the major-axis orientation, and $\eta$ sets the signed minor-to-major semiaxis ratio $\tan\eta$. First note directly from $Q(\theta)$ that
$$
Q(\theta)\mathbf J_L\ \propto\ \begin{pmatrix}\cos(\theta-\pi/4)\\\sin(\theta-\pi/4)\end{pmatrix}.
$$
Choose the two fast-axis angles as
$$
\boxed{\theta_1=\beta+\eta+\frac\pi4,\qquad\theta_2=\beta.}
$$
The first <quarter-wave plate> produces linear light at $\beta+\eta$. In the axes of the second, its components are $(\cos\eta,\sin\eta)$; the second gives precisely $\mathbf J_{\rm target}$. This proves <pure-polarization preparation with two quarter-wave plates>, including the linear and circular limiting cases. A <half-wave plate> can also rotate the axes, but is not required. The phrase about left-handed input is understood as left-circular light in this construction; the explicit vector fixes the convention.
\b[Switchable retardation.] The <electro-optic Kerr effect> induces <linear birefringence> quadratic in a static transverse <electric field>. Defining the coefficient $K$ by $\Delta n=\lambda_0 K E_{\rm dc}^2$, its <phase retardation> is
$$
\delta_K=\frac{2\pi L\Delta n}{\lambda_0}=2\pi KLE_{\rm dc}^2.
$$
The two eigenpolarizations are linear, parallel and perpendicular to the applied field. Thus an appropriate field produces a switchable <quarter-wave plate>:
$$
\boxed{E_{\rm dc}^2=\frac{1}{4|K|L}\quad\text{for the smallest }|\delta_K|=\pi/2.}
$$
The sign of $K$ determines which axis is fast. Switching the field off removes the induced retardation, assuming no residual birefringence.
The <Faraday effect> instead gives <circular birefringence> for a longitudinal <magnetic field>. The two circular eigenmodes acquire different phases, rotating a <linear polarization> through
$$
\theta_F=VBL=\frac{\pi L}{\lambda_0}(n_+-n_-),
$$
with sign set by the handedness convention and the <Verdet constant> $V$. Its <Jones matrix> in the linear basis is a real rotation, up to a common <phase>. It preserves the ellipticity of a pure state, so a linear input remains linear. A $\pi/2$ <phase> difference between circular eigenmodes rotates a linear state by $\pi/4$; it does not make it circular. \b[The Kerr device supplies the switchable <quarter-wave plate>; a Faraday rotator alone does not.]
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