Past exam of the optics and electrodynamics course of the University of Cambridge / 2012 / Paper 1 / 3

Use $\mathbf{E}(z,t)=\mathrm{Re}[\mathbf{J}{e}^{i(kz-\omega t)}]$. A Jones vector $\mathbf{J}=(J_x,J_y{)}^{\mathsf{T}}$ records the two transverse complex amplitudes. The amplitude ratio and relative phase determine the polarization ellipse; a common nonzero scalar does not change the polarization. A fixed Jones vector describes fully polarized monochromatic light, not an unpolarized statistical mixture.

Linear birefringence gives orthogonal principal polarizations different refractive indices. Put $n_y-n_x=\Delta n>0$, so $x$ is the fast axis. A plate of thickness $d$ gives slow-minus-fast phase retardationThe thinnest quarter-wave plate hasHere ${\lambda }_0=2\pi c/{\omega }_0$ is the vacuum wavelength. For linear input at angle $\theta$ to the fast axis, the output at design frequency is $(\cos \theta ,i\sin \theta {)}^{\mathsf{T}}$ up to common phase. Axis-aligned input remains linear; equal nonzero components at $45^{\circ }$ modulo $90^{\circ }$ give circular polarization; other input angles give elliptical polarization. There is no need to spatially separate the two components.^[ref]

For waveplate retardance detuning, defineNeglecting material dispersion reduces this to $ϵ=(\pi /2)(\omega -{\omega }_0)/{\omega }_0$. Remove the common fast-axis phase. Then $Q=\mathrm{diag}(1,i{e}^{iϵ})$. A similar half-wave plate of twice the thickness has twice the retardation, giving $H=\mathrm{diag}(1,-e^{2iϵ})$. The shared error parameter presumes the same material and matched thickness ratio.

For the half-wave and quarter-wave achromatic circularizer, the first fast-axis angle is $\alpha =15^{\circ }$ and the second is $\gamma =\alpha +\beta =75^{\circ }$. With the stated passive rotation convention,An explicit expression valid for the given shared error is obtained by putting $z=e^{iϵ}$ and definingAt $ϵ=0$, the half-wave plate rotates the linear input to $30^{\circ }$, which is $-45^{\circ }$ relative to the second fast axis. ThusThis is circular polarization. With our time convention its fixed-point electric vector is proportional, after a time-origin shift, to $(\cos \omega t,-\sin \omega t)$; this specifies the rotation without relying on conflicting handedness conventions.

Expanding the exact product to first order givesThese coefficients are rounded to three significant figures. Before rounding, the derivative is exactly ${\mathbf{J}}_0^{\prime }=i\kappa {\mathbf{J}}_0$, with $\kappa =(3-\sqrt{3})/2$. The Jones-component ratio sensitivity therefore vanishes:Only common phase changes at first order. This proves compensation for the prescribed input near the design frequency, not exact achromatic performance throughout the visible spectrum for arbitrary input polarization.

A transverse static electric field can induce linear birefringence through the electro-optic Kerr effect. Define its coefficient by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, with the principal axes parallel and perpendicular to the applied field. A length $d_K$ then gives ${\delta }_K=2\pi K{d}_K{E}_{\mathrm{dc}}^2$.

For input polarized at $45^{\circ }$ to those axes, switching from zero retardation to $|{\delta }_K|=\pi /2$ switches linear to circular polarization. Other input angles generally become elliptical, and axis-aligned input stays linear. The required field magnitude for the lowest-order setting isA Kerr cell is suitable as a switchable quarter-wave plate. The field can be changed electronically without mechanically rotating the axes; actual switching time depends on the material response and driving circuit. This is induced linear birefringence, not polarization-selective absorption.^[ref]

The longitudinal Faraday effect produces circular birefringence: right- and left-circular components have different phase velocities. Linear input contains equal amplitudes of both circular eigenpolarizations. A relative phase ${\delta }_F$ rotates its linear polarization through ${\delta }_F/2$, preserving linearity in an ideal transparent sample. Conventionally the signed rotation is ${\vartheta }_F=V{B}_{\parallel }{d}_F$, with Verdet constant $V$.

A pure Faraday rotator cannot serve as the required switchable quarter-wave plate. It rotates a line into another line rather than producing the required relative phase between fixed orthogonal linear components. Applying a field that gives a nominal quarter-cycle circular-eigenmode retardation simply rotates linear polarization by $45^{\circ }$. Circular dichroism or a separate linear retarder would change the device, and is not the stated pure Faraday mechanism.