= Solution
\b[<Birefringence> is polarization-dependent propagation in an optically anisotropic medium.] Its <permittivity tensor> relates $\mathbf D$ to $\mathbf E$. In a lossless uniaxial <dielectric>, take the <optic axis> along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a <wave vector> making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the <extraordinary phase-index angular dependence> is
$$
\boxed{\frac1{n_{\rm ex}(\vartheta)^2}=\frac{\cos^2\vartheta}{n_o^2}+\frac{\sin^2\vartheta}{n_e^2}.}
$$
The ordinary <electric field> is perpendicular to the plane containing the <wave vector> and <optic axis>. The extraordinary <electric displacement field> lies in that plane and is perpendicular to the <wave vector>, but its <electric field> need not be. Its energy-flow direction can therefore differ from the wave-normal direction: <birefringent walk-off>. Along the <optic axis> the two indices coincide. <Refraction> generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2016/paper-1-birefringent-indices.png]
{title=Illustrative positive-uniaxial ordinary and extraordinary phase indices coincide along the optic axis}
{source=paper-1-birefringent-indices.py}
A <waveplate> instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the <retardance> is $\Delta=2\pi L(n_2-n_1)/\lambda_0$. A <quarter-wave plate> gives $|\Delta|=\pi/2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and <circular polarization>. A <half-wave plate> gives $|\Delta|=\pi$ and rotates a <linear polarization>. This is a phase effect; unequal absorption is a separate property.<Polarisation by Sander Konijnenberg, Aurèle J.L. Adam and H. Paul Urbach>[Creating Different Polarisation States; wave plates and relative phase]{ref}
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