Extraordinary wave
= Extraordinary wave
{title2=$n_{\mathrm{ex}}(\theta)$}
The <extraordinary wave>'s <electric displacement field> lies in that plane and is perpendicular to the <wavevector>; its <electric field> need not be. For <angle> $\theta$ between the <wavevector> and <optic axis>,
$$
\frac{1}{n_{\mathrm{ex}}(\theta)^2}=\frac{\cos^2\theta}{n_o^2}+\frac{\sin^2\theta}{n_e^2}.
$$
Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi/2)=n_e$.<Fundamentals of Photonics and Quantum Electronics lecture notes by Franz X. Kärtner>[§§2.8.2–2.8.3, printed p. 156, equations 2.410–2.413 and longitudinal electric-field discussion]{ref}
A short derivation starts from the existing <plane wave in an anisotropic dielectric> equation. Put $\mathbf k=(k_x,0,k_z)$ and $k_0=\omega/c$, with <angular frequency> $\omega$ and vacuum <speed of light> $c$. The $y$ component gives the <ordinary wave> relation $k_x^2+k_z^2=n_o^2k_0^2$. The <determinant> of the $x,z$ system gives
$$
\frac{k_x^2}{n_e^2}+\frac{k_z^2}{n_o^2}=k_0^2.
$$
Substituting $k_x=n_{\mathrm{ex}}k_0\sin\theta$ and $k_z=n_{\mathrm{ex}}k_0\cos\theta$ yields the displayed directional <refractive index>.