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The extraordinary wave's Poynting vector can differ in direction from its wavevector. The latter specifies wave phase propagation, while the former specifies energy flux. This angular separation is walk-off.^[ref]

In the lossless, nondispersive model above, take the gradient of $\omega =c\sqrt{k_x^2/n_e^2+k_z^2/n_o^2}$ to obtain the group velocity, parallel to the Poynting vector. If its angle to the optic axis is $\varphi$, this gives $\tan \varphi =(n_o^2/n_e^2)\tan \theta$. Consequently the extraordinary wave generally carries energy away from its wavevector direction. An internal wavevector normal to a slab face therefore does not guarantee a normal energy path. Birefringent walk-off vanishes at $\theta =0$ and $\theta =\pi /2$; the latter direction still has unequal refractive indices. This is a useful waveplate geometry: overlapping components can acquire phase retardation without spatial splitting. A thin specimen, insufficient imaging resolution, or excitation of only one linear polarization can also hide a double image, for different reasons.

Place a clear calcite specimen on a transparency bearing printed words on an overhead projector. The projected readout is two copies of the lettering. Put a rotatable linear polarizer after the specimen: suitable orientations suppress one copy or the other. These are the actual apparatus and observations in the University of Iowa demonstration.^[ref]

For a useful control, remove the specimen while keeping the transparency, projector and screen fixed: the doubling should disappear. Replace it, hold it stationary and rotate only the polarizer; complementary brightness changes distinguish two linear polarizations from an accidental duplicate transparency. These are suggested controls, rather than additional reported measurements. Ideally the two copies obey complementary Malus's law curves, $I_o(\alpha )=I_{o,\max }{\cos }^2\alpha$ and $I_e(\alpha )=I_{e,\max }{\sin }^2\alpha$, with angle $\alpha$ measured from the exiting ordinary wave's linear polarization. Unequal maxima do not invalidate the test. Finite polarizer extinction, poor focus and specimen defects limit visibility; image separation alone does not determine either refractive index without the specimen geometry.

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 isThe 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.

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.^[ref]

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.