 Solution

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.