= 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 $\mathbf D\parallel\mathbf E$. Thus
$$
\boxed{\alpha_o=0.}
$$
Its <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 $\Delta=0.1\,\mathrm{mm}$,
$$
\boxed{\alpha_e=\arctan(0.1/10)=\arctan(0.010)\simeq0.0100\,\mathrm{rad}=0.573^\circ.}
$$
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.
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