= Solution
A charge moving uniformly through a transparent dielectric can emit <Cherenkov radiation> if its speed exceeds the <phase velocity> of light in that medium:
$$
\boxed{v>\frac{c}{n(\omega)},\qquad \beta n(\omega)>1,\quad\beta=\frac vc.}
$$
It need not exceed the vacuum speed $c$. The passing charge polarizes the medium; the induced response produces radiation that adds coherently along selected directions. Equivalently, <phase> matching along the trajectory requires $\mathbf k\cdot\mathbf v=\omega$. Since $|\mathbf k|=n(\omega)\omega/c$,
$$
\boxed{\cos\theta_C=\frac{1}{\beta n(\omega)}}.
$$
Here $\theta_C$ is the angle of the emitted light rays to the forward particle <velocity>: the rays form a cone around the track. With <dispersion>, both the threshold and cone angle depend on <frequency>; only transparent <frequency> bands satisfying the threshold contribute escaping radiation.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2017-q2-cherenkov-cone.png]
{title=Wavefront envelope and photon direction for Cherenkov radiation}
In the wavefront construction, a disturbance emitted a time $t$ earlier has radius $ct/n$, while the particle has moved $vt$. Tangents from the present particle position to these wavefronts make a backward-opening wavefront cone of half-angle $\alpha$, where $\sin\alpha=1/(\beta n)$. The light rays are normal to this cone. Consequently $\alpha+\theta_C=\pi/2$: the <Cherenkov photon-cone and wavefront-cone angles> are complementary, rather than identical.
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