 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: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 $|\mathbf{k}|=n(\omega )\omega /c$,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.

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.