Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / 1 / c / Solution

First obtain the particle speed from kinetic energy rather than treating $1\mathrm{GeV}$ as its total energy. With Electron rest energy $m_e{c}^2=0.510999\mathrm{MeV}$,For Cherenkov radiation, the photon propagation direction makes angle ${\theta }_C$ to the particle trajectory, with $\cos {\theta }_C=1/(n\beta )$. The speed correction is negligible at three significant digits, but the Cherenkov photon-cone and wavefront-cone angles must be distinguished.

If the given opening $\alpha$ is the full cone of photon directions, ${\theta }_C=\alpha /2=48.75^{\circ }$. The result isIf instead $\alpha$ denotes the full opening of the conical wavefront, its half-angle is complementary to the photon direction: ${\theta }_C=90^{\circ }-\alpha /2=41.25^{\circ }$. That convention givesThe original PDF supplies no cone diagram or definition distinguishing these geometries. Thus the literal full photon-cone interpretation gives $1.52$, while $1.33$ follows from the wavefront interpretation. It would be incorrect to interchange sine and cosine without stating which cone is meant.