Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section A / 1 / Solution

The emission mechanism is Cherenkov radiation. An Electron moving faster than the phase velocity $c/n$ of light in the material excites electric polarization disturbances whose radiation adds coherently on a cone. This does not require the Electron to exceed the speed of light in vacuum. Taking $n=1.4$ as constant over the detected optical band, the Cherenkov cone angle satisfiesEmission requires $\beta >1/n$. The detector accepts ${\theta }_C<20^{\circ }$, soNumerically,The lower endpoint has vanishing Cherenkov radiation yield, and the upper endpoint is excluded by the stated strict angular acceptance. Faster Electrons emit at larger cone angles: at $v\simeq c$, ${\theta }_C\simeq {\cos }^{-1}(1/1.4)=44.4^{\circ }$, outside this detector's acceptance. The angular condition gives the accepted speed interval conditional on rays reaching the detector; no collection geometry or total signal yield is specified.