 Solution

At $p=600\mathrm{MeV}/c$, the velocities are $\beta =p/\sqrt{p^2+m^2{c}^2}$, giving approximately $1.0000$, $0.98485$ and $0.53874$ for electrons, muons and protons. The glass threshold is $\beta >1/1.55=0.64516$. Thus the first two produce Cherenkov radiation, with internal angleswhereas a primary proton at this momentum is below threshold.

If an optical arrangement transports those cone angles without changing them, a plane $L=2\mathrm{m}$ downstream would intersect rings of radiusThese are conditional geometric estimates, not the ray trace for a flat slab in air. The $0.1\mathrm{m}$ emission depth alone spreads the radius by order $0.12\mathrm{m}$, already larger than the approximately $0.062\mathrm{m}$ separation. The slab is one radiation length: electrons lose appreciable energy by bremsstrahlung and begin an electromagnetic shower, producing a broader and less clean pattern. Muons mainly ionize and retain a comparatively clean cone, with some Multiple Coulomb scattering. A below-threshold primary proton gives no direct cone, although energetic delta electrons and other secondary interactions can produce light.

There is an essential optical qualification. For a normal beam through a parallel-faced glass slab followed by air, the internal critical angle is ${\theta }_{\mathrm{crit}}=\arcsin (1/1.55)\simeq 40.18^{\circ }$. Both emission angles exceed it. Snell's law would require $\sin {\theta }_{\mathrm{air}}=1.55\sin {\theta }_C>1$, so the ideal photons undergo total internal reflection at the downstream face and do not form those direct downstream rings. Optical coupling, a prism or different surface geometry must be specified to extract them. The stated material parameters alone do not define a unique detected pattern; this flat-slab geometry is a concrete counterexample to silently assuming straight transmitted rays.