Question

Find the kinetic enegry of electrons emitting light in a medium with refractive index `n = 1.50` at an angle `theta = 30^(@)` to their propagation direction.

Answer 1

From `cos theta = (v)/(V)`
we get `V = V sec theta`
so `(V)/(c ) = (v)/(c )sec theta = (sec theta)/(n) = (sec 30^(@))/(1.5) = (2//sqrt(3))/(3//2) = (4)/(3sqrt(3))`
Thus for electrons
`T_(e) = 0.511 [(1)/(sqrt(1-(16)/(27)))-1] = 0.511 [sqrt((27)/(11))-1] = 0.289 MeV`
Generally `T = mc^(2)[(1)/(sqrt(1-(1)/(n^(2)cos^(2)theta)))-1]`