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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» 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]` |
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