Teaching Modern Physics - QuarkNet - Fermilab

4. A common particle detector technique involves optical scintillating fiber, which emits
light when crossed by a charged particle. If the fiber consists of a core with index of
refraction 1.59 and a cladding of index of refraction 1.49, what fraction of the light is
trapped in the fiber and transported to an end? Assume all light is emitted at the center of
the fiber (i.e., on the fiber axis) and that the light emission is isotropic.
This problem is fairly tricky, as it involves light emission
into three dimensions. Thus, one must use calculus and
solid-angle calculations. First, one must find the critical
angle of reflection, then convert that to limits of integration.
Finally, the integrated solid angle must be divided by the total
solid angle.
sin θc = 1.49/1.59 → θc = 69.6°
θ = 90° – θc = 20.43°
Light is emitted into all solid angles.
2π
π
Ω = ∫∫ sinθ
dθ
dφ
= 4π
0 0
Only the light emitted with angle smaller than 20.43° (= 0.36 radians) will be trapped. Thus,
2π
0.
36
Ωtrapped = ∫ ∫ sinθ
dθ
dφ
= 0.
396
0
0
The fraction trapped is Ωtrapped/Ω = 3.15%.
A common error is to ignore the azimuthal angle and use a simple proportion. This error yields
11.3%.
5. Cerenkov light is a form of light emitted when a charged particle travels in a transparent
medium faster than light travels in the same medium. Calculate the minimum velocity an
electron must have to emit Cerenkov light in water.
This is a simple problem. Use n = c/v. The index of refraction of water is 1.33.
So v = 2.26 × 10 8 m/s.
53
θc
θ