Suppose our detector used aerogel (n = 1.03) for the Cerenkov material and the p

Question

Suppose our detector used aerogel (n = 1.03) for the Cerenkov material and the photomultiplier tube array had a resolution of Delta theta = 1.2 degrees. This means for instance, that the detector can distinguish between Cerenkov light emitted at an angle of 15 degrees and that emitted at 13.8 degrees but can't tell the difference between Cerenkov light emitted at 15 degrees and that emitted at 13.9 degrees. What is the highest velocity v_max at which a charged particle can be accurately measured to be below the speed of light in vacuum (c)? Express your answer as a multiple of c to three significant figures.

Explanation / Answer

= cos^-1(c/(nv))
= cos^-1(2.998 * 10^8 / (1.03 * 2.998 * 10^8))
= cos^-1(0.970873786)
= 13.86243 degrees

As the speed of the particle decreases, this angle will also decrease. If we want to find the maximum *detectable* velocity of a particle traveling below the speed of light, we subtract the instrument’s resolution from this angle. Since the instrument has a resolution of 1.2 degrees, we want to find the velocity at 12.66243 degrees (note: either use our calculator or be VERY careful that yours has the correct units – e.g. make sure you’re not inadvertently switching between degrees and radians. Use one or the other, don’t mix them):

12.66243 = cos^-1(c/(nv))
12.66243 = cos^-1(2.998 * 10^8 / (1.03 * v))
cos(12.66243) = (2.998 * 10^8 / (1.03 * v))
0.9756784837 = 291067961 / v
v = 298232644.5 m/s

Divide by the speed of light and you get:

v = 0.995 c

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