Mastering Physics Solutions: Cerenkov Radiation II

Cerenkov Radiation II

Part A = θ = cos^-1(c/(nv))
Part B = aerogel
Part C = v = 0.995 c Click to use the calculator/solver for this part of the problem
Part D = 0.97 c

Solution Below:

Electromagnetic radiation (known as Cerenkov radiation) is emitted when a charged particle moves through a medium faster then the local speed of light. It should be stressed that the particle is never going faster then the speed of light in vacuum (or c), just faster than the speed of light in the material (which is always less than c).

When a charged particle passes straight through a medium faster than the local speed of light, it will emit Cerenkov radiation in a cone. Let’s see how the cone angle is correlated to the speed of the particle.

Part A

If a particle is traveling straight through a material with index of refraction n at a speed v, what is the angle θ between the vector of the propagating Cerenkov radiation and the vector in the direction of the propagating particle?
Express your answer in terms of v, c, and n.

θ = cos^-1(c/(nv))

Part B

Suppose you wish to accurately measure the speed of high-energy particles with velocities greater than 98% the speed of light in vacuum. You can use a ring-imaging Cerenkov detector consisting of a thin slab of material separated from an array of photomultiplier tubes by an arbitrary open space. (Photomultiplier tubes, or PMTs, are devices used to detect weak light signals.) The detector works on the principle that the Cerenkov radiation emitted in the thin slab will be a cone of light that can be measured with the array of PMTs. The PMTs, having a finite width, can only resolve a finite change in the angle of the ring created by the Cerenkov radiation. Use these constraints and the equation for θ from Part A to determine which of the following substrate materials is best suited to giving you the greatest precision in determining particle velocity.

  • diamond (n = 2.417)
  • crown glass (n = 1.52)
  • ice (n = 1.3)
  • aerogel (n = 1.03)
  • vacuum (n = 1)

The aerogel is the only material that will keep light at >= 0.98c

aerogel

Part C

Suppose our detector used aerogel (n = 1.03) for the Cerenkov material and the photomultiplier tube array had a resolution of Δ = 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 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.

This one has the potential to be confusing. Let’s go through step by step. Start by finding the angle you’ll get for a particle traveling at the speed of light (use the formula from Part A):

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

v = 0.995 c

Part D

What is the lowest velocity vmin that a charged particle can have and still emit Cerenkov radiation in the aerogel?
Express your answer as a multiple of c to two significant figures.

Remember that Cerenkov radiation is emitted when a particle moves faster than the local speed of light in a medium. For instance, light moves slower in water than it does air. Water’s refractive index is about 1.33, so light moves 33% slower: 0.667 c. This means if a particle in water is moving faster than 0.667 c, it will emit Cerenkov radiation. The problem told us that the aerogel has a refractive index of 1.03. So just divide 1.0 c by 1.03, which gives you 0.97 c. Above this speed, a particle traveling through the aerogel will emit Cerenkov radiation.

0.97 c

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4 Responses to Mastering Physics Solutions: Cerenkov Radiation II

  1. Anna says:

    Part D says:
    What is the lowest velocity vmin that a charged particle can have and still emit Cerenkov radiation in the aerogel?
    Express your answer as a multiple of c to two significant figures.

  2. Olie says:

    is there a potability that part d will be added soon?

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