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GR9277 #69
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{69}
A fast charged particle passes perpendicularly through a thin glass sheet of index of refraction 1.5. The particle emits light in the glass. The minimum speed of the particle is

1. c/3
2. 4c/9
3. 5c/9
4. 2c/3
5. c

Optics$\Rightarrow$}Speed of Light

The speed of light is related to the index of refraction by $n=c/v$. Thus, the minimal velocity the particle must have is $v=c/n=2/3c$, since $n=3/2$.

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Ryry013
2018-10-25 12:05:41
My half-BS solution: I tried to figure out which of the answers could reasonably come from 1.5. I thought 1.5 = 3/2, saw 2/3, and then picked it. A guessing strategy for if you know zero Physics.
natec
2013-09-03 10:52:30
when I first read this problem, I thought the particle was entering the glass. The word perpendicular lead me to that assumption. I guess the particle was already in the glass and emitting the Cherenkov Radiation. Since I thought the particle was entering the glass, I thought this was a question about Bremsstrahlung (breaking) radiation.
hjq1990
2012-10-02 23:16:55
Seems that I misunderstood this question at first, wondering the relation of the initial speed of particle and the energy of the emitted light. After reading your notes, I think what matters is that the particle has to be moving in a velocity bigger than the medium speed in order to emit. This speed is a minimal and we need not consider any rigid conservation here.
Tritium
2010-09-19 21:50:13
It seem like this question is improperly formed. The question states only that a fast particle travels through a medium with an index of refraction n = 1.5 and emits light. Why must the particle's MINIMUM speed equal the speed of light in the medium (c/n = 0.67c)? What is to prevent the particle from traveling at half the speed of light, while emitting light that travels within the medium at the speed c/n = 0.67c? This question would make sense if it said the particle emits a CONE of light. Then we would know the particle was giving off Cerenkov radiation, and must be traveling faster than c/n. Am I misunderstanding the question?
dicerandom
2006-10-25 18:19:22
Wait... what? The minimum speed the particle can have is the maximum speed it could possibly obtain? And it emits light because it actually went faster than that?

When I read this I figured light was emitted because the thing knocked into an atom and excited an electron which then dropped down a level or two and radiated.
 Shoshe2006-11-03 15:33:19 Your assumption of "The minimum speed the particle can have is the maximum speed it could possibly obtain" isn't right. The maximum speed the particle could possibly obtain is $c$, the speed of light in a vacuum. Nothing prevents a particle traveling through a medium from going faster than the ${\it speed of light in that medium}\ (v=c/n)$. Cherenkov radiation is emitted when a charged particle's velocity in a medium is greater than the speed of light in that medium. So the charged particle in this question must be going faster than $v=c/n = c/1.5 = \frac{2}{3}c$.
poop
2005-11-23 15:42:12
FYI, this phenomenon is called "Cherenkov Radiation"
 yosun2005-11-23 15:56:55 FYI: Cherenkov Radiation occurs when the particle travels faster than the speed of light in the medium. A particle emits light when it makes transitions from level to level, but since this is a free particle, the only way it can emit light is through Cherenkov Radiation.
 gt20092009-05-20 10:37:56 Why are there so many people named poop on here?
 Albert2009-10-21 08:05:09 Did you not think that there could be just one person named poop who is writing all those posts? Use it, its more than just a hat rack :)
 his dudeness2010-09-04 19:17:48 my personal favorite is "poop loops"

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$