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If we take glass fiber with refraction index n=2 and make it
move with speed c/2 in oppoosite direction?

If we do succeed in stopping the light, where is polarizaion Sx = 0?

2007-05-16 04:42:44 · 2 answers · asked by Alexander 6 in Science & Mathematics Physics

2 answers

c/2 will not be fast enough to do it I think, but you might stop it if you go a little faster.

"Light" in an optically dense medium is partly light and partly charged particles shaking. The honest-to-god light can't be slowed down, but if you can get the medium going fast enough the other way, you can get the apparent speed down to zero.

Bonus problem obviously presents itself: just how fast DO you have to go to stop the "light" in a medium of index n?

Model the "light" as a system where sometimes you have light propagating and sometimes you have an excited electron just sitting there. You can calculate the ratio of the times sitting and moving as a function of n.

When you Lorentz transform the light with a velocity v, nothing happens. When you Lorentz transform the excited electron, it's going backwards at velocity v. So you can calculate a net speed of the transformed "light" that can be set to zero for any n>1 with a sufficiently high v.

And if the polarization direction should not change unless the medium messes with it (as sugars do).

2007-05-16 04:49:15 · answer #1 · answered by Anonymous · 0 0

The light is still travelling down the fiber. I'm not sure what you want to stop it with respect to

2007-05-16 04:49:00 · answer #2 · answered by Gene 7 · 0 0

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