law of reciprocity at critical angle

[geezus05]

I know it is a pretty elementary problem but i was just wondering if anyone could tell me if the law of reciprocity is still in order during refraction at the critical angle. It seems as if you had a light source at the other end of the ray and tried to go the opposite path, then it would not refract the same way, but rather continue to travel down the boundary of the two media.Plus there is also distance to take into consideration. If the incident angle was at the critical angle in one direction it would travel indefinitely but if it was reversed it would have to travel through the boundary without regard for distance and then exit somehow at what would be the incident/critical angle if it was reversed. Which seems improbable but wouldn't that violate the law of reciprocity if it didn't ? I've been trying to figure this out all day and can't seem to figure this one out. I'm not sure if this is the best area to post it under but any help is appreciated.

 

[cev]

All light at angles greater than *or equal to* the critical angle is completely reflected, Therefore there is no problem with reciprocity.

Best,
Curtis

 

[geezus05]

well yes but when it is at an angle equal to the critical angle then it is reflected 90 degrees to the normal and will travel indefinitely down the boundary. It seems as if you could not reverse this path, which would violate the law of reciprocity.

 

[IRstuff]

The law of reciprocity does not apply to nonlinear phenomenon, of which there are plenty.

Faraday isolators:

http://usna.edu/Users/physics/mungan/Scholarship/FaradayIsolators.pdf

Optical parametric oscillators:

http://en.wikipedia.org/wiki/Optical_parametric_oscillator

TTFN

FAQ731-376