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Snell's law with complex n, interpret complex angle?

  1. Sep 4, 2010 #1
    Reference for this topic:
    https://www.physicsforums.com/showthread.php?t=221973

    This post discussed the topic, but the last comment was exactly the question I am trying to figure out, and no one responded:

    I've been teaching myself optics from Klein and Furtak, and spent a long time on page 75 deriving the laws of reflection and refraction from matching the boundary conditions. I finally got the equations to come out right, but I'm still having trouble interpreting what it says on page 76: that with complex n, n', the equation n sin theta = n' sin theta' still holds, but that theta' is no longer the direction of propagation. That makes NO sense to me, because theta' was *defined* to be the direction of propagation in the new medium. Now, the equation seems to be true because I derived it from the boundary conditions, but unless the real and imaginary parts of the indices of refraction satisfy a strict constraint, (which I have no reason to believe that they do) the angle of propagation turns out to be complex. How am I to interpret a complex angle?

    In other words, if I know n, k (absorption coefficient), theta (angle of incidence), n', and k', what is that formula telling me about theta' ? Which way will the beam go? Is the direction of absorption *different* from the direction of propagation? I thought I had just figured out that the complex vector wave number K was not of the most general form A+Bi, but rather (a+bi)*A. But this seems to be contradicting that. Any ideas?
     
  2. jcsd
  3. Sep 5, 2010 #2

    Andy Resnick

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    My copy of Klein and Furtak is in my office, but I'll take a crack:

    If n and n' are complex, than [itex]\theta[/itex] and [itex]\theta '[/itex] are complex numbers as well. That's not really a problem, since the sine function can still be evaluated:

    [tex] sin(z)=(e^{iz}-e^{-iz})/(2i), [/tex]

    Physically, one interpretation is that as the light propogates and is attenuated, the surfaces of constant phase are no longer coincident with the direction of propagation, similar to evanescent waves.
     
  4. Sep 12, 2010 #3
    Thank you very much! That was the detail I was forgetting. A couple more pages of algebra and I got expressions for the actual directions of propagation and attenuation.
     
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