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Total internal reflection question

  1. Dec 8, 2006 #1
    How do the angle of incidence relate mathematically with the angle of reflection after the critical angle, that is, in a TIR?
    Say we have a person standing at the bottom of a swimming pool, if we take the n for water as 1.33 and for air as 1, we get the critical angle of 48.75 degrees. When the person sees the surface of the water that's further away, the angle of incidence (eye to surface) increases, but how does the other angle change?
    The equation n1 sin(theta1) = n2 sin(theta2) doesn't seem to work anymore when the criticle angle is passed...
     
  2. jcsd
  3. Dec 8, 2006 #2

    cepheid

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    It's not clear what your question is. The last sentence you wrote is the whole point of T.I.R. Indeed, the equation breaks down because sine can't be > 1. Which means that there is NO transmitted ray (hence total internal reflection). If you have already calculated the criticAL angle, then the only part of the problem that remains is the qualitative question. What happens when the person looks a the surface of the water at a point far away from him (in the direction // to the surface) rather than at the point directly above him?
     
  4. Dec 8, 2006 #3
    Yeah I know, but what I want is a relation between the angle the person is looking by to the surface and the angle of the reflecting ray. I don't know if I'm making myself clear...

    Once the critical angle is passed, how do you relate the angles???
     
  5. Dec 9, 2006 #4

    cepheid

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    Oh, the reflecting ray? The totally internally reflected ray still obeys the law of reflection:

    angle of incidence = angle of reflection

    Look at a proof of this law, and you'll that there's no reason why it wouldn't also be true at an interface where a ray is (in general) partially reflected and partially transmitted.
     
  6. Dec 9, 2006 #5
    Thanks, I was thinking that but somehow something didn't make sense.
    Thank you.
     
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