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Is critical angle diagram realistic?

  1. Feb 8, 2010 #1
    as per diagrams in textbooks, if incidence angle = critical angle then the ray should travel straight until it reaches the surface of the denser medium and then go tangentially to the boundary something like the following

    http://www.antonine-education.co.uk/Physics%20A%20level/Unit_2/Waves/Refraction/Refract_8.gif [Broken]

    if i now place a mirror vertically against this ray (which is tangentially traveling) , will it exactly trace its path? As per 'principle of reversibility' it should. However, it cannot 'know' where exactly it had bent.

    therefore it seems to me that the behavior of ray with i=90' is somewhat undefined ( like tan 90'). please tell me experts where i am unclear.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 8, 2010 #2


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    It's more complicated than a ray diagram. Waves have to be used.
    This is done in an optics or EM textbook.
  4. Feb 9, 2010 #3
    A situation is usually undefined at a boundary condition.
    That's what the boundary is here; a point where one law ceases to apply and another takes over.
    Actually, there should also be a reflected ray in your diagram; and this ray can be reversed.
  5. Feb 9, 2010 #4
    The principle of reversibilty is preserved. In the derivation of cosec(ic) =µ, light is made incident at 90 degrees, but that does not make sense.To be incident, it has to be actually a little lesser than 90
  6. Feb 9, 2010 #5


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    Isn't it just a 'limiting case' though? Go to 89.999 degrees and it will work ok.
    Stonebridge. Yes, they often miss out the reflected ray which is very sloppy and doesn't help in the understanding.
  7. Feb 9, 2010 #6
    There is something missing from the picture: the intensity of the refracted ray.
    This decreases as the angle goes towards the critical value and it's exactly zero at the critical angle. So the red line tangent to the surface represents a "ray" of zero intensity and if you put a mirror in its path there is actually nothing to reflect back.

    So the diagram is just an aid to calculate the critical angle (set the angle of refraction to 90 degrees and apply Snell's law).
  8. Feb 11, 2010 #7
    And at 89.999 degrees there isn't an issue with the principle of reversibility
  9. Feb 11, 2010 #8


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    I agree.
    But is it worth losing any sleep over? Mine was an Engineer's response and a Mathematician's response would probably be to worry a lot about it.
    But a Physicist should be aware that there is a finite thickness involved to the interface and diffraction and quantum physics and lots of other things which really imply that 90 degrees exactly can be ignored.
  10. Feb 12, 2010 #9
    thanks nasu for the most convincing answer. Thank you all

    Well I looked for a textbook-writer's response. Had the author made a brief note of indefiniteness( also the decreased intensity or the presence of additional refracted ray), why would i post the question?
  11. Feb 12, 2010 #10


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    It can't be the first bit if inadequacy that you've seen in a text book. There are drop offs in so many textbooks and many students get confused. They can get very stroppy when told the book's wrong.
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