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Acceptance Angle in Optical fibre

  1. Apr 28, 2013 #1
    According to the derivation of acceptance angle the first step is,
    The condition for Total Internal reflection(TIR) is sin(i)≥n2/n1,
    where i is angle of incidence in the core,
    n2= refractive index of the cladding and n1=refractive index of the core.
    sin(i(c))=n2/n1, where i(c) is the critical angle.
    For TIR to take place the angle HAS TO BE ONLY GREATER than critical angle.
    But the condition above contradicts the statement.
    which one is true>
     
  2. jcsd
  3. Apr 28, 2013 #2
    That's a very good point.

    IMHO the critical angle is a boundary between two states and should not be included in either state.
    Kind of like cutting a piece of paper along a line. Neither piece is part of the cut line.

    The contradiction is present even in the text describing the formula and the formula itself here:


    Wikipedia link
     
    Last edited: Apr 28, 2013
  4. May 8, 2013 #3
    So basically you assume here that TIR is greater than or equal to critical angle?
     
  5. May 8, 2013 #4

    sophiecentaur

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    I think you're letting 'pure maths' get in the way of common sense Engineering here. Engineering doesn't tend to distinguish between what, I seem to remember, are referred to as 'closed intervals' and 'open intervals'.
     
  6. May 8, 2013 #5
    No,I would state that TIR occurs at angles greater than the critical angle.
    For beams incident at critical angle, since no beam can be perfectly parallel, the portion at greater than critical angle is totally reflected. The portion at less than the critical angle is partially refracted. The critical angle is the 0 width boundary between the 2 portions.
     
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