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Critical angle = brewster angle?

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data
    when the critical angle=brewster angle, what is the refractive index, if incident n is air=1.

    arcsin(1/n)=arctan(n)=>n=1.272....????

    how is this? i tried differentiating both side and solving but i just get into a big mess, can anyone help me if i am missing something?
     
    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 4, 2009 #2
    ive tried pythagoras, and trig idendenties, but all i come up with is x=x??
     
  4. Jun 4, 2009 #3

    Cyosis

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    Homework Helper

    If you want to use identities you can take the sin on both sides and then use [itex]\cos^2x+\sin^2x=1,\;1+\tan^2x=\sec^2x[/itex] to simplify the right hand side.
     
  5. Jun 4, 2009 #4
    ok thanks i think i got it now... just confused about the algebra
     
  6. Jun 4, 2009 #5
    all solved, thanks very much
     
  7. Feb 27, 2011 #6
    I'm still at a loss as to how n = 1.272.

    so getting to:
    1/n = sin(arctan(n))

    then squaring and adding cos^2 gets:

    1/n^2 - cos^2(arctan(n)) = 1

    but any way I rearrange the above formula to get 1/cos^2 for sec^2 ends up resulting in n=1?
     
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