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Relationship between Polarizing Angle and Angle of internal reflection

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    The critical angle for total internal reflection for a substance is 58.0°. What is the polarizing angle for this substance?


    2. Relevant equations
    Index of refraction: n = 1 / sin(θ[itex]_{c}[/itex])

    and n = tan (θ[itex]_{p}[/itex])


    3. The attempt at a solution So I plugged in the numbers for the desired angles and came to the final formula of:

    Polarizing Angle = tan [itex]^{-1}[/itex] (n)

    which equals to about 49.7°, but I get that as the wrong answer. I really don't know the true relationship between the polarizing angle and critical angle of total internal reflection. Can someone please help me in that area and see where I went wrong with my calculations?
     
  2. jcsd
  3. Mar 8, 2013 #2

    TSny

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    Your answer looks correct to me (assuming that the polarizing angle is for light incident on the material from the outside).
     
  4. Mar 8, 2013 #3

    rude man

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    I agree, without adding the qualification which I don't understand. (Can you get a polarizing angle going from a denser to a less dense medium?).
     
  5. Mar 8, 2013 #4
    Hi! Thanks for your guy's replies to my question. I was lucky to find out that I used the formulas incorrectly, and therefore, got the wrong answer. I was able to find the correct one and properly use it to get the right answer this time around. Thanks once again!
     
  6. Mar 8, 2013 #5

    TSny

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    Yes, you just need to satisfy tan(θ1) = n2/n1. n2 can be greater or less than n1.
     
  7. Mar 8, 2013 #6

    rude man

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    OK.
    Can you figure out what we did wrong? The OP says he got it right after modifying his computations ... ?
     
  8. Mar 8, 2013 #7

    TSny

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    Nope, beats me!
     
  9. Mar 8, 2013 #8
    I was able to find another resource that was similar to my problem and was able to solve it by using snell's law under critical angle conditions:

    n[itex]_{1}[/itex]sinθ[itex]_{c}[/itex] = n[itex]_{2}[/itex]

    and Snell's law under polarization conditions:

    n[itex]_{1}[/itex]sinθ[itex]_{p}[/itex] = n[itex]_{2}[/itex]cosθ[itex]_{p}[/itex]

    which simplified down to tanθ[itex]_{p}[/itex] = [itex]\frac{n_{2}}{n_{1}}[/itex]

    Setting them to equation 1 & 2, and then plugging first equation into 2nd I was able to get the right answer from that.
     
  10. Mar 8, 2013 #9

    TSny

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    Did you use the same value of n1 in both equations and the same value of n2 in both equations? If so, then the question must have wanted you to find the polarization angle θp for the case where the light is starting out inside the material.
     
  11. Mar 9, 2013 #10
    Yes, I used the same values of n1 and n2 for both.
     
  12. Mar 9, 2013 #11

    TSny

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    OK, thanks. That accounts for why rude man and I were getting a different result. Usually when you speak of the polarizing angle for a substance, such as glass, it is assumed that you are considering light incident on the glass (from air, say). But your question was apparently asking for the polarizing angle for light internally reflecting inside the glass.
     
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