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Snell's Law x2: Double-checking trig?

  1. Aug 1, 2007 #1


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    1. The problem statement, all variables and given/known data

    In this drawing:


    Index of refraction for glass: 1.52
    Index of refraction for surrounding carbon disulfide: 1.63
    Incident angle at point A: 43.0°
    At what angle does the ray leave the glass at point B?

    2. Relevant equations

    [tex]n_{1}sin\theta_{1} = n_{2}sin\theta_{2}[/tex] (twice)

    3. The attempt at a solution

    Please double-check my conceptual understanding of this.

    Part 1, as light travels through the carbon disulfide into the glass:

    [tex]1.63sin43 = 1.52sin\theta_{2}[/tex]

    [tex]\theta_{2} = sin^{-1}\frac{1.63sin43}{1.52}[/tex]

    So Θ within the glass is 47°.

    Drawn out and making a triangle with the upper left hand corner of the glass, a triangle forms with point B as one of its own points, meaning that the angle with a normal line at that point would be 180 - (90 + 47) = 43°.

    Part 2, as light travels through the glass back into the carbon disulfide:

    [tex]1.52sin43 = 1.63sin\theta_{2}[/tex]

    [tex]\theta_{2} = sin^{-1}\frac{1.52sin43}{1.63}[/tex]

    Which yields 39.4923°.

    Am I going about this correctly?
  2. jcsd
  3. Aug 1, 2007 #2


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    Homework Helper

    Looks OK to me.
  4. Aug 1, 2007 #3


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    Thanks for taking a peek - turned out to be correct.
  5. Aug 1, 2007 #4


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    Science Advisor
    Homework Helper
    Gold Member

    Looks correct to me too.
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