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Finding the Velocity of Light through an Unknown substance

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    32-005.gif
    In the above figure, a ray of light is incident upon the interface between two media having indices of refraction n1 = 1.2 and n2, which is unknown. The known angles are f2 = 60° and f3 = 29.9°. (Note that the angles in the diagram may not be accurate.) What is the speed of light in the second medium?


    2. Relevant equations
    Snell's Law n[itex]_{1}[/itex]sin θ[itex]_{1}[/itex] = n[itex]_{2}[/itex]sin θ[itex]_{2}[/itex]

    And the velocity of light equation:

    n = [itex]\frac{c}{v_{n}}[/itex]

    3. The attempt at a solution
    Since the Angles given aren't the "true" angles that are used in Snell's Law

    I took each angle and subtract them from 90° to get θ[itex]_{2}[/itex] = 30° and θ[itex]_{3}[/itex] = 60.1°

    From there, I calculated that θ[itex]_{1}[/itex] = 29.9°

    I took the variables n[itex]_{1}[/itex] = 1.2 and θ[itex]_{1}[/itex] & θ[itex]_{2}[/itex] and plugged them into the Snell's Formula to get n[itex]_{2}[/itex] = 1.196 as the index of refraction of the unknown substance.

    I then plugged that calculation into the velocity formula to get 2.5 x 10[itex]^{8}[/itex], but I get that as the wrong answer.

    I don't know where I went wrong with this. Thank you for your help!
     
  2. jcsd
  3. Mar 9, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    What does the "law of reflection" tell you about θ1 and θ3.
     
  4. Mar 10, 2013 #3
    The Law's of reflection would state that θ[itex]_{1}[/itex] = θ[itex]_{3}[/itex]. I then got the angle as 60.1 degrees. I was looking at the picture incorrectly and used the wrong angles that were required for snell's Law. I got it now!

    Thanks TSny!
     
    Last edited: Mar 10, 2013
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