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Snell's law, critical angle & refraction

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Given a three layer model


    Assume a ray goes through layer 1 and hits the interface between layer 1 and layer 2. What is the critical angle?

    2. Relevant equations

    Snells law
    [tex]\frac{\sin \theta_1}{\sin \theta_2}=\frac{v_1}{v_2}[/tex]

    3. The attempt at a solution

    To find the critical angle, you normally take [tex]\sin \theta_c = \frac{v_1}{v_2}=\frac{1.5}{1.3}[/tex]. But in this case that means I have to take [tex]\sin^{-1}[/tex] of a value that is over 1! How do I solve this?
  2. jcsd
  3. May 14, 2010 #2


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    Homework Helper

    According to Snell's law
    n1sin(θ1) = n2sin(θ2)

    If θ1 is θc, then θ2 = 90 degrees.

    So sin(θc) = n2/n1
  4. May 14, 2010 #3
    When I look up Snell's law on Wikipedia it says

    \frac{\sin \theta_1}{\sin \theta_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}

    Why does the subscript change in the [tex]n_n[/tex] ? Isnt [tex]v_1=n_1[/tex] and [tex]v_2=n_2[/tex]?

    Thanks for answering
  5. May 15, 2010 #4


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    Homework Helper

    According to the definition,
    refractive index n = c/v. where c is the velocity of light in vacuum and v is the velocity in the refracting medium.
    So v = c/n
    Or v1 = c/n1 and v2 = c/n2
    then v1/v2 = .....?
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