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Reflection and Refraction, much index of refraction

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    1. The index of refraction for water is 1.33 and that of glass is 1.50.

    a. What is the critical angle for a glass-water interface?


    b. In which medium is the light ray incident for total internal reflection?


    2. Relevant equations
    nisin[tex]\vartheta[/tex]i=nrsin[tex]\vartheta[/tex]r


    3. The attempt at a solution
    a. I think the answer to a. is 62.46 degrees, but I am not sure.
    b. I think the answer is glass, simply because it is going to be moving from a less dense area to a more dense area.
     
    Last edited: Jan 28, 2009
  2. jcsd
  3. Jan 28, 2009 #2
    Use Snell's law:

    [tex]n_1\sin\theta_1 = n_2\sin\theta_2\ .[/tex]

    Note: It's not additive like you suggested.

    At the critical angle, [tex]\theta_2[/tex] is 90 degrees (the light refracts along the boundary). That is to say, it's sin is 1. We can therefore rearrange for the critical angle:

    [tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)[/tex]

    Now we have two refractive indices, the glass and the water. If you plug them in incorrectly, you're going to end up with trying to find the inverse sin of a value > 1, which isn't possible as this has no solution.
     
  4. Jan 28, 2009 #3
    Quite sorry for the mistake in Snell's Law. I knew what it meant, I just messed up when writing it in the forum.

    Anyway, I appreciate the input, but if you are using
    [tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)[/tex], would it not be possible to put 1.33 (n2) over 1.5 (n1)? This would look like
    [tex]\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{1.33}{1.5} \right)[/tex] and if plugged into a calculator, would return 62.45732485 degrees.
     
  5. Jan 28, 2009 #4
    For this problem I calculated 62.4 degrees, the same thing you got.

    As for B, I put water.
     
  6. Jan 28, 2009 #5
    Yes, that's correct as you've stated.

    It also gives you the answer to your second question as [tex]n_2[/tex] represents the refractive index of the medium that the light travelling in medium [tex]n_1[/tex] is incident on.
     
  7. Jan 28, 2009 #6
    Ah, thank you. I was actually thinking that it would be glass because my book details the fact that "Total internal reflection occurs when light passes from a more optically dense medium to a less optically dense medium at an angle so great that there is no refracted ray." This would mean that glass-water would be a more optically dense to a less optically dense scenario.
     
  8. Jan 28, 2009 #7
    Thank you very much. You have been a big help today, astrorob.
     
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