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Optics; Deriving the index of refraction from Snell's law

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data

    There is a diagram in the problem statement so here is a link to the image of the problem:
    http://imgur.com/KDrRsyO

    2. Relevant equations

    Snell's Law:

    [itex]n_{1} * sin(\theta_{1}) = n_{2} * sin(\theta_{2})[/itex]

    3. The attempt at a solution

    My attempt using Snell's law fails because I can't see how it is possible that n ever becomes squared. It looks like either some sort of vector addition or some use of Pythagorean theorem but I haven't been able to determine how to use those ideas to solve the problem.

    I can see some use of special triangles given the angles are 60 degrees, but it's a bit overwhelming and my trig skills are a bit rusty.

    Thanks.
     
  2. jcsd
  3. Jan 26, 2013 #2

    TSny

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    Start by writing out Snell's law for each of the two refractions.
     
  4. Jan 26, 2013 #3

    tiny-tim

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    hi steve233! :smile:

    hint: θ2 + θ3 = … ? :wink:
     
  5. Jan 26, 2013 #4
    Hmm... So here is another attempt:

    1st refraction:
    [itex] n_{1} * sin(\theta_{1}) = n * sin(\theta_{2}) [/itex]

    2nd refraction:
    [itex] n * sin(\theta_{3}) = n_{1} * sin(\theta) [/itex]

    Here I assume that the refractive index is the same when the light is outside of the prism and inside of the prism.

    I think I'm wrong here but, [itex] \theta_{2} + \theta_{3} = 60^{\circ} [/itex].
    I think this is wrong because it seems like [itex] 90^{\circ} - \theta_{2} [/itex] = [itex] 60^{\circ} \Longrightarrow \theta_{2} = 30^{\circ}[/itex].

    I'm still stuck on how things become squared though...
     
    Last edited: Jan 26, 2013
  6. Jan 26, 2013 #5

    TSny

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    Ok. You can assume that outside the prism you just have air where ##n_1 ≈ 1.00##. You also know the value of ##\theta_1##. So, you have three unknowns: ##\theta_2##, ##\theta_3## and ##n##. [Edit: That is, you need to eliminate ##\theta_2## and ##\theta_3## so that you can express ##n## in terms of ##\theta##.]

    I'm not following you here. ##n_1## is for air and ##n## is for the prism material.

    Not sure how you got that, but I think it's right! So, that's your third equation.
    I don't see how you are deducing that. In fact, I don't think there is any way to find the values of ##\theta_2## and ##\theta_3## individually since they would depend on ##n##.

    The square will show up in doing the algebra of solving for ##n## in terms of ##\theta##.
     
    Last edited: Jan 26, 2013
  7. Jan 26, 2013 #6
    Ah ok, very helpful. Thanks.

    [itex] n_{1} [/itex] is indeed the refractive index of air, I should have mentioned that.

    So the equations become:

    (1) [itex] \sqrt{3} / 2 = n * sin(\theta_{2}) [/itex]
    (2) [itex] n * sin(\theta_{3}) = sin(\theta) [/itex]
    (3) [itex] \theta_{2} + \theta_{3} = 60^{\circ} [/itex]

    [itex] \theta_{3} = 60^{\circ} - \theta_{2} [/itex]
    [itex] n * sin(\theta_{3}) = n * sin(60^{\circ} - \theta_{2}) = n * (sin(60)cos(\theta_{2}) - cos(60)sin(\theta_{2})) [/itex]

    [itex] \theta_{2} = arcsin(\sqrt{3}/(2 * n)) [/itex]

    So,
    [itex] n * (sin(60)cos(\theta_{2}) - cos(60)sin(\theta_{2})) [/itex]
    [itex] n * ((\sqrt{3} / 2) * cos(arcsin(\sqrt{3}/(2 * n)) - (1/2) * sin(arcsin(\sqrt{3}/(2 * n))) = sin(\theta)[/itex]

    This seems a little complicated... I know there is a simple replacement for cos(arcsin x) but I don't know anything that simply replaces sin(arcsin x). Am I on the wrong track here?

    Thanks again for the help.
     
  8. Jan 26, 2013 #7

    tiny-tim

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    hi steve233! :smile:

    sin(arcsin(x)) = x

    cos(arcsin(x)) = √(1-x2) :wink:
     
  9. Jan 26, 2013 #8
    Ah yes... more evidence that I need to brush up on my trig :)
    I'll give this a shot and post the update soon.

    Thanks!

    Edit: Success! Thanks for all of the help TSny and tiny-tim.
    I'm a bit new to these forums, is there some way to "upvote"?
     
    Last edited: Jan 26, 2013
  10. Jan 27, 2013 #9

    tiny-tim

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    (just got up :zzz:)

    not needed … saying "thanks" is enough! :smile:
     
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