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Total internal reflection question

Discussion in 'MCAT Study Question Q&A' started by dorjiako, Aug 18, 2011.

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  1. dorjiako

    dorjiako
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    The question says: Light from air strikes a translucent plastic material. If the light strikes at an angle of 30 degrees to the material and bends by 15 degress, what is the index of refraction of the plastic?

    Looks pretty easy on the surface but is unable to get their answer 1.22. Any input on how they arrived at their answer is highly welcomed.
     
    #1 dorjiako, Aug 18, 2011
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  2. tmh

    tmh

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    This is sort of weirdly written, but I got the answer.

    First, if light is entering at 30 degrees to the material, it's 60 degrees to the normal. When it says it bends by 15 degrees, I think it means that the angle of refraction is 15 degrees smaller (45 degrees).

    So 1.00sin60 = nsin45
    n = 1.22
     
    #2 tmh, Aug 18, 2011
  3. FutureDoctor503

    FutureDoctor503 A.W.E.S.O.M.-O
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    n1*sin theta1 = n2*sin theta2
    n1 = 1
    sin theta1 = 60 degrees because the questions says light strikes at an angle of 30 degrees to the material (in the equation, we have to use the angle from the normal (90-30=60)
    sin theta2 = 45 degrees because the question says the light bends by 15 degrees (60-15=45 degrees from the normal)
    If you plug in these numbers, you should be able to get the answer. :)
     
    #3 FutureDoctor503, Aug 18, 2011
  4. dorjiako

    dorjiako
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    Thanks again. I was using n1sin 60degree = n2 sin 15 degrees (instead of substracting 15 from 60). Thanks again.
     
    #4 dorjiako, Aug 18, 2011
  5. tmh

    tmh

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    Also, this is not total internal reflection.
     
    #5 tmh, Aug 18, 2011
  6. dorjiako

    dorjiako
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    You are right. However, this problem was asked under Kaplan's mcat high yield total internal reflection section. It was probably while I was tring to figure out how internal reflection applied to the question as I tried to solve it.
     
    #6 dorjiako, Aug 18, 2011
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