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Angles and refraction

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data

    A 1.0-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 60 degrees from the normal.

    What is the ray's direction of travel in the glass?

    2. Relevant equations

    n΄sinθ΄ = n¹sinθ¹

    or, if that is confusing,

    n1sinθ1 = n2sinθ2

    3. The attempt at a solution

    The index of refraction for water is 1.33, and the index of refraction for air is 1

    therefore i set up the problem as:

    (1)sin(60) = (1.33)sinθ

    θ should = about 40.628 degrees, but the computer says it is incorrect... where am I going wrong?

    thanks
     
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2

    learningphysics

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    What exactly does the question ask for?
     
  4. Sep 11, 2007 #3
    oh sorry about that, it asks this

    'What is the ray's direction of travel in the glass?'

    i just edited it also
     
  5. Sep 11, 2007 #4

    learningphysics

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    So you did the first part to get the angle across the air/water interface... but now you need to handle the water/glass interface... actually I'm thinking since:

    n1sin(theta1) = n2sin(theta2)... n2sin(theta2) = n3sin(theta3)...

    You can simply ignore the water part...

    n1sin(theta1) = n3sin(theta3)

    unless I'm missing something...
     
  6. Sep 11, 2007 #5
    Lol!!!!!!!!!!!

    LOL oh i see!!!!!!!! hahaha i was thinking that the water itself was the reflective surface and ignored the word 'glass'!! haha this should work
     
  7. Sep 11, 2007 #6
    yay!

    yes you were correct, it works!!

    using the other index as 1.5 (ignoring the water entirely), i get 1sin(60)=1.5sin(x), and then i just had to solver for x, which gave me the correct value of ~35.3 deg

    thanks!
     
  8. Sep 11, 2007 #7

    learningphysics

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    no prob! you're welcome!
     
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