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Optics question (Snell's law)

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data

    An underwater scuba diver sees the Sun at an apparent angle of 42.0° above the horizontal. What is the actual elevation angle of the Sun above the horizontal? (Use 1.333 for the index of refraction of water.)

    The correct answer must be 7.86° above the horizon.

    2. Relevant equations

    Snell's law

    3. The attempt at a solution

    [tex]sin \theta = 1.333 \ sin (42)[/tex]

    [tex]sin^{-1} (0.89) = 62.87[/tex]

    Now to find the actual elevation angle [tex]90- 62.87= 27.13[/tex]

    But this is wrong since the correct answer must be 7.86°. What's wrong?
     
  2. jcsd
  3. Aug 8, 2011 #2

    Doc Al

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    Staff: Mentor

    So what's the angle of incidence? (You're using the wrong angles in Snell's law.)
     
  4. Aug 9, 2011 #3
    Okay, I see. Here's what I did now:

    [tex]sin(42) = 1.333 \ sin (\theta)[/tex]

    [tex]\theta = sin^{-1} (\frac{sin(42)}{1.333}) = 30.13[/tex]

    90 - 30.13=59.86

    Why is it still not right? :frown: (it should be 7.86°)
     
  5. Aug 9, 2011 #4

    Doc Al

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    Staff: Mentor

    No. You were closer before.

    Please draw a diagram for yourself. The angle of incidence is with respect to the normal, not the horizontal.
     
  6. Aug 9, 2011 #5
    Yes, here's a diagram:

    [PLAIN]http://img534.imageshack.us/img534/909/diverh.jpg [Broken]

    What I've marked in red is the elevation angle of the Sun above the horizontal I'm required to find.

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

    [tex]sin \theta_1 = 1.333 sin 42[/tex]

    [tex]\theta_1 = 63.11[/tex]

    Then to get the marked angle 90-63.11=26.88. But why is this wrong?
     
    Last edited by a moderator: May 5, 2017
  7. Aug 9, 2011 #6

    Doc Al

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    Staff: Mentor

    Good.

    Good.

    Here's your mistake. 42 degrees is the apparent angle with the horizon, not the angle of refraction (θ2). (The angles of incidence and refraction are with respect to the normal, not the horizontal.)
     
    Last edited by a moderator: May 5, 2017
  8. Aug 9, 2011 #7
    Ohhh... So [tex]\theta_1 = sin^{-1} (1.33 \ sin (90-42))= 81.3[/tex], and 90-81.3=8.7. Thank you very much for the explanation. :)
     
  9. Aug 9, 2011 #8

    Doc Al

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    Staff: Mentor

    Now you've got it. :approve:
     
  10. Aug 9, 2011 #9

    PeterO

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    Homework Helper

    A general suggestion/guide:
    Application of Snell's Law is a fairly straight forward activity. Once you have done the early standard questions, the only way spice up the application is to not tell you the angle of incidence/refraction - at least not directly. Instead you are given some other angle(s) which you can use to calculate the angle you need for the formula.
    Always read a question carefully to check how to interpret which angle is which.
     
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