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Water Refraction

  1. Apr 23, 2008 #1
    [SOLVED] Water Refraction

    1. The problem statement, all variables and given/known data

    An opaque stick which is d = 1.93 meters tall stands vertically upright on the bottom of a pool of clear water. Parallel rays of sunlight making an angle of 45.0 degrees with the water?s surface illuminate the pool as shown. What is the length of the shadow the stick casts on the bottom of the pool?



    2. Relevant equations

    [tex]n_{a}[/tex]=1.00
    [tex]n_{b}[/tex]=1.33


    [​IMG]

    3. The attempt at a solution

    So I just took

    [tex]arctan(\frac{1.33}{1.00})=53.06[/tex]

    And I need the angle that will form a triangle with the rod, so

    [tex]90-53.06=36.94[/tex]

    Then, since the shadow forms the base of a triangle, I solved for it:

    [tex]tan(36.94)=\frac{opp}{adj}=\frac{opp}{1.93m}[/tex]

    Adn I got that the shadow equals 1.45 m, which was incorrect. What am I doing wrong?
     
  2. jcsd
  3. Apr 23, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Why?

    Hint: Use Snell's law to find the angle the light makes upon refraction.
     
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