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Apparent depth and snells law

  1. Aug 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish? (For water, n = 1.33.)

    2. Relevant equations

    Snell's law: [tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]

    3. The attempt at a solution

    So far I have found the angle of refraction in the water using Snell's law. Since they are asking for "a position almost directly above the fish", I took [tex]\theta_1 = 90[/tex].

    [tex]1= 1.33 sin \theta_2[/tex]

    [tex]\theta_2 = 48.75[/tex]

    But I what else can I do?? I'm really stuck! I need to find the difference between he apparent and the real depth x, 80-x=apparent length. But how? Here's a diagram which shows a similar situation:


    The correct answer is 60 cm.
  2. jcsd
  3. Aug 16, 2010 #2


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    The angle in Snell's law is measured from the normal of the water surface, and it is near zero now.

  4. Aug 16, 2010 #3
    If this angle is zero, then what do I need to do to find the apparent depth?
  5. Aug 16, 2010 #4


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    It is not zero, just small.

  6. Aug 16, 2010 #5
    Oh, okay. How does this help us to determine the apparent depth?
  7. Aug 16, 2010 #6


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    Look at the figure and find some right triangles.

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