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Optics, Snell's law

  1. Mar 11, 2015 #1
    1. The problem statement, all variables and given/known data
    MGxXqsJ.png
    A stone lies to the very edge at the bottom of a pool. The pool is filled with water to the top. The person standing three meters away from the pool is 1 meter tall and he can see exactly the half of the stone. Calculate the depth of the pool.


    2. Relevant equations
    Snell's law: n * sin(x) = n2 * sin(y)

    3. The attempt at a solution
    QHBwIM4.png
    Okay.. Since I can see half of the stone, the angle should be 45 degrees, right? I should be able to calculate x degrees and therefore the bottom of the triangle (where he stands). With the bottom, I can calculate the bottom line of the triangle in the pool and then the depth. So I do that.

    sin(45) * 1.33 = 1.00 * sin(x)
    1.33 is n for water. 1 is n for air. x = 70.05 degrees.

    So the inside of the triangle is approx 20 degrees. I use tan.

    tan 70 = 1/adjacent => 1/(tan 70) = adjacent => 0.36397. So the bottom of that triangle is 0.36397 meters, but that doesn't make sense, it should be atleast 3 meters because he is standing that far away from the pool. What am I doing wrong?
     
  2. jcsd
  3. Mar 11, 2015 #2

    Bystander

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    Angles of incidence and refraction in Snell's law are measured from the normal to the refracting surface.
     
  4. Mar 11, 2015 #3
    Hello! Thank you for your answer. I did draw a normal perpendicular to the water's surface. But if is measured from the normal to the refracting surface aren't both angles 90 degrees then?
     
  5. Mar 11, 2015 #4

    SammyS

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    Seeing half the stone has nothing to do with a 45° angle.

    Basically, a ray from the center of the stone should just graze the edge of the pool as it exits the water, then continue on to the eye.
     

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    Last edited: Mar 11, 2015
  6. Mar 11, 2015 #5
    Hello and thank you for your answer.

    But if the ray comes from the very center of the stone and it is truly in the edge of the pool, shouldn't it be "like" 45 degrees from the pool's wall down to the center of the stone?

    And if the ray just grazes the edge of the pool, I then get this: QeeP5gu.png

    which would make by original attempt not work. Basically the only thing I need now is either x or y (degrees) and with it I can calculate the side (depth). But using Snell's law, now I only know the n's, not the angles. So I have two unknowns. How should I proceed?
     
  7. Mar 11, 2015 #6

    SammyS

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    Do you know trigonometry?

    You can get sin(z) by using the Pythagorean theorem. Otherwise, use a scientific calculator to find the angle z

    Then use Snell's Law , etc.
     
  8. Mar 11, 2015 #7
    Hello, yes I forgot that! I calculated z degrees and then got x to be 45 degrees (making y also 45). So the depth is the same as the base of the triangle, approx 4 meters.
     
  9. Mar 11, 2015 #8

    SammyS

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    Yes, approximately 45° , so approximately 4 meters deep.
     
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