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Determine the depth of a swimming pool filled with water

  1. Oct 13, 2009 #1
    We wish to determine the depth of a swimming pool filled with water without getting wet. We measure the width of the pool, which is 5.50m. We then note that the bottom edge is just visible when we stand and look at an angle of 14.0 degrees above the horizontal line.

    a)Calculate the depth of the pool.

    b)If the pool were filled with a different clear liquid, with index of refraction n=1.65. Would you be able to see the bottom of the pool from the side now?

    c)The bottom of the pool is made of a clear solid material. What would be the maximum value of index of refraction of thisclear floor so that total internal reflection occurs? (It is filled with the n=1.65 liquid).
     
  2. jcsd
  3. Oct 13, 2009 #2

    rl.bhat

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    Hi Sammiebaby, welcome to PF.
    Please go through the rules of PF.
    You have to your attempts before you seek our help.
    At least find out the relevant equation required to solve the given problem.
    Can you draw the ray diagram? can you state the Snell's law? What is the critical angle?
     
  4. Oct 13, 2009 #3
    Sin[tex]\theta[/tex]1n1=Sin[tex]\theta[/tex]2n2

    I got the angle of the index of refraction.

    Sin[tex]\theta[/tex]2=(1.00/1.33)Sin(14')=.182
    [tex]\theta[/tex]2=Sin^-1(.182)
    [tex]\theta[/tex]2=10.5'

    Im not sure how to find the depth tho.
     
  5. Oct 13, 2009 #4

    rl.bhat

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    The given angle is above the horizon. So it is not the angle of incidence.
    Draw the ray diagram and identify the different angles.
    Find the angle of refraction. If you see the diagram, you can see that tanθ2 = width/depth.
     
  6. Oct 13, 2009 #5
    So,

    tan(10.5')=5.50m/x

    x=5.50m/tan(10.5')

    x=29.7m

    So the depth is 29.7m? Im confused because that seems large.
     
  7. Oct 13, 2009 #6

    rl.bhat

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    Your angle of refraction is wrong. The angle of incidence is (90 - 14) degrees.
     
  8. Oct 13, 2009 #7
    Oh wow, Im having a slow night sorry!
    Sooo lets try this again:

    Sin[tex]\theta[/tex]2=(1.00/1.33)Sin(76)=.730
    [tex]\theta[/tex]2=Sin^-1(.730)
    [tex]\theta[/tex]2=46.9'

    Sooo,

    tan(46.9')=(5.50m)/depth

    Depth=(5.50m)/tan(46.9')

    Depth=5.15m

    That sounds better. Did I mess up again or did I get it right?
     
  9. Oct 13, 2009 #8

    rl.bhat

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    Yes. You are right.
     
  10. Oct 13, 2009 #9
    Thank you so much! =]
     
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