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You are looking up from underwater in a swimming pool radius of circle?

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data

    You are looking up from under the water in a swimming pool. If you are 2m below the surface, what is the radius of the "hole" at the water surface through which you can see out of the pool?

    2. Relevant equations

    n.sin(theta)1 = n.sin(theta)2

    3. The attempt at a solution


    Could it be assumed that looking straight up through the water? So, sin(theta)1 = sin theta(90) = 1 ?

    I'm assuming a hole would have an area of pi.r2, but don't know how to include that in the former equation (Snell's Law).

    I also know n = c/v
    c = 3x10^8 m/s (speed of light in a vacuum)
    v = velocity

    These values weren't given, but if applicable:
    n air = 1.00
    n water = 1.33
     
  2. jcsd
  3. Apr 17, 2010 #2

    mgb_phys

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    I suspect they are looking for what angle from straight up will the ray from your eye hit the surface and be reflected back into the water (total internal reflection)
     
  4. Apr 18, 2010 #3
    So that deals with critical angle then?
    sin(theta)c = n1/n2

    where n1 = lesser indice
    n2 = greater indice

    Hows does that relate to radius of a circle??
     
  5. Apr 18, 2010 #4
    You can only move your head so far until the only light reaching your eye is light TIRing from the swimming pool's bottom.

    Find what angle you can tilt your head until this happens and find the corresponding horizontal distance moved from a initial position of looking straight up.
     
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