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Refraction problem

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A penny lies on the bottom of a pool 0.75m from the edge of the pool and 1.5m below the surface. A flashlight beam is shone over the edge of the pool to illuminate the penny. At what angle to the pool deck should the flashlight be aimed?

    2. Relevant equations

    ??

    3. The attempt at a solution
    ??
     
  2. jcsd
  3. Apr 25, 2009 #2

    LowlyPion

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    You're going to need the index of refraction aren't you?

    Water is 1.33. Is that what you are supposed to use?

    If you don't know it already, then maybe better look at Snell's Law.
     
  4. Apr 25, 2009 #3

    tiny-tim

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    Hi roxxyroxx! :wink:
    (I assume it means that the light enters the pool exactly at the edge?)

    Use the sin/sin rule …

    show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
     
  5. Apr 25, 2009 #4
    ok well i did sin(0.75/1.5) --> sin^-1(0.75/1.5) giving me 30 degrees so:
    (1.00)(sin 30) = (1.33)(sin theta)
    sin theta = 0.3759...
    theta = 22 degrees
    but the answer should be 54 degrees
     
  6. Apr 25, 2009 #5

    LowlyPion

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    Isn't the angle you want from entering the water given by

    tan-1(.75/1.5) = θ
     
  7. Apr 25, 2009 #6
    ok so tan^-1(0.75/1.5) = 26.565...
    (1.33)(sin 26.565...) = (1.00)(sin theta)
    sin theta = 0.595
    theta = 37 degrees
    but the answer is 54 ..
     
  8. Apr 25, 2009 #7

    rl.bhat

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    This angle is the angle which the light makes with the normal to the pool. But the answer required is the angle to the pool deck.
    So the required angle is 90 - theta
     
  9. Apr 25, 2009 #8
    ookk thank you! >.<
     
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