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Search light !

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A search light is being used at the edge of a tank to find a lost key at the bottom of the tank. The tank is filled to the top with a fluid for which the index of refraction is 1.68. The key is 8.176 m below the surface and 6.84 m from the edge of the tank. At what angle to the normal will the searchlight have to point in order to locate the key if the light enters the water 1 m from the edge of the tank?

    2. Relevant equations

    snell's law
    n1 sin theta1 = n1 sin theta2


    3. The attempt at a solution

    inverse sin of 6.84/8.176

    inverse sign of 7.84/8.176

    and then i added those two together..

    Im totally lost and dont understand what to do . Any help will be much appreciated =]
     
  2. jcsd
  3. Apr 2, 2009 #2

    LowlyPion

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    Draw a picture first off.

    The θ2 is given by tan-1 of (6.84 - 1)/8.176

    The rest should be easy peasy.
     
  4. Apr 2, 2009 #3
    so everything i did was ok except for I use inverse of tan and not sin to find the angles?
     
  5. Apr 2, 2009 #4

    LowlyPion

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    Not exactly. You failed to account for the light entering the water at 1m.

    In the drawing then your net x displacement over the y depth describes θ with respect to the vertical.

    To determine the θ1 of course you use sin-1
     
  6. Apr 2, 2009 #5
    im lost. sorry . so what is wrong with the way i was calculating theta 1?
     
  7. Apr 2, 2009 #6

    LowlyPion

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    You use Snell's law. So knowing θ2 from the tan-1 and ...

    n1*sinθ1 = n2*sinθ 2

    1*sinθ1 = 1.68*sinθ2

    To determine θ1 you take the sin-1.

    Isn't that what you were doing before?
     
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