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Boat light distance when it strikes bottom.

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    A spotlight on a boat is y = 2.5 m above the water, and the light strikes the water at a point that is x = 9.5 m horizontally displaced from the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.

    http://img529.imageshack.us/img529/6...6100altza6.gif [Broken]

    2. Relevant equations

    3. The attempt at a solution

    I believe this is just a simple geometry problem? Or will it run into trouble.

    Maybe some help to get it going and the idea of what to do.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 3, 2008 #2


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    The physics is Snell's law. Yes, the rest is just geometry.
  4. Mar 3, 2008 #3


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    From my understanding I think you will need to take into consideration the bending (refraction) of light as it passes through the water.
    The equation is:
    N1*sin(A1) = N2*sin(A2)
    The N's are indexes of refraction. I pulled these numbers off a website so I don't know how reliable they are but it says the index of refraction for air is 1.00029 and water is 1.33. But the problem should probably have those numbers somewhere maybe in the chapter if it's from a book.
    So you should just plug in the index for air for N1 and water for N2 and then you will need trigonometry to solve for the angle above water (A1), and then you should be able to find the distance.
  5. Mar 3, 2008 #4
    Ya i got it thanks, used the trig to get the angles above water, then used the snell's law of refraction to get the angle below water between the unknown distance points. Then used more trig to find the unknown distance, then just added the unknown distance to the rest of the known distance, and got an answer.

    about 13.7 meters.

    thanks alot.
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