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  1. #1
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    spotlight

    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.
    m

    http://img529.imageshack.us/img529/6...6100altza6.gif

    I believe this is just a trig problem?

    is this correct?
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  2. #2
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    Quote Originally Posted by rcmango View Post
    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.
    m

    http://img529.imageshack.us/img529/6...6100altza6.gif

    I believe this is just a trig problem?

    is this correct? Yes
    1. Calculate \alpha:

    \tan(\alpha)=\frac{9.5}{2.5}~\implies~\alpha\appro  x 75.26^\circ

    I use the refraction index n_{vacuum \rightarrow water} = 1.33

    2. Calculate \beta:

    \frac{\sin(\alpha)}{\sin(\beta)}= 1.33~\implies~\beta \approx 46.65^\circ

    3. Calculate y:

    y = 4 \cdot \tan(46.65^\circ) \approx 4.24\ m

    4. Calculate d:

    d = x+y+ßimplies~d = 13.74\ m
    Attached Thumbnails Attached Thumbnails spotlight-licht_inwasser.gif  
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  3. #3
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    Yes, Thanks for the good explanation and diagram. I 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 the answer.

    very similiar to what you've confirmed to my solution.

    I appreciate all the help.
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