A spotlight on a boat is y = 2.3 m above the water, and the light strikes the wa

Question

A spotlight on a boat is y = 2.3 m above the water, and the light strikes the water at a point that is x = 6.5 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

A spotlight on a boat is y = 2.3 m above the water, and the light strikes the water at a point that is x = 6.5 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

Explanation / Answer

Let the index of refraction in air be 1, in water 1.333. Then sin air = 1.333 sin water, so sin water = sin air/1.333.

We know sin air = 6.5/(2.32 + 6.52)½ = 0.942722.

Then sin water = 0.942722/1.333 = 0.707219. (this is almost a 45 degree angle)

Let s be the horizontal distance from the point where the light strikes the water to where it strikes the bottom; then sin water = s/(s2 + 4.02)½ which, as we solve for s becomes:

(0.707219)2(s2 + 4.02) = s2

0.500158s2 - s2 = -8.002528

s2 = (-8.002528)/(-0.499842) = 16.01012

s = 4.001264 and since d = s + x = s + 6.5,

d = 10.5013 m

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