A spotlight on a boat is y = 3.0 m above the water, and the light strikes the wa

Question

A spotlight on a boat is y = 3.0 m above the water, and the light strikes the water at a point that is x = 7.7 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.

Explanation / Answer

The angle between water and / light going in is ARCTAN (3.0/7.7) which is 21.285deg So, from the normal for the incident ray being 68.715deg, use Snell's law (REMEMBER SNELL'S LAW REQUIRES USE OF "NORMAL" RAYS) -> sin I / sin R = refractive index, n n for air/water = 4/3 So, (sin (68.715....)) / (4/3) is sin R sin R = 0.69884... and R = 44.334...deg. The inner angle of a triangle under the water (to the bottom of the water, 4m lower) is then 45.666...deg Calculating using trig again, the "adjacent" is then 4m / (tan 45.666..) = 3.908m So the total distance is from the boat (7.7) + the horizontal light travel under the water (3.908) =11.608m

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