\"A coin lies on the bottom of a pool under 1.45 meters of water and 0.75 meters

Question

"A coin lies on the bottom of a pool under 1.45 meters of water and 0.75 meters from the side wall. If a light beam is incident on the water surface at the wall, at what angle relative to the wall must the beam be directed so that it will illuminate the coin?" I've got an idea to use snells law, and I know my first angle is 90 degrees. I figure you take the tangent inverse of 1.45/0.75 to get 65.62 degrees, but do I subtract this value from 90 degrees to find the angle? Help please! I will rate! Thanks!!

Explanation / Answer

The angle the beam makes with the wall, in the water is ; Tan() = .9/1.5 = .6 so () = arctan(.6) = 31 deg But this beam was refracted by the water. Use Snell's Law to figure the incident angle in air, before entering the water; (1)Sin() = (1.33)Sin(31) Solve this for the angle () and you're done. The angle is w.r.t. the vertical ,which is the wall, so its the angle you want.

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