Question

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, h = 1.3 m above the water level, onto the surface of the water at a point L = 2.7 m from his foot at the edge of the pool Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 2.1 m deep?

Answer 1

First we need to find what angle the light hits the surface of the water. To do this we construct a right triangle with 1.2 m as its height and 3 m as its length. We find the angle θa to be equal to arctan(1.2/3) = .381 radians.

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Next we use the equation for refraction, using the index of refraction (Na) of air to be about 1.000 and index of refraction of water (Nw) to be about 1.333.

sin(θa) x Na = sin(θw) x Nw

sin(.381) x 1.000 = sin(θw) x 1.333

θw = .279 Radians from the perpendicular.

Now we construct yet another right triangle using this angle as well as the 2.1 meters given to us in the problem, which we'll use as the height.

tan(.279) = 2.1/d

d = 2.1 / tan(.770) = 7.331 m from where the light entered the water originally. So the distance from the edge of the pool is 3m + 7.331m = 10.331 m

Answer 2

Yes, good reply but the height of the flashlight is h = 1.3 m my answer is 5.05 meters from the edge of the pool.
incident angle = 64.29 degrees
refraction angle = 48.33 degrees
total distance from the edge = 2.7+2.1*tan(48.33*pi/180)= 5.05 meters