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When a man stands near the edge of an empty drainage ditch of depth 2.80 m, he can barely see the boundary between the opposite wall and bottom of the ditch as in Figure P22.47a. The distance from his eyes to the ground is 1.85 m. (a) What is the horizontal distance $d$ from the man to the edge of the drainage ditch? (b) After the drainage ditch is filled with water as in Figure P22.47b, what is the maximum distance $x$ the man can stand from the edge and still see the same boundary?

a. $1.07 \mathrm{m}$

b. 1.65 m

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Rutgers, The State University of New Jersey

Numerade Educator

University of Washington

Hope College

in this problem. On the topic of reflection and refraction of light, a man is standing near the edge of a ditch which has a depth of 2.8 m, and he can barely see the boundary between the opposite wall and the bottom of the ditch. Now the distance from his eyes to the ground is given, and we want to calculate the horizontal distance D from the man to the edge of the ditch and after the ditch is filled with water, we want to find the maximum distance X that the men can stand from the edge to still be able to see the same boundary. Now our diagram on the left is the first situation, and given that the angle data in the figure is that he degrees the maximum distance, the observer can be from the pool and continue to see the lower edge on the opposite side of the pool is given as follows. You can see that D is equal to one 0.85 m times the 10 of 30 degrees which gives us this distance D that he needs to be standing from the edge of the ditch to be 1.7 meters. Now for Part B, the pool is completely filled with water, and the light ray coming to the observer's eye from the lower opposite edge of the pool would reflect at the surface of the water, as we can see as it moves from water to air. And the angle of refraction, which we've labeled as Phi, is equal to using stars Law, the ox sign of the refractive index of water in water times the sign of 30 degrees divided by the defective index of air. And so if we calculate this angle phi by substituting the values, we get this to be the oxide of 1.333 which is the effective index of water and the sign of 30 degrees divided by the refractive index of air, which is just one. We get this angle of refraction from water to air to be 41.8 degrees, so the maximum distance the observer can now be from the pool to still see the same boundary we calculate as follows. The prime is one 0.85 m times the 10 of this angle phi, and this is one 0.85 m, times the 10 of 41.8 degrees which gives us the new distance of one 0.65 meters.

University of Kwazulu-Natal