A fisherman spots a fish underneath the water. It appears that the fish is d0 = 0.51 m under the water surface at an angle of θa = 62 degrees with respect to the normal to the surface of the water. The index of refraction of water is nw = 1.3 and the index of refraction of air is na = 1. Solve for the numerical value of θw in degrees. Now assume that the real position of the fish is directly below the apparent position, as shown in the figure. Express the real depth of the fish, d, in terms of L and tanθw.
A fisherman spots a fish underneath the water. It appears that the fish is d0 = 0.51 m under the water surface at an angle of θa = 62 degrees with respect to the normal to the surface of the water. The index of refraction of water is nw = 1.3 and the index of refraction of air is na = 1. Solve for the numerical value of θw in degrees. Now assume that the real position of the fish is directly below the apparent position, as shown in the figure. Express the real depth of the fish, d, in terms of L and tanθw.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A fisherman spots a fish underneath the water. It appears that the fish is d0 = 0.51 m under the water surface at an angle of θa = 62 degrees with respect to the normal to the surface of the water. The index of refraction of water is nw = 1.3 and the index of refraction of air is na = 1.
Solve for the numerical value of θw in degrees.
Now assume that the real position of the fish is directly below the apparent position, as shown in the figure. Express the real depth of the fish, d, in terms of L and tanθw.
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