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.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter25: Reflection And Refraction Of Light
Section: Chapter Questions
Problem 7OQ: Light traveling in a medium of index of refraction n1 is incident on another medium having an index...
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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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