Question: A fish is swimming 2m below the surface of a calm lake. At what angle to the horizontal must the fish look to observe a fisherman sitting ~ 50 m away at the edge of a lake ( the refractive index n= 1.33 for water). Solution: In this case the way the fish can see horizontally across the lake is when the angle of incidence is nearly 90 degrees in the air. This corresponds to the light ray inside the water being refracted close to the critical angle for total internal reflection where sin θc = 1/n, giving θ = 48.6 degrees. This is with respect to the normal, so the horizontal angle the fish needs to be looking at is 41.4 degrees.   Here, I understand the critical angle is 48.6 degree. However, what does it mean by "This is with respect to the normal, so the horizontal angle the fish needs to be looking at is 41.4 degrees." ?

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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Question: A fish is swimming 2m below the surface of a calm lake. At what angle to the horizontal must the fish look to observe a fisherman sitting ~ 50 m away at the edge of a lake ( the refractive index n= 1.33 for water).

Solution: In this case the way the fish can see horizontally across the lake is when the angle of incidence is nearly 90 degrees in the air. This corresponds to the light ray inside the water being refracted close to the critical angle for total internal reflection where sin θc = 1/n, giving θ = 48.6 degrees. This is with respect to the normal, so the horizontal angle the fish needs to be looking at is 41.4 degrees.

 

Here, I understand the critical angle is 48.6 degree. However, what does it mean by "This is with respect to the normal, so the horizontal angle the fish needs to be looking at is 41.4 degrees." ?

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