2 internal reflection questions

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SUMMARY

The discussion focuses on two physics problems involving total internal reflection and trigonometry. The first problem requires determining the maximum angle for total internal reflection in a pipe with an index of refraction of 1.40, while the second problem involves calculating the depth of a cylindrical tank filled with water when sunlight at a 27.0° angle fails to illuminate the bottom. The correct approach for the first problem involves using Snell's Law, specifically n1sin(theta) = n2sin(90), and for the second problem, applying right triangle trigonometry with a base of 2.80 m and an angle of 27 degrees.

PREREQUISITES
  • Understanding of Snell's Law of Refraction
  • Knowledge of total internal reflection principles
  • Basic trigonometry, specifically right triangle relationships
  • Familiarity with the concept of angles of incidence and refraction
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  • Study Snell's Law and its applications in optics
  • Learn about total internal reflection and its conditions
  • Review right triangle trigonometry and its practical applications
  • Explore problems involving angles of elevation and their geometric interpretations
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Students in physics, educators teaching optics and trigonometry, and anyone looking to enhance their problem-solving skills in geometric applications of light and angles.

Pepsi24chevy
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One of my problems reads,
1. Determine the maximum angle for which the light rays incident on the end of the pipe in Figure P35.38 are subject to total internal reflection along the walls of the pipe. Assume that the pipe has an index of refraction of 1.40 and the outside medium is air.

http://www.webassign.net/pse/p35-38.gif

Now i thought using snells law of refraction would give me the answer by, n1sin(theta)= n2sin(90) and solve for the other sin but this isn't correct. I know that the index of refraction of the air is 1.


My other problem goes like, An opaque cylindrical tank with an open top has a diameter of 2.80 m and is completely filled with water. When the afternoon Sun reaches an angle of 27.0° above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank?


For this one I have no idea of how to set it up. My book doesn't even have a section that factors in any type of length measurement into this chapter.:frown:
 
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You sound like you're on the right track with the first problem.

As for the second, you're given a side and an angle. Draw a picture and think about right triangle trig.
 
so for the second one, i am going to get a triangle with a base of 2.8 and a angle of 27 degrees. So I do tan(27)*2.8 and the H of the triangle? I am sure i am probably just setting this up wrong, but I dont' know.
 

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