2 internal reflection questions

AI Thread Summary
The discussion focuses on two physics problems related to light and geometry. The first problem involves determining the maximum angle for total internal reflection in a pipe with an index of refraction of 1.40, where the user initially misapplies Snell's law. The second problem concerns calculating the depth of an opaque cylindrical tank filled with water, where sunlight stops illuminating the bottom at a 27° angle above the horizon. Participants suggest using right triangle trigonometry to solve the second problem, specifically by drawing a triangle with a base of 2.8 m and applying the tangent function. Overall, the conversation emphasizes the importance of proper setup and understanding of trigonometric relationships in solving these physics problems.
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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