Total internal reflection questions

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SUMMARY

The discussion focuses on solving a problem related to total internal reflection (TIR) in a glass prism. The user, twiztidmxcn, successfully calculated the critical angle at 41.81 degrees using Snell's Law with refractive indices of glass (n1 = 1.5) and air (n2 = 1.0). The user encountered confusion regarding the subtraction of angles in the context of similar triangles, specifically how to determine the exit angle of the laser beam after reflecting off the hypotenuse. Suggestions included redrawing the triangle with exaggerated angles to clarify the relationships between the angles involved.

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twiztidmxcn
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hey

so I'm doing this question on total internal reflection and have got myself totally lost

here is the question:

what is the smallest angle \theta(1) for which a laser beam will undergo TIR on the hypotenus of this glass prism?

after reflecting off the hypotenuse at \theta(c) the laser exits the prism through the bottom face. does it exit to the right or left of the normal? at what angle?

http://photos-603.facebook.com/n11/161/86/37900227/n37900227_30061603_2132.jpg

that is a picture of what is going on here

Now, i realize that at some point I am going to be using snells law as well as the equation for critical angle. My question is.. what do I do?

I am totally lost as you can probably see, and any help would be much appreciated
thank you
twiztidmxcn

*edit*
i've used the critical angle equation and found that the critical angle is 41.81 degrees (using n1 = 1.5 for glass, n2 = 1.0 for air) and didn't know what to do with it... i ended up subtracting the triangle angle of 30 degrees and setting 1.5*sin11.81 = 1*sin theta and solved for theta. i got the right answer, but am not sure how the 30 degrees is subtracted. i know it involves similar triangles but am just not seeing it i guess.

same for the second part, i just subtracted 60-41.81 and then setup the same equation to find my angle coming out, which again i got correct, but am stumped as to how that works.
 
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Good job solving the middle part first, that's how I would approach it as well. Now just draw the two 41.8 degree beams coming off the hypotenuse side (one toward the left face and one toward the bottom face). Then label the angles where the beams exit out the two sides. That should help you keep the Snell's law beam angle changes straight at the two sides.
 
Here's a helpful hint. Redraw the triangle such that the two angles are very different (like 20 degrees and 70 degrees). Then draw your ray tracing going backward, and the angles you need will be much more obvious as you look at them (you will see whether to subtract 60 or 30 degrees because you will see one angle much larger than the other).
 

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