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GR9677 #56
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Alternate Solutions |
tensorwhat
2009-04-03 07:42:49 | Easier way to think about this using critical angles for total internal reflection......
For a piece of plastic or glass with  you have a critical angle of  (look it up), so by decreasing the index of refraction (eg. water 1.33) you would slightly increase the critical angle so there about 
ramparts
2009-11-03 18:06:23 |
Yeah. I'll be sure to look that one up on the test. Thanks a lot.
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Comments |
torturedbabycow
2010-03-27 19:54:37 | As far as solving that annoying inverse sine, I think the easiest way by far is to just draw out the triangle - one side is 1, and the hypotenuse is 1.33. So, since 1.33 is pretty close to  , the angle should be pretty close to 45 degrees. Stare at the triangle a few more seconds, and it becomes obvious (at least to me) that it should be a little more than 45, so voila, answer (C)! |  | jmason86
2009-10-01 19:33:10 | This is probably one to just have memorized. Stupid ETS trying to make us solve that inverse sign of 1/1.33. I hate 'em. | | f4hy
2009-04-03 17:32:47 | I only got this one because I remembered doing this exact problem in an optics class and knew that for water the angle is 50 degrees | | tensorwhat
2009-04-03 07:42:49 | Easier way to think about this using critical angles for total internal reflection......
For a piece of plastic or glass with you have a critical angle of (look it up), so by decreasing the index of refraction (eg. water 1.33) you would slightly increase the critical angle so there about
ramparts
2009-11-03 18:06:23 |
Yeah. I'll be sure to look that one up on the test. Thanks a lot.
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