GR9677 #56



Alternate Solutions 
calcuttj 20140920 06:48:40  1/1.33 > 3/4 > (3/2)/2 > 1.5/2
sin(45) = /2, is about 1.4
so our angle is slightly above 45   tensorwhat 20090403 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 20091103 18:06:23 
Yeah. I'll be sure to look that one up on the test. Thanks a lot.

 

Comments 
calcuttj 20140920 06:48:40  1/1.33 > 3/4 > (3/2)/2 > 1.5/2
sin(45) = /2, is about 1.4
so our angle is slightly above 45   walczyk 20120929 21:53:57  Just figured out a good approximate expansion people might want to memorize, arcsin in degrees: arcsin(z) ~ 60z + 10z^3, you get about 49.2 for this answer. You just have to be quick with your fractions.   mistaj 20110825 11:19:41  This is Brewster's Law: where t is the transmission medium (air n = 1.00) and i is the incident medium (water n = 1.33). Dividing 1 by 1.33, you get about 0.7. Now, (which is really good for this problem). So, in radians we have . Now, we can get an idea of this in terms of by solving for which is roughly which is roughly half of 90 degrees. Choice C is the only option.
mistaj 20110825 11:38:57 
Woops, misunderstood this. It still works though! But forget Brewster's Law and use sin instead of tan.

  walczyk 20110225 16:42:21  so here's how i figured out the hard part, arcsin(1/1.33): 1/1.33 is close to 3/4 (remember the inverse is 1.33!!) . sin(45) is 1/sqrt(2) so its bigger than 45 degrees (sqrt(2) is like 1.414.. so its bigger than 1.33!!) now sin(60) is sqrt(3)/2, which is like .8 or something so its less than 60 degrees. the only option left is 50 degrees, and we're done. the first time i did it i used the fact that (sqrt(3)/2)^2 is 3/4 so its obviously greater than 3/4.   torturedbabycow 20100327 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 20091001 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.
lrichey 20111104 17:40:57 
I agree... but here is a way I figured out....
1.33~1 1/3, hence 4/3
1/(4/3)=3/4. 3/4 is the square of a 306090 triangle relation, giving inversesine(3/4)~ 60.
Which is closer to 50 degrees than 75 degrees
don't know if this works or helps at all

  f4hy 20090403 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 20090403 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 20091103 18:06:23 
Yeah. I'll be sure to look that one up on the test. Thanks a lot.

 

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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)!

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