 GR 8677927796770177 | # Login | Register

GR9677 #56
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Optics}Total Internal Reflections

Total internal reflection is when one has a beam of light having all of the incident wave reflected. Going through a bit of formalism in electromagnetism one can derive Snell's Law for Total Internal Reflection,

where , and one assumes that the surface has for air.

One must solve the equation . One can immediately throw out choices (A) and (E). From the unit circle, one recalls that and . Since , one deduces that the angle must be choice (C).  Alternate Solutions
calcuttj
2014-09-20 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
2009-04-03 07:42:49

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
 ramparts2009-11-03 18:06:23 Yeah. I'll be sure to look that one up on the test. Thanks a lot. calcuttj
2014-09-20 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
2012-09-29 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
2011-08-25 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.
 mistaj2011-08-25 11:38:57 Woops, misunderstood this. It still works though! But forget Brewster's Law and use sin instead of tan. walczyk
2011-02-25 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
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.
 lrichey2011-11-04 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 30-60-90 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
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

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
 ramparts2009-11-03 18:06:23 Yeah. I'll be sure to look that one up on the test. Thanks a lot.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$