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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 n_{inside}=1.33, and one assumes that the surface has n_{outside}=1 for air.

One must solve the equation \theta = \sin^{-1}(1/1.33). One can immediately throw out choices (A) and (E). From the unit circle, one recalls that \sin(30^{\deg})=1/2 and \sin(60^{\deg})=1.7/2=0.85. Since 1/1.33 \approx 0.7, one deduces that the angle must be choice (C).


See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
calcuttj
2014-09-20 06:48:40
1/1.33 -> 3/4 -> (3/2)/2 -> 1.5/2

sin(45) = \sqrt{2}/2, \sqrt{2} is about 1.4

so our angle is slightly above 45
Alternate Solution - Unverified
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 n=1.5 you have a critical angle of 41.5^o (look it up), so by decreasing the index of refraction (eg. water 1.33) you would slightly increase the critical angle so there about 50^o
ramparts
2009-11-03 18:06:23
Yeah. I'll be sure to look that one up on the test. Thanks a lot.
Alternate Solution - Unverified
Comments
calcuttj
2014-09-20 06:48:40
1/1.33 -> 3/4 -> (3/2)/2 -> 1.5/2

sin(45) = \sqrt{2}/2, \sqrt{2} is about 1.4

so our angle is slightly above 45
Alternate Solution - Unverified
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.NEC
mistaj
2011-08-25 11:19:41
This is Brewster's Law: \rm tan\theta = \frac{n_t}{n_i} 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, \rm tan\theta \simeq \theta (which is really good for this problem). So, in radians we have \theta = 0.7. Now, we can get an idea of this in terms of \pi by solving \frac{\pi}{x} = 0.7 for x which is roughly x = \frac{\pi}{4} which is roughly half of 90 degrees. Choice C is the only option.
mistaj
2011-08-25 11:38:57
Woops, misunderstood this. It still works though! But forget Brewster's Law and use sin instead of tan.
NEC
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.NEC
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 \sqrt{2}, 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)!NEC
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.
lrichey
2011-11-04 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
NEC
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 degreesNEC
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 n=1.5 you have a critical angle of 41.5^o (look it up), so by decreasing the index of refraction (eg. water 1.33) you would slightly increase the critical angle so there about 50^o
ramparts
2009-11-03 18:06:23
Yeah. I'll be sure to look that one up on the test. Thanks a lot.
Alternate Solution - Unverified

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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.

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