GREPhysics.NET: GR9677 #56

GR9677 #56

Problem

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Optics $\Rightarrow$ }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,

$n_{inside} \sin\theta = n_{outside},<br />$

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).

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

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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.

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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

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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.

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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.

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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.

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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)!

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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

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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

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tensorwhat
2009-04-03 07:42:49

ramparts
2009-11-03 18:06:23

Yeah. I'll be sure to look that one up on the test. Thanks a lot.

Reply to this comment

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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

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