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Problem
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Optics}Refraction

From Snell's Law, one obtains \sin\theta = n(\lambda) \sin\theta^{'}, since the index of refraction of air is about 1.

Now, differentiate both sides with respect to \lambda.

which gives choice (E) to be the angular spread.

This solution is due to ShyamSunder Regunathan.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
jsdillon
2008-04-07 20:27:09
This problem can be solved using the limiting case of \theta=0.

We also know that \delta\theta' must go to 0 at \theta=0 since normally incident rays don't refract (think Snell's Law with \theta=0). That leaves only (E), the correct answer. (D) doesn't work because as \theta goes to 0, \theta' also goes to 0.

(I thought, since I'm now almost all the way through all four tests with the exam now 6 days away, that I'd leave a small contribution. Thanks Yosun!)
Alternate Solution - Unverified
Comments
solidstate
2011-11-05 22:43:21
TRICK TRICK: only E has right unit.
Yurlungur
2012-04-08 20:04:32
Actually (B), (D) and (E) are unitless. However, eliminating (A) and (C) this way is nice. :)
msdec
2016-09-08 04:33:49
Yes but only E is in the proper units of radians.
NEC
mistaj
2011-08-24 08:35:17
You can eliminate all but B and E by noting n(\lambda) needs to be part of the final solution. This is the case because if n = 1 for all wavelengths, the solution would be 0 since the angular spread is constant. Now note, that the transmitted angle and the incident index of refraction is only in E. Since n will impact our final answer (imagine if both n(\lambda) and n are the same - there will be no refraction), the only complete answer is E.NEC
bluelight
2011-02-24 20:38:07
i like Yosun's solution but how does the last step come about? can sombody explain it?
Da Broglie
2011-04-08 17:04:12
In the last step Yosun moved n(\lambda)cos(\theta)^\prime \frac{d\theta^\prime}{d\lambda} to the left side of the equal sign and divided through by n(\lambda)cos(\theta)^\prime converting \frac{sin(\theta)^\prime}{cos(\theta)^\prime} to tan(\theta)^\prime. Multiply through by \delta\lambda to solve for \delta\theta^\prime.
NEC
barson
2009-10-18 04:39:38
How to define the theta which depends on lamda?
I got some confuse about this.
NEC
realcomfy
2008-10-24 14:16:00
Just curious, but how does the n \left( \lambda \right) turn into just plain n in that final solution?
nobel
2008-11-01 07:34:16
n(lambda) implies n is a function of lambda. its not
n *lambda
Answered Question!
jsdillon
2008-04-07 20:27:09
This problem can be solved using the limiting case of \theta=0.

We also know that \delta\theta' must go to 0 at \theta=0 since normally incident rays don't refract (think Snell's Law with \theta=0). That leaves only (E), the correct answer. (D) doesn't work because as \theta goes to 0, \theta' also goes to 0.

(I thought, since I'm now almost all the way through all four tests with the exam now 6 days away, that I'd leave a small contribution. Thanks Yosun!)
Daw6
2009-11-05 10:43:05
We can also think about the limiting case n(\lambda)=1, i.e. no change in medium is taking place. In this case, \partial\theta'= 0 also. The only two choices that provide this contain a factor of dn/d\theta, and only (E) provides the aforementioned limiting case as well.
Just a thought.
apr2010
2010-04-09 16:56:08
Following your argumentation, D is possible.

I would just say, as Snells law contains also \sin \theta choose D over E.
Yurlungur
2012-04-08 20:11:40
I think this is the best solution.
Alternate Solution - Unverified
Mexicana
2007-10-02 18:23:52
Another funky way of doing this (somehow less abstractly than yosun) is to use the general formula for combination of errors which would give for this case (\delta\theta')^2=(\frac{\partial \theta'}{\partial\lambda})^2(\delta\lambda)^2. Then you just need to get (\frac{\partial \theta'}{\partial\lambda})^2 using Snell's Law. Whenever you see the word 'spread' or 'uncertainty' or just 'error', always remember the pretty formula for combination of errors!
Richard
2007-10-30 22:20:00
Personally, I find this to be more correct.
There seems to be a bit of hand-waving in dear Yosun's quoted solution. In particular, I am thinking of the last step.
NEC
scottopoly
2006-10-29 23:57:28
Really minor, but the problem says it's in a vacuum, so your comment that it's in air is incorrect.Typo Alert!

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