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GR0177 #97
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Optics$\Rightarrow$}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 $\lambda$.

$\begin{eqnarray} \frac{d}{d\lambda}\sin\theta &=& \frac{d}{d\lambda}(n(\lambda) \sin\theta^{'})\\ 0&=& \frac{dn(\lambda)}{d\lambda} \sin\theta^{'}+(n(\lambda) \cos\theta^{'}) \frac{d\theta^{'}}{d\lambda}\\ \delta \theta^{'} &=& |\tan\theta^{'}/n \frac{dn(\lambda)}{d\lambda} \delta \lambda|,
\end{eqnarray}$

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

This solution is due to ShyamSunder Regunathan.

Alternate Solutions
 jsdillon2008-04-07 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!)Reply to this comment
solidstate
2011-11-05 22:43:21
TRICK TRICK: only E has right unit.
 Yurlungur2012-04-08 20:04:32 Actually (B), (D) and (E) are unitless. However, eliminating (A) and (C) this way is nice. :)
 msdec2016-09-08 04:33:49 Yes but only E is in the proper units of radians.
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.
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 Broglie2011-04-08 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 $\delta\theta^\prime$.
barson
2009-10-18 04:39:38
How to define the theta which depends on lamda?
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?
 nobel2008-11-01 07:34:16 n(lambda) implies n is a function of lambda. its not n *lambda
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!)
 Daw62009-11-05 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.
 apr20102010-04-09 16:56:08 Following your argumentation, D is possible. I would just say, as Snells law contains also $\sin \theta$ choose D over E.
 Yurlungur2012-04-08 20:11:40 I think this is the best solution.
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!
 Richard2007-10-30 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.
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.

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