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GR0177 #97
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Problem
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This problem is still being typed. |
Optics }Refraction
From Snell's Law, one obtains  \sin\theta^{'}) , since the index of refraction of air is about 1.
Now, differentiate both sides with respect to  .
 \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|, <br />
\end{eqnarray})
which gives choice (E) to be the angular spread.
This solution is due to ShyamSunder Regunathan.
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Alternate Solutions |
jsdillon
2008-04-07 20:27:09 | This problem can be solved using the limiting case of  =0.
We also know that  ' must go to 0 at =0 since normally incident rays don't refract (think Snell's Law with =0). That leaves only (E), the correct answer. (D) doesn't work because as goes to 0, ' 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!) |  |
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Comments |
f
2019-10-15 21:12:02 | Il a remporté deux antennes, achevés depuis plus de 80 pour cent de ses passes et a fait deux passes clés, y compris son assist.\r\nf http://www.lindbaek.eu/tobuyes13.asp |  | 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. :)
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msdec
2016-09-08 04:33:49 |
Yes but only E is in the proper units of radians.
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| | mistaj
2011-08-24 08:35:17 | You can eliminate all but B and E by noting ) 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 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 Broglie
2011-04-08 17:04:12 |
In the last step Yosun moved cos(\theta)^\prime \frac{d\theta^\prime}{d\lambda}) to the left side of the equal sign and divided through by cos(\theta)^\prime) converting ^\prime}{cos(\theta)^\prime}) to ^\prime) . Multiply through by  to solve for  .
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| | barson
2009-10-18 04:39:38 | How to define the theta which depends on lamda?
I got some confuse about this. | | realcomfy
2008-10-24 14:16:00 | Just curious, but how does the ) turn into just plain  in that final solution?
nobel
2008-11-01 07:34:16 |
n(lambda) implies n is a function of lambda. its not
n *lambda
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|  | jsdillon
2008-04-07 20:27:09 | This problem can be solved using the limiting case of =0.
We also know that ' must go to 0 at =0 since normally incident rays don't refract (think Snell's Law with =0). That leaves only (E), the correct answer. (D) doesn't work because as goes to 0, ' 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()=1, i.e. no change in medium is taking place. In this case,  '= 0 also. The only two choices that provide this contain a factor of dn/d, and only (E) provides the aforementioned limiting case as well.
Just a thought.
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apr2010
2010-04-09 16:56:08 |
Following your argumentation, D is possible.
I would just say, as Snells law contains also  choose D over E.
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Yurlungur
2012-04-08 20:11:40 |
I think this is the best solution.
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| | 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 ^2=(\frac{\partial \theta'}{\partial\lambda})^2(\delta\lambda)^2) . Then you just need to get ^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.
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| | 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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