GR0177 #97


Problem


This problem is still being typed. 
Optics}Refraction
From Snell's Law, one obtains , since the index of refraction of air is about 1.
Now, differentiate both sides with respect to .
which gives choice (E) to be the angular spread.
This solution is due to ShyamSunder Regunathan.


Alternate Solutions 
jsdillon 20080407 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!)  

Comments 
f 20191015 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 20111105 22:43:21  TRICK TRICK: only E has right unit.
Yurlungur 20120408 20:04:32 
Actually (B), (D) and (E) are unitless. However, eliminating (A) and (C) this way is nice. :)

msdec 20160908 04:33:49 
Yes but only E is in the proper units of radians.

  mistaj 20110824 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 20110224 20:38:07  i like Yosun's solution but how does the last step come about? can sombody explain it?
Da Broglie 20110408 17:04:12 
In the last step Yosun moved to the left side of the equal sign and divided through by converting to . Multiply through by to solve for .

  barson 20091018 04:39:38  How to define the theta which depends on lamda?
I got some confuse about this.   realcomfy 20081024 14:16:00  Just curious, but how does the turn into just plain in that final solution?
nobel 20081101 07:34:16 
n(lambda) implies n is a function of lambda. its not
n *lambda

  jsdillon 20080407 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 20091105 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.

apr2010 20100409 16:56:08 
Following your argumentation, D is possible.
I would just say, as Snells law contains also choose D over E.

Yurlungur 20120408 20:11:40 
I think this is the best solution.

  Mexicana 20071002 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 . Then you just need to get 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 20071030 22:20:00 
Personally, I find this to be more correct.
There seems to be a bit of handwaving in dear Yosun's quoted solution. In particular, I am thinking of the last step.

  scottopoly 20061029 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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