Physics Help Forum Phase and fringes
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 Nov 13th 2009, 01:09 PM #1 Junior Member   Join Date: Nov 2009 Posts: 11 Phase and fringes Hi all Please take a look at problem 35.61 in this PDF (it is on page 5): http://physics.wustl.edu/classes/SP2...ework/YF35.pdf What I can do is to find the following quantity: (phase gained per minute) - (phase gained at T=20 degrees). But problem is that I don't know how to go from this to the number of fringes. Can you guys help me?
 Nov 14th 2009, 12:09 AM #2 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 is the answer 15 fringes/ minute or 23 fringes/ minute.?? if any of above is correct, please reply. i will post the solution. many thanks for posting such a beautiful problem.
 Nov 14th 2009, 01:13 AM #3 Junior Member   Join Date: Nov 2009 Posts: 11 The answer is 23 fringes/minute, but can you explain why? Last edited by Niles_64; Nov 14th 2009 at 01:18 AM.
 Nov 14th 2009, 01:26 AM #4 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 actually light traverses through the rod twice in one arm. so no of extra wavelengths traversed is 2L/(lambda/n)=2L n/lambda= N (say) where n is refrac index. differentiate the above equation to get dN= 2/lambda(ndL + Ldn) now in one minute , temperature increases by 5 degrees. so let dT=5 for dL use L *(coeff of linear expansion)*dT ie. change in length for dn=n*(coeff of increase of refrac index)dT ie change in refractive index put these values in dN equation and u will get the answer. physics involved is that with temp change, both length and refractive index is changing , so phase diff will change due to both, resulting in fringe crossing. and for each change in wavelength, there will one fringe shift. Last edited by r.samanta; Nov 14th 2009 at 01:30 AM.
 Nov 14th 2009, 01:33 AM #5 Junior Member   Join Date: Nov 2009 Posts: 11 I am sorry, but I do not understand the solution at all. How is it that we can find the number of extra wavelengths from that equation? Also, how do we go from the extra number of wavelengths to a number of fringes? Also, we are talking about a Michelson-interferometer. So there is another wave going through the same path, but without a glass-rod in the path. Why are we not working with this light also? Last edited by Niles_64; Nov 14th 2009 at 01:38 AM.
 Nov 14th 2009, 01:42 AM #6 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 please see if this helps you are correct that there is another path without glass rod. that is why i used the term 'extra wavelength: the path difference arising due to air will get cancelled during differentiation since it is a constant( think about this) and please note this i wavelength path diff= 2 pi phase difference= 1fringe shift(think about young experiment now)
Nov 14th 2009, 01:59 AM   #7
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 Originally Posted by r.samanta please see if this helps you are correct that there is another path without glass rod. that is why i used the term 'extra wavelength: the path difference arising due to air will get cancelled during differentiation since it is a constant( think about this) and please note this i wavelength path diff= 2 pi phase difference= 1fringe shift(think about young experiment now)
Thank you. But let us look at the light coming from the other arm, where there is no glass rod. What we must do is to compare the two lengths, so the extra wavelengths that light gets here is

2L(1+alpha dT)/lambda0,

where lambda0 is wavelength in air (vacuum) and alpha is coefficient of linear expansion. This is not constant?

Also, I have read about Youngs experiment many times now. I still can't see where it says that 2 pi phase difference= 1fringe shift.

Last edited by Niles_64; Nov 14th 2009 at 02:12 AM.

Nov 14th 2009, 02:31 AM   #8
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it would have been great if i had pen and paper
still, when u are applying linear expansion for the other arm also, have you considered that you need to use the coefficient of linear expansion of air and also refractive index of air?? as these are not supplied, i guess the heating arrengement is only for the arm with glass rod.
as for fringe shift, try to think it this way - what will happen if an extra 2 pi phase diff appeared in youngs experiment?? the central bright band will be replacedby the next bright band. remember the condition of bright fringes phase diff = 2 n pi,now add 2 pi on both sides , u see that right side is replaced by 2(n+1)pi which means n th order fringes are replaced by (n+1) th order fringe which is equivalent to saying there has been one fringe shift.
however, a question that i have for you is 'does refractive index decrease or increase with temperature", if it decreases(which i thought from common sense) then the ans would have been 15 fringes/min. but the correct answer appears only if n increases with temperature. what do u think, friend?
a great deal of research is going on in this area- it is found that the the temp coefficient of refractive index generally positive for most type of glasses, whereas for liquids it is negative. so your book has given correct answer.
Temperature and Refractive Index and also have a look at the attached pdf.
Attached Files
 tie-19_temperature_coefficient_of_refractive_index_v2d.pdf (566.4 KB, 3 views)

Last edited by r.samanta; Nov 14th 2009 at 03:01 AM.

Nov 14th 2009, 03:14 AM   #9
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 Originally Posted by r.samanta as for fringe shift, try to think it this way - what will happen if an extra 2 pi phase diff appeared in youngs experiment?? the central bright band will be replacedby the next bright band. remember the condition of bright fringes phase diff = 2 n pi,now add 2 pi on both sides , u see that right side is replaced by 2(n+1)pi which means n th order fringes are replaced by (n+1) th order fringe which is equivalent to saying there has been one fringe shift.
That is a good explanation. But isn't n just a random integer (1, 2, 3, ...) or is n always equal to the number of fringes?

Ok, so I can also solve this problem by finding the phase differnece per minute in the arm with the glass rod. Let us say now as a funny experiment that I put another glass rod in arm number 2, and we still have the "old" glass rod in arm number 1. Then I can find the phase difference in arm #1 per minute and the phase difference in arm #2 per minut.

Do I add or subtract these two phase differences to find the number of fringes per minute?

Regarding the refractive index: That is actually a very good question. Unfortunately, I do not know.

 Nov 14th 2009, 03:59 AM #10 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 yes . u can take n as random integer , u can say that this is fringe order number once you fix a value to n eg 5th order fringe etc. regarding funny experiment (is it funny??), there will be no fringe shift if glass rods are of same length. if they are diff length, then only u find phase difference. i dont know if this is what you asked. (please note- i am going offline for today)

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