Physics Help Forum Phase and fringes

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Nov 14th 2009, 04:06 AM   #11
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Posts: 11
 Originally Posted by r.samanta 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)
No, it is not particularly funny

But you say "find phase difference". My question is how to find this phase difference. The way I would do it is

1) Find phase difference in arm 1
2) Find phase difference in arm 2
3) Add (+) the two phase difference to get total difference
4) Divide (/) the phase difference by 2Pi to find total number of fringes.

Is this correct? See you tomorrow (or sooner, peraps?)

 Nov 14th 2009, 04:30 AM #12 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 i do not know what you mean by phase difference in arm 1? are u referring to the pi phase change due to reflection at mirror? we dont have to worry about that. that will be cancelled by pi phase change in the other mirror also. in michelson, the two waves that interfere come from two arms by separate reflections. it is their phase difference that i need to find,
 Nov 14th 2009, 04:38 AM #13 Junior Member   Join Date: Nov 2009 Posts: 11 Ok, what I meant was that now we are looking at an experiment like the first one, but this time there is also a glass-rod in arm #1 (the rods are not identical). What I want to do is to find the number of fringes crossing the field of view every minute. This is my approach: 1) Find phase difference in arm 1 2) Find phase difference in arm 2 3) Add (+) the two phase difference to get total difference 4) Divide (/) the phase difference by 2Pi to find total number of fringes. Is this approach correct? I have another question regarding the 23 fringes/minute: Does this mean that there are 23 new fringes each minute? So first 23 fringes on the wall, then 46 the next minute, ... ? So if the phase difference is zero, then this means that the number of fringes is constant? Also (so this is the third question in this post): Now we look at the original setup, where there is only 1 glass-rod. If I want to find the 23 fringes/minute using phase-differences, then should I find the phase difference per minute between the two arms? Or the phase difference per minute in the arm containing the glass-rod? Last edited by Niles_64; Nov 14th 2009 at 05:15 AM.
 Nov 14th 2009, 09:11 AM #14 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 oh no! did my posts imply that? sorry but no extra fringes are appearing here , its just the fringe SHIFT that is taking place per minute. also, the two interfering waves are coming from the 2 arms. i have to calculate Phase difference BETWEEN THEM.I am still in the dark about what u are implying by phase difference in a single arm.
 Nov 14th 2009, 09:24 AM #15 Junior Member   Join Date: Nov 2009 Posts: 11 Ahh ok, so what we have found (i.e. the 23 fringes/minute) is the number of fringes that shift places. Ok, good to know. My goal is to calculate the 23 fringes/minute using the phase difference that occurs because of the glass rod is expanding. The phase of the light in the rod is given by 2*k*L(1+alpha*dT) = 2*k0*n(1+beta*dT)*L*(1+alpha*dT) where alpha is linear expansion coefficient and beta is coefficient for refractive index. Now what do I have to do from here in order to find the 23 fringes/min? - thanks for being patient
 Nov 14th 2009, 09:41 AM #16 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 i think i have understood your problem. u are trying to solve the problem by phase difference instead of N(no of wavelengths) in that case , look at my earlier solution , i think u have understood by now that you simply need to multiply the equation by 2 pi(remember 1 wavelength= 2 pi?)to get the corresponding phase difference. so phi(phase difference)=2 pi(2L glass n/lambda)- 2 pi(2L air/ lambda) when you differentiate, L air is constant(no expansion for air as i told u earlier) so right part is zero. finally 2 pi phase difference creates 1 fringe shift. after differentiating , divide by 2 pi -u get fringe shifts and will get the same answer.
Nov 14th 2009, 10:11 AM   #17
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Join Date: Nov 2009
Posts: 11
 Originally Posted by r.samanta i think i have understood your problem. u are trying to solve the problem by phase difference instead of N(no of wavelengths) in that case , look at my earlier solution , i think u have understood by now that you simply need to multiply the equation by 2 pi(remember 1 wavelength= 2 pi?)to get the corresponding phase difference. so phi(phase difference)=2 pi(2L glass n/lambda)- 2 pi(2L air/ lambda) when you differentiate, L air is constant(no expansion for air as i told u earlier) so right part is zero. finally 2 pi phase difference creates 1 fringe shift. after differentiating , divide by 2 pi -u get fringe shifts and will get the same answer.
But why do I have to differentiate the phase difference? We said that for each 2Pi difference in phase, there is 1 fringe shift, so why are we differentiating the phase difference, which gives us the change in the phase difference / minute?

 Nov 14th 2009, 10:37 AM #18 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 now, this is really my last.forgive me but i am already starting to see fringes in my eyes. i have to differentiate phi because i need to calculate the CHANGE in phase difference due to linear expansion and change of refractive index.the phase difference phi results in fringe formation. but 'd phi ' is making the SHIFT of the fringes.
 Nov 14th 2009, 10:56 AM #19 Junior Member   Join Date: Nov 2009 Posts: 11 Thank you very much, sir. You have really helped me alot, and you don't have to ask for forgiveness. I think I understand it now. Again, thank you.
 Nov 14th 2009, 11:00 AM #20 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 a little correction- i am not 'sir' , i am in final year in school(12 grade) , its making me blush.

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