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?

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.

The answer is 23 fringes/minute, but can you explain why?

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.

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?

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.

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)

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.

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)

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?)

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,

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?

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.

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

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?

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.

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.

a little correction- i am not 'sir' , i am in final year in school(12 grade) , its making me blush.