# MM Experiment: Fringe Shift

1. Apr 26, 2006

### Living_Dog

I am reading Paul Tipler's Fundamentals of Modern Physics, (Worth Pub., 1973) and would like to understand how ΔN (Eq. 1-5, pg. 6) )is arrived at. Here is the derivation as I understand it up to that point:

Michelson devised an experiment to determine the difference in the number of fringes observed when his interferometer is rotated from 0o to 90o.

Let T1 be the time of flight (tof) along the parallel (//) arm and
let T2 be the tof along the perpendicular (+) arm.

Assuming that the // arm moves with the ether wind, then
T1 = (2L/c)(1 + V2/c2).
and
T2 = (2L/c)[ 1 + V2/2c2 ].

The total tof difference = T1 - T2 = LV2/c2.

Now you rotate the apparatus (90o) and note a fringe shift from the original orientation (0o) of the device. Tipler then states that the total path difference between (0o and 90o) created is 2cΔT which shows a fringe shift given by:

ΔN = path difference/λ = 2cΔT/λ = 2LV2/c3 = (2L/λ)*(V2/c2).

It is this last expression, for ΔN, with which I am having trouble and woiuld like to understand how it is arrived at.

tia!

-LD

2. Apr 26, 2006

### clj4

2ΔT is twice the time difference induced by rotating the device
c2ΔT is twice the distance difference induced by rotating the device
ΔN = path difference/λ = 2cΔT/λ shows how many wavelengths fit into the distance difference

3. Apr 26, 2006

### Tom Mattson

Staff Emeritus
This looks like it could be 3 different questions.

1.) Why is $\Delta N$ equal to the path length difference over the wavelength?

Any path length difference in the two arms of the interferometer is going to cause a phase shift in the two beams when they return to the interference region. If the beams are an integral number of wavelengths out of phase then it's as if there was no phase shift. The interference is purely constructive, and you get a bright spot. But if the beams are a half-integral number of wavelengths out of phase then they exactly cancel each other out, and you get a dark spot. This can be generalized as follows. (PLD=Path Length Difference)

Constructive Interference: $PLD=(n+1/2)\lambda$ ($n=0,1,2,...$)
Destructive Interference: $PLD=n\lambda$ ($n=0,1,2,...$)

You can generalize this to non-integers and come up with a unified formula by replacing $n$ with a real number $\Delta N$, as follows.

$$PLD=\Delta N \lambda$$

$$\Delta N=\frac{PLD}{\lambda}$$

2.) Why is the path length difference equal to $2c\Delta T$?

Use $T_1$ and $T_2$ to derive expressions for the distances $L_1$ and $L_2$ traveled by each beam. Then subtract: $PLD=|L_1-L_2|$.

3.) Why does $\Delta T=Lv^2/c^2$?

$\Delta T=|T_1-T_2|$. Just carry out the subtraction.

4. Apr 26, 2006

### Living_Dog

Thanks! This is what I needed to see. I'm sorry I didn't articulate the specific question, but your reply is even more helpful than I hoped for - so thanks again! :)

-LD

5. Apr 27, 2006

### Living_Dog

To Tom Mattson:

I read over your reply and still have some questions - I hope you don't mind and are willing to help me through this rough spot.

Question 1: I just want to check - but isn't that a typo? You said it correctly but switched the two around, i.e.
Constructive Interference = n$$\lambda$$ and
Destructive Interference = (n + $$\frac{1}{2}$$)$$\lambda$$.

$$\Delta$$N is just some symbol for any number, where the difference ($$\Delta$$) doesn't really mean that two integers are being subtracted... that would contradict the idea that this is any number.

So it could be thought of as replace n with a real number 'a', as follows:

PLD = a/$$\lambda$$

Therefore, a = PLD$$\cdot\lambda$$.

Question 2: This one is rather long b/c I have to show you what I did to NOT get the 2c$$\Delta$$T result (I get c$$\Delta$$T/2):

Clearly, PLD = |L1 - L2|.

Using the expressions for T1 and T2 in Tipler, namely:

T1 = $$\frac{2L_1}{c}(1 + \frac{v^2}{c^2})$$

T2 = $$\frac{2L_2}{c}(1 + \frac{v^2}{2c^2})$$

Then

L1 = $$\frac{cT_1}{2(1 + \frac{v^2}{c^2})}$$

and

L2 = $$\frac{cT_2}{2(1 + \frac{v^2}{2c^2})}$$

Substituting and simplifying I get:

PLD = $$\frac{cT_1}{2(1 + \frac{v^2}{c^2})} - \frac{cT_2}{2(1 + \frac{v^2}{2c^2})}$$

= $$(\frac{c}{2})\frac{[T_1(1 + \frac{v^2}{c^2}) - T_2(1 + \frac{v^2}{2c^2})]}{(1 + \frac{v^2}{c^2})(1 + \frac{v^2}{2c^2})}$$,

which, when you drop all small factors $$\frac{v^2}{c^2}$$, and higher order, yields my result:

PLD = c$$\Delta$$T/2.

But since the experiment is rotated, and performed again, the net result will be twice the result for only one measurment. Therefore:

PLD = c$$\Delta$$T

...ok, where'd I go wrong? :/

Question 3: I believe this is another typo, since $$\Delta$$T = L$$\frac{v^2}{c^3}$$.

phew! That's it. The 2nd question is critical here since I think I got the rest of it ok - thanks to you.

-LD

PS: Once I read your post I realized that my confusion was about the nature of the fringes and what $$\Delta$$N means for them. I thought that they were going to shift so many places to the right. That's not it at all... there are going to be a different NUMBER OF THEM - not a different position for them! That's what I didn't see - how could these fringes slide to the right if all I do is change the path difference.

That was the major thing blocking my progess. Now if you could only answer at least question 2 I would be all set to move on...

thx again! :)

-LD

Last edited: Apr 27, 2006
6. May 1, 2006

### clj4

With the risk of confusing you, I need to correct you:

$$\Delta$$N is the SHIFT of fringes expressed in FRACTIONS of ONE fringe. (hence the $$\Delta$$) . The number of fringes doesn't change, the position of the fringes slides by a very small amount. In that respect the presence of "N" is misleading. We don't even need it.

7. May 1, 2006

### Tom Mattson

Staff Emeritus
Sorry, I had forgotten about this thread before clj4 brought it back. I hope my lateness hasn't hurt you. You were right about my typos.

I think this is related to another one of my sloppy typing.

I said...

That's not right. The MMX is an two way speed of light experiment. The distance traveled by the two beams is not $L_1$ and $L_2$, but rather $2L_1$ and $2L_2$. So the path length difference is: $PLD=|2L_1-2L_2|=2|L_1-L_2|$.

Sorry for my carelessness.

8. May 1, 2006

9. May 6, 2006

### Living_Dog

Oh man! what a twist! :)

Yeah, that explains it. I knew the light travelled round trip, but totally forgot to apply that here. Thus I get a factor of 2 and everything is now settled.

thx!

BTW, I am not in classes. I was many years ago and, by God's grace, have rekindled my desire to get a PhD. Compared to my initial attempt I am learning so much more now that I feel comfortable about the outcome. :)

thx again!

-LD

Last edited: May 6, 2006
10. May 6, 2006

### Living_Dog

11. May 6, 2006

### Living_Dog

i c - so the fringes actually shift, but don't change 'N' at all.

This is an annoying trend in physics - to not call a spade a spade. They always do that right in the middle of complicated material: change nomenclature, use bad, misleading, or confusing nomenclature, etc. Someday a physicist should write a book scolding physics text writters for perpetuating old, cumbersome, and misleading nomenclature. ...

One of my pet peaves is the term centripetal "force." But that's for another thread. :)

thx for coming to the rescue clj4!

-LD