How does fringe shift relate to velocity in the Michelson-morley experiment?

[Benzoate]

Homework Statement

A shift of one fringe in the Michelson-morley experiment would result from a difference of one wavelength or a change of one period of vibration in the round trip travel of the light when the interferometer is rotated by 90 degrees. What speed would the Michelson have computed for Earth's motion through the ether had the experiment seen a shift of one fringe?

Homework Equations

delta(t)=(L*v^2)/(c^3)

The Attempt at a Solution

delta=(L*v^2)/(c^3) => v=sqrt((c^3*delta(t))/(L)) , where L is the length of the wavelength andd c is the speed of light.

I don't understand how shifting the fringe would effect the velocity.

My professor says I'm have the right equation for finding the velocitiy of the Earth's motion, But he says the velocity is also related to the fringe where the fringe is represented by light over one period per wavelenght and light oscillates over that period.

 

[Dick]

I think the L in your formula is not the wavelength of the light. It's the total path length of each interferometer arm. The wavelength of light enters via the delta(t). delta(t)=w/c, where w in the wavelength of the light. And you shouldn't be thinking the fringe affects the velocity, the velocity affects the shift in the fringe. If light were traveling in a medium, then along the interferometer arm pointing in the direction of the Earth's motion the velocity of light would be c+v in one direction and c-v in the other. The average velocity along that path of the interferometer is then less than c - can you show this?

 

[Benzoate]

I think the L in your formula is not the wavelength of the light. It's the total path length of each interferometer arm. The wavelength of light enters via the delta(t). delta(t)=w/c, where w in the wavelength of the light. And you shouldn't be thinking the fringe affects the velocity, the velocity affects the shift in the fringe. If light were traveling in a medium, then along the interferometer arm pointing in the direction of the Earth's motion the velocity of light would be c+v in one direction and c-v in the other. The average velocity along that path of the interferometer is then less than c - can you show this?

Yes you are correct in that L is the path length of the interferometer.
I thought the wavelength was equal to c/v. Perhaps you are saying that I need to find the wavelength of light , in order to find the speed of the Earth motion



What does the average velocity being less than c have to do with the finding the velocitity

v= sqrt(((c^3)*delta(t))/(L)) . How do I find delta t and L?

in order to find the velocity

 

[Dick]

If a signal travels down one path of length L with speed v1 and another with speed v2 then the time difference is L/v1-L/v2. If this difference is half the period of the light wave then they will interfere destructively and cause a dark band. You don't find L. It's given by the design of the interferometer. Nor do you find delta(t), that's follows from the wavelength of the light you use in the interferometer as I've already explained. You find only v.

 

[Benzoate]

If a signal travels down one path of length L with speed v1 and another with speed v2 then the time difference is L/v1-L/v2. If this difference is half the period of the light wave then they will interfere destructively and cause a dark band. You don't find L. It's given by the design of the interferometer. Nor do you find delta(t), that's follows from the wavelength of the light you use in the interferometer as I've already explained. You find only v.

I think I understand now. L/v1-L/v2=delta(t)=L*v^2/c^3 , now it will be easier to cancel out the L to find v(1) and v(2) because don only variable in the equation for v would be c, and c is of course a constant. There would be no more unkown variables.

 

[learningphysics]

So is L/2 the one way path and L is the round-trip path?

The fringe shift is just \delta{t}/T where T = \lambda/c

 

[Dick]

So is L/2 the one way path and L is the round-trip path?

The fringe shift is just \delta{t}/T where T = \lambda/c

Yes to all. I think I did say "total path length".