# Michelson inteferometer

1. Sep 20, 2008

### Physicsissuef

Hello people!

I got big problem.

$$l=\frac{ct_n}{2}$$ - the light in one way.

Here are the pictures, I think you'll see what I am talking about.

http://img218.imageshack.us/my.php?image=picture001id5.jpg

http://img135.imageshack.us/my.php?image=picture002tx5.jpg

Thank you!

Last edited by a moderator: Apr 23, 2017
2. Sep 20, 2008

### Redbelly98

Staff Emeritus
This equations says (apparently) that light travels a distance 2l. A-to-M2, and M2-to-A1, are each a distance l.
Perhaps they are assuming very small angles, so that M2-to-B is approximately l also. You're right, they can't all be exactly equal to l.

3. Sep 21, 2008

### Physicsissuef

Why they aren't using Pitagorean theorem? If they use Pitagorean theorem, "x" or "l" (as they say) would be $$\frac{t_nc\sqrt{1-\frac{v^2}{c^2}}}{2}$$, right?

4. Sep 21, 2008

### yuiop

$$l = \frac{t_nc\sqrt{1-\frac{v^2}{c^2}}}{2} \quad _{(1)}$$

to get

$$t_n = \frac{2l}{c}\left(1-\frac{v^2}{c^2} \right)^{-1/2} \quad _{(2)}$$

then that is exactly what I see in the photographed text at the bottom of the your first image.

Last edited: Sep 21, 2008
5. Sep 21, 2008

### Redbelly98

Staff Emeritus
Also: the red "l"s in the first image are incorrect. Those distances are greater than l.

edit:
This contradicts my earlier post. I've since realized that the figure is in the reference frame of the "ether", in which those distances (A-to-M1) are greater than l. They are equal to l in the lab reference frame only.

Last edited: Sep 21, 2008
6. Sep 21, 2008

### Physicsissuef

I think the second picture is where the Michelson interferometer is in the ether. So the real distances are "l", but when the Michelson interferometer is on the earth, also there is effect from the ground, and the light "turns" by some angle, so those "l"-s on the first pictures are actually greater then "l", and the normal distance is "l".