Michelson Interferometer and white light fringes

1. Jan 30, 2005

Baggio

Hi,

I conducted an experiment with the Michelson Inteferometer to obeserve white light fringes but I cannot explain the presence of a central black fringes. I understand a phase shift of pi is introduced in one of the beams but where/how does this shift occur?

thanks

2. Jan 30, 2005

mathman

The phase shift is simply a result of the fact that the distances to different parts of the screen from the two slits are different. When this difference is n+1/2 (n integer)wavelengths the waves cancel and it shows as black.

3. Jan 30, 2005

Baggio

Michelson interferometer doesn't use slits.. Its uses two mirrors and a beam splitter. Apparently the pi shift is introduced due to reflections but i don't see which reflection causes it

4. Jan 31, 2005

Gonzolo

When you vary the length of an arm, the central fringe should alter between dark and bright.

What kind of beamsplitter do you have? Do you use a compensator?

5. Feb 1, 2005

Baggio

Yes we use a compensator

6. Feb 1, 2005

Gonzolo

What kind of beamsplitter do you use? It is probably either 1) a film on substrate, 2) a film framed by a ring, or 3) a cube.

If it is 1), your compensator has to be the same thickness and material as the beamsplitter substrate, and it has to be in the arm of the film (not the backside).

The length of the arms must also be adjusted of course.

7. Apr 2, 2007

Stainsor

Baggio,

A pi phase shift is introduced into an electromagnetic wave anytime it is traveling through a medium of one index of refraction and reflects off of a surface with a higher index of refraction.

Beamsplitters are made of thin pieces of glass with a coating on each surface that has an index of refraction halfway between that of glass and air. It is this coating off of which the beam reflects. In the Michelson interferometer, the beamsplitter is typically coated such that one side will not reflect light (an Anti-Reflective coating), but the other will reflect light.

Let's assume that you put the reflective side first and we'll examine the two light paths.

Path 1:
The beam will travel through the air and half will reflect off of your beamsplitter. Since Nair < Ncoating there will be a pi phase shift. This half then reflects off of a mirror. This gives another pi phase shift because Nair < Nmirror. To complete its path, this half of the beam will travel directly through the beamsplitter at your screen with a total of a 2*pi phase shift.

Path 2:
The beam will travel through the air and half will pass through your beamsplitter. This half then reflects off of a mirror with a pi phase shift. Then it returns to the beamsplitter from the back side. It enters the beamsplitter, travels through the glass, reflects off the front side and goes to your screen. This time it's travelling in glass and reflecting off air. Nglass > Ncoating so there is no pi phase shift. This half of the beam has a total of a 1*pi phase shift.

Note:
One fact was overlooked in the above discussion: Actually the beamsplitter splits the beams a second time when the return to it. The result is that half of your input beam returns in the direction it originally came from.

8. Apr 3, 2007

lpfr

The Michelson interferometers I know, do not produce fringes but concentric circles when alignment is perfect. If you see fringes, the interferometer is not perfectly aligned and you just see a partial view of the circles. What puzzles me is that you see "a central black fringes" (one or several?). If you see the center, it alternates from dark to light as the arm length varies. In fact, as the arm length varies, you see concentric circles originating in the center that enlarges, or circles that converge and disappear at the center. I do not understand what you saw.

9. Apr 3, 2007

Baggio

Hi, thanks Stainsor... Yep thanks for that if only you replied two years ago :)

Now I have more complicated problems to deal with now I'd doing a PhD in quantum optics :S

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