# Explanation of Michelson Interferometer Fringe Shift

I have been working on an experiment where 2 glass microscope slides are pinched together at one end (so that there is a "wedge" of air between them) and placed in the path of a laser in one leg of a Michelson interferometer. When I move the glass slides (fractions of a mm at a time) so that the path of the laser is closer or further from the place where the slides are pinched, a fringe shift occurs. I cannot seem to explain why this is happening! Any help with explaining this phenomenon would be greatly appreciated! If any more specifics about the setup or dimensions of the slides are needed, please let me know.Also, a full "light to dark" fringe shift occured roughly every 4mm of moving the slides.

• What did you do to solve your homework? – CuriousOne Jun 8 '15 at 1:05
• Are you using a beam that makes multiple reflections inside the wedge? – DJohnM Jun 8 '15 at 6:44
• How is the flatness of your glass plates; if you use a single plate (instead of a pinched pair) do you see the same effect? Do you see a similar effect if you tilt the glass plates? It isn't impossible that the constrained layer of gas between the glass plates has a slightly different index of refraction, but a back of the envelope estimation puts it at $10^{-8}$ m OPL change which is probably at the edge of your instrument's sensitivity. – Chris Mueller Jun 8 '15 at 12:31
• Actually, the fringe shift didn't seem to occur with many pairs of slides, but with some of them we got a near constant fringe shift as we slowly moved the slides in the beam. Im also not sure about the flatness of the slides. How would that create a fringe pattern that was nearly constant? i.e for every mm we moved the slides the pattern would move by 1/4 fringe or so – lander Jun 9 '15 at 2:10
• Are your fringes linear or circular? – WhatRoughBeast Jun 11 '15 at 1:39

This is speculating - but if your slides are of non-uniform thickness, or they are bent as a result of the pinching, they will present a different path length in one leg of the interferometer (and therefore give rise to a shift in the fringe pattern). This may become clear by looking at this diagram:

In the diagram on the left, the total path length is independent of the position of the ray - in all cases the light bends by the same amount as it interacts with the different surfaces. In the diagram on the right, the rays closer to the "pinch point" will traverse less glass than the ones that are further away (which intersect the glass at a greater angle). This means that the path length will change as you move the slides left to right.

It is not clear whether there is a spacer as part of your "wedge" (I imagine there must be one, but I can't see it in your photo). If there is, then the big clip you use will surely bend the slides; and a Michelson interferometer is very, very sensitive to path length differences...

• There is actually not a spacer, but I imagine the slides could still be bent by the pressure and maybe some grime or particles stuck to the inside of the slides. Is there any way to estimate the amount the slides would need to bend in order to see this quantity of fringe shift? Sorry I don't have much info about how the slides are pinched. The slides are each about 1mm thick with refractive index around 1.51-1.54 – lander Jun 11 '15 at 3:57

At a guess, the effect rises from the fact that your interferometer is not properly aligned. The presence of linear, rather than circular, fringes suggests that there is an angular misalignment. Then moving the wedge causes a lateral shift in the intersection point of the beam and the angled slide, which results in a shift in the apparent position of the beam at the target.

Try aligning the system to produce a bull's eye fringe pattern, and see if the anomaly persists.

• I am pretty sure the interferometer was aligned, since we tried very carefully to get the fringe pattern to show up clearly, and there was never any alignment that made a circular pattern. Now that I think of it, one setup with a lens made a circular pattern. Why is it necessary to have a circular pattern though? – lander Jun 11 '15 at 16:53
• Getting the fringes clear is not alignment. For off-center (vertical) fringes, think of two sources which are laterally offset. If the offset changes, the location of the fringe (same path length from both sources) will change as well. And if the wedge beam is not perfectly perpendicular to its mirror, the location of the return bean will vary with wedge location. See, for instance, en.wikipedia.org/wiki/Michelson_interferometer, fig 3. – WhatRoughBeast Jun 11 '15 at 18:15
• Yeah, the bullseye pattern occurs when you have gotten the angles on the mirrors right. You may want to adjust those and the angle of the beam to produce the bullseye, then put the slides in preserving the bullseye. – Alan Jun 11 '15 at 18:28

This answer is based on Floris' insight that the slides might be bent.

Let's say the laser hits the slide at an angle of $\theta$ and travels through the panel at an angle of $\theta'=\sin^{-1}({n_a\over n_s}\sin(\theta))$. Let's assume the curvature is light enough that the laser essentially exits parallel to how it entered. I am also going to assume you are using a $\lambda=633\mathrm{nm}$ laser. We can express the change in phase from having no slides as $\phi={4\pi D\over\lambda}\sec\theta'\left({n_s\over n_a}-\sec(\theta'-\theta)\right)$ using a little trigonometry, where $D$ is the thickness of a slide. We want the curvature which is ${d\theta\over ds}$, where $ds=\sec\theta\,dx$. Let's express ${d\theta\over ds} = \left({d\phi\over d\theta}\right)^{-1} \left({d\phi\over dx}\right) \cos\theta$. You measured $d\phi\over dx$ to be ${2\pi\over 4\mathrm{mm}}$, and specified $n_s=1.52={n_s\over n_a}$. Let's say our angle of incidence is 10 degrees. We will now give this mess to Wolfram.

So at this point, we have to have a curvature of .06 degrees per mm to observe this fringe effect.

• Or perhaps I should say this is maybe a nonanswer? .6 degrees per cm is pretty observable to the naked eye aided by ruler, so the fringe effect may be due to something else (assuming my analysis wasn't faulty) – Alan Jun 11 '15 at 9:37
• Hmmm, very interesting. It definitely wasn't visibly bent, and i could barely fit a razor blade edge into the widest part between the glass slides. – lander Jun 11 '15 at 13:35

In the 2 glasses there are 4 surfaces, i.e interface air/glass, and 8 surface orientations (a..h) and plenty room for interference between reflections and the main beam.

LASER (air) a1b (glass) c2d (air) e3f (glass) g4h

At each interface the is reflection that will be reflected forward again (self-interference) look for iridiscence in thin materials.

previous answer, not important now: Evaluate the dimensions of the screw thread that moves the clamp (distance, pitch) and how many turns are required for a given deviation. I do not see the screw, but I imagine that it exists and influences the outcome.

• It was set up on a linear translation stage, so I already know it took about 2mm of movement to get a full fringe shift. I was wondering why the fringes are even shifting at all – lander Jun 13 '15 at 8:13
• A photo would be nice. I'll change my answer to include interferences. – Helder Velez Jun 13 '15 at 10:24
• What would you like a photo of? – lander Jun 14 '15 at 23:01
• of the set up (at right) that holds the glasses. ( what is the distance beween the glasses ? ) – Helder Velez Jun 15 '15 at 1:13
• I didn't measure the distance between the glasses, and unfortunately I don't have another good picture of the clamp holding them. It was just a binder clip from the store that was squeezing one end together – lander Jul 3 '15 at 2:53