# Average number of fringes

Context: Trying to comprehend the Frequency ratio method(fringe counting method) which employs a Michelson interferometer, used for calibrating accelerometers.
Problem:

A Michelson interferometer is depicted above. Its components are the laser source, a fixed mirror, a beam splitter (not depicted, but assume it's there), a detector (photo-diode) and a mobile mirror that's glued to the arm of an shm shaker that produces a sinusoidal displacement (ξ(t)=ξsin(2πft)); the accelerometer is not relevant for my questions.
The concern is to derive the number of averaged fringes that land on the detector. The following formula has been suggested:

Where λ is the wavelength of the laser.

Question: Where do all the multiplicative constants originate from (ξx2x2x2)?

Further info: More details are described in the following reference: G. S. Pineda,L. F. Argote,Vibration measurement using laser interferometry,pp 6-8. It's available online for free.

Thank you!

• What is $R_F$ in your equation? Jun 9 '18 at 15:31
• RF is the number of averaged fringes during several cycles of the smh. ξ is the amplitude of the sinusoidal displacement produced by the shaker. Jun 9 '18 at 15:33
• Just glancing at the diagram and equation, without looking at the reference, I'd guess that the 2's are there because bright fringes occur for both the (+,+) and (-,-) constructive interference of the beams. That is, it's not the amplitude that's detected, but the square of the light amplitude, which doubles the frequency. The xi probably comes from the amplitude of the oscillation of the driven mirror on the accelerometer, and probably depends on the kinematics of the accelerometer: the mass of the mirror, the spring constant, etc. Jun 10 '18 at 14:04

## 2 Answers

Even though I'm not particularly sure, I think this might be an answer.

First let's start with an example: The arm of the shaker, from an equilibrium position, starts to move and spans a distance of 2λ and then halts. Since the periodicity of fringe bright-to-bright or dark-to-dark transition occurs at a mirror path difference of λ/2, we would count 4 fringe transitions.

Let's state this as a formula for this specific case (i.e. the shaker stops, moves and then stops again): $$R_f=\frac{ξ}{\frac{λ}{2}}=\frac{2ξ}{λ}$$ where ξ represents the distance spanned by the moving mirror.

If the shaker moves the mirror back from the distance ξ to the equilibrium position, we would count the same fringes again, therefore we can state that the number of fringes, for this particular case is: $$R_f=2\frac{2ξ}{λ}=\frac{4ξ}{λ}$$ or twice the amount measured in the first scenario.

Now considering the fact that the shaker produces a SHM displacement, we know that it also has to move from the equilibrium position back and span a distance of ξ (or -ξ as a signed distance). Therefore the number of fringes that we would count will be doubled the number of the counted fringes from the second scenario(considering that it moved for a full period), hence the total number of fringes for this scenario would be : $$R_f=2\frac{4ξ}{λ}=\frac{8ξ}{λ}$$

Please correct me if I'm wrong.

NOTE: $R_f$ represents the number of counted fringes under a particular scenario. This notation is consistent with the one used in question post.

During one complete cycle of the shaker, the shaker mirror moves through a distance of $4\xi$, where $\xi$ is the amplitude of the mirror. (From rest at one extreme it moves forward by $2\xi$, stops, reveres direction and moves back by $2\xi$ to its initial position.) However, because the beam is reflected from this moving mirror, the change in path difference between the two beams is twice as much as the movement of this mirror - ie $8\xi$.

The two paths to the detector are in phase and produce a bright fringe when the path difference is a multiple of $\lambda$ (not $\lambda /2$). So the number of fringes passed through during the mirror movement is $$R_F=\frac{8\xi}{\lambda}$$

• Sorry for the error, but I followed youtube video which stated that a λ/2 'path difference' will produce identical patterns . youtube.com/watch?v=j-u3IEgcTiQ . I guess I got 'path difference' confused with 'mirror path'. Jun 10 '18 at 19:16
• That's right. If the mirror moves through a distance of $\lambda /2$ then the optical path difference between the two arms changes by $\lambda$ - so this is a change from eg one bright fringe to the next bright fringe. Jun 10 '18 at 19:29