# How is a one position shift of an interferometer fringe pattern defined?

When Michelson and Morley conducted their 1887 interferometer experiment, they were expecting a fringe pattern shift of 0.4 (see the chart at http://en.wikipedia.org/wiki/Michelson-Morley_experiment). I understand that fringe patterns are represented as series of alternating dark and light bands, but I don't understand what exactly one fringe is. In particular, would a fringe shift of 1.0 be defined as a dark band moving (shifting) to where its adjacent dark band had previously been?

To check my understanding:

• If so, then if a laser with a 532 nanometer wavelength were used as the light source in a Michelson-Morley style interferometer, would adjusting the position of the movable mirror a total of 266 nanometers (266 rather than 532 because the beam reflects back off the mirror) result in a 1.0 fringe shift as previously defined?
• Or, since light waves are sinusoidal, would the dark bands occur at every crossing of the x-axis (and the light bands at the peaks and valleys), such that the 1.0 fringe shift would occur after moving the mirror only 133 nanometers - rather than the 266 nanometers of the previous question?
• Not sure what's with all the close votes. Do we not allow experimental questions any more? – user10851 Jun 10 '13 at 15:46

The electromagnetic field of a single-wavelength beam (e.g. that of a laser) is sinusoidal with period $\lambda$.

e.g. for one dimension E field can be described as follows (ignoring time evolution): $$E = \cos\left(\frac{2\pi}{\lambda}z\right)$$

If you have a second beam that is mutually coherent with the first, the two E-fields can be described as follows:

$$E_1 = \cos\left(\frac{2\pi}{\lambda}z\right) = \frac{e^{2\pi i z/\lambda} + e^{-2\pi i z/\lambda }}{2}$$ $$E_2 = \cos\left(\frac{2\pi}{\lambda}z + \phi \right) = \frac{e^{i(2\pi z/\lambda + \phi)} + e^{-i(2\pi z/\lambda + \phi)}}{2}$$

The combined electric field is thus $$E = E_1 + E_2 = \frac{e^{2\pi i z/\lambda} + e^{-2\pi i z/\lambda }}{2} + \frac{e^{2\pi i z/\lambda}e^{+i\phi} + e^{-2\pi i z/\lambda }e^{-i\phi}}{2}$$

This can be regrouped as $$E = \frac{1}{2} \left(e^{2\pi i z/\lambda} (1+e^{i\phi}) + e^{-2\pi i z/\lambda}(1+e^{-i\phi}) \right)$$

In this last form, it is evident that if your phase shift is $\pi$ you will have $E = 0$ ($e^{i\pi} = -1$). If your phase shift is $2\pi$ the magnitude of the $E$ field will be that of the combined $E$ fields ($e^{2\pi i} = 1$).

In the above formulation, $\phi$ represents a phase difference between the two beams, which results from a path length difference. It is this phase difference that you care about.

The phase difference is a direct function of the path length difference $\Delta z$: $$\phi = \frac{2\pi\Delta z}{\lambda}$$

A phase difference of $\phi = 2 \pi$ (resulting from $\Delta z = \lambda$ will be the same as if $\Delta z = 0$. Thus, a path length difference of $\lambda$ will result in a full cycle of your fringe pattern: starting bright, then dark, then bright again at $2\pi$.

In summary, a shift of one wavelength will correspond to one fringe.

Michelson and Morley used both partially monochromatic interferometry (using Sodium lines) as well as white-light interferometry. Things become tricky with white-light interferometry because each wavelength of the spectrum will have a difference fringe spacing (because $\lambda$ is different).

• Thank you. I did also find the following link which, although not mathematical, corroborates your explanation that each full positional fringe band shift corresponds to one full wavelength shift. Therefore, adjusting the movable mirror in a Michelson-Morley style interferometer (which is using a 532 nanometer laser) by a total of 266 nanometers would result in a 1.0 fringe pattern shift. – Carl Reiff Jun 5 '13 at 4:28