Optical path

A diagram of the interferometer is in Figure 5.1. Diffuse light from a monochromatic source illuminates a beamsplitter, which is a partially reflecting mirror. About 40 percent of the the light reflects from the first surface of the beamsplitter, travels upward through the compensation plate, reflects again off of mirror 1, passes back through the compensation plate, through beamsplitter and out of the interferometer into the lens of your eye. Light from the source also passes directly through the beamsplitter, through the gas cell, reflects back from mirror 2, passes again through the gas cell, through the glass of the beamsplitter, reflects off the face of the beamsplitter, once more through the glass of the beamsplitter and out of the interferometer into the lens of your eye.

The beam which travels in arm 2 to mirror 2 and back passes through glass in the beamsplitter three times. The compensation plate in arm 1 adds two extra passes through glass for the beam to mirror 1 and back. If it were not for the gas cell, the two arms of the interferometer would provide completely equivalent paths for the light. The reflected image of mirror 2 appears close to mirror 1. Imagine that the path length in arm 1 is $\ell_1$ and that the path length in arm 2 is $\ell_2$. If $\ell_2 > \ell_1$ the image of mirror 2 is above mirror 1 in the figure. In that case, when the tilts of the mirrors are adjusted so that the image of 2 is parallel to 1, there will be a circular dark interference rings whenever the path lengths permit destructive interference. Dark rings will appear wherever

$$m \lambda = 2 (\ell_2 - \ell_1) \cos(\theta_m)$$ (51)

is satisfied. The wavelength of the light is $\lambda$. The order of interference (the number of wavelengths in the path length difference) is $m$, and the angular radius of the ring is $\theta_m$.

As an aside, the reason this equation describes dark rings, rather than bright rings, is that there is a phase change of $\pi$ whenever light reflects off a surface of a material with an index of refraction greater than that for the material the light is traveling in. So there is a phase change of $\pi$ for the beam in arm 1 at its reflection from the beamsplitter and mirror 1 for a total change of $2\pi$. The beam in arm 2 changes phase only on reflection at mirror 2, and not at the beamsplitter because there it is traveling inside the glass at the time of reflection. The total change in arm 2 is $\pi$, and the difference in phase between the two arms is just $\pi$. This means that if the physical path lengths are identical, the waves will still arrive out of phase and there will be destructive interference.

Figure 5.1: A cell containing pressurized air changes the optical path along one arm.

In one arm we have a cell of length $x$, filled with gas (in our case it is air) of index of refraction $n$. The number of wavelengths of light which the path through the cell represents is $x/\lambda_{cell}$ where $\lambda_{cell}$ is the wavelength of light in the cell. Since the speed of light is less in the cell than in empty space, the wavelength in the cell is $\lambda_{cell} = \lambda/n$ where $n$ is the index of refraction. As a consequence, the cell pathlength measured in wavelengths of light is $x / \lambda_{cell} = x / (\lambda/n) = (nx) / \lambda$. We refer to the quantity $nx$ as the optical path length in the cell.

The index of refraction of a gas depends primarily on its composition and its density. For air at a constant temperature and humidity, a good approximation is

$$n = 1 + k p$$ (52)

where $p$ is the pressure, and $k$ is a constant.

At the center of the ring pattern there will be a dark spot with order of interference $m$ whenever, from Eq. 5.1 with $\theta =0$,

$$m \lambda = 2 (\ell_2 - \ell_1)$$ (53)

It follows that as the pressure in the cell is changed, its optical path length will change, and according to either Eq. 5.1 or Eq. 5.3 the fringe pattern will change.

The difference in optical path lengths for the two arms is

$$\ell_2 - \ell_1 = nx + x_0$$ (54)

where $x_0$ is a constant that depends on the position of the mirrors, and the optical properties of the beamsplitter and compensation plate. By substituting Eq. 5.4 and Eq. 5.2 into Eq. 5.3 we get an expression for the dependence of the interference on pressure

$$m \lambda = 2 ((1 + k p)x + x_0)$$ (55)

From the measurement of the changes in the interference pattern when the pressure is changed we will be able to determine the index of refraction constant $k$, and thus the index of refraction of air.