Suppose a transparent vessel \(30.0 \mathrm{cm}\) long is placed in one arm of a Michelson interferometer, as in Example 25.2. The vessel initially contains air at \(0^{\circ} \mathrm{C}\) and 1.00 atm. With light of vacuum wavelength \(633 \mathrm{nm}\), the mirrors are arranged so that a bright spot appears at the center of the screen. As air is slowly pumped out of the vessel, one of the mirrors is gradually moved to keep the center region of the screen bright. The distance the mirror moves is measured to determine the value of the index of refraction of air, \(n .\) Assume that, outside of the vessel, the light travels through vacuum. Calculate the distance that the mirror would be moved as the container is emptied of air.

First find the initial path difference in the air-filled container using the given index of refraction of air \(n\), the vessel length \(L\), and the vacuum wavelength \(\lambda\). The path difference in air can be found by multiplying the length \(L\) by the index of refraction \(n\). Initially, the container is filled with air at \(0^{\circ} \mathrm{C}\) and 1.00 atm, which gives \(n \approx 1.000292\). So, the initial path difference in air will be: $$ \Delta x_{air} = nL $$

To find the number of wavelengths that fit into the path difference we found in Step 1, we'll divide the path difference in air by the vacuum wavelength \(\lambda\). The integer number of wavelengths \(m\) is given by: $$ m = \frac{\Delta x_{air}}{\lambda} $$

Now we'll calculate the path difference in vacuum after the air is pumped out. As we know, the path difference should still be an integer number of wavelengths. Therefore, we can write the following equation: $$ \Delta x_{vacuum} = m\lambda $$

Finally, we want to find the difference between the initial path difference in air and the final path difference in vacuum. This difference will be the distance the mirror must be moved in order to maintain the same number of wavelengths. The distance the mirror should be moved \(d\) can be found by: $$ d = \Delta x_{air} - \Delta x_{vacuum} $$ Now let's calculate the distance using the provided values: \(L = 30.0 cm = 0.3 m \), \(\lambda = 633 \mathrm{nm} = 633 \times 10^{-9} m \), and \(n \approx 1.000292\).

1. First, calculate the initial path difference in air: $$\Delta x_{air} = nL = (1.000292)(0.3 m) \approx 0.3000876 m$$ 2. Find the integer number of wavelengths in the initial path difference: $$m = \frac{\Delta x_{air}}{\lambda} \approx \frac{0.3000876}{633 \times 10^{-9}} \approx 474147$$ Here, we'll round \(m\) up to the nearest integer since we cannot have a non-integer number of wavelengths. 3. Calculate the final path difference in vacuum: $$\Delta x_{vacuum} = m\lambda = (474147)(633 \times 10^{-9} m) \approx 0.300079071 m$$ 4. Finally, calculate the distance the mirror would be moved: $$d = \Delta x_{air} - \Delta x_{vacuum} = 0.3000876 m - 0.300079071 m \approx 0.000008529 m$$ So, the distance the mirror would be moved as the container is emptied of air is approximately \(8.529 \times 10^{-6} m\) or \(8.529 \mathrm{\mu m}\).