What is the index of refraction of a gas in an interferometer?

[Evergreen]

Interference question...

One of the beams of an interferometer, as seen in the figure below, passes through a small glass container containing a cavity D = 1.19 cm deep.

http://capa1.physics.sunysb.edu/CAPA/Pictures/gian2450.gif

When a gas is allowed to slowly fill the container, a total of 209 dark fringes are counted to move past a reference line. The light used has a wavelength of 594 nm. Calculate the index of refraction of the gas, assuming that the interferometer is in vacuum.


??some help

 

[Doc Al]

phase difference

As you add the gas, since it has an index of refraction greater than that of the empty cavity (assume a vacuum), you are essentially phase shifting the light in that leg of the interferometer. For every λ of shift, you will seen a fringe move past the reference line. To figure this out, you must calculate how much of a shift you get by replacing the vacuum with the gas.

Hint: compare the number of wavelengths in the cavity when it is filled with gas to the number when it is empty (filled with vacuum). Remember that the light passes through the cavity twice.

 

[M98Ranger]

The index of refraction of the gas can be calculated using the formula n = (N + ΔN)/D, where n is the index of refraction, N is the number of fringes counted, ΔN is the change in number of fringes, and D is the depth of the cavity. In this case, N = 209, ΔN = 0 (since the reference line is not moving), and D = 1.19 cm. Plugging in these values, we get n = (209 + 0)/1.19 = 175.63. This means that the index of refraction of the gas is 175.63 times that of vacuum. Since the index of refraction of vacuum is 1, the index of refraction of the gas would be approximately 175.63. This value may vary depending on the type of gas used in the experiment.