Calculating the index of refraction of a gas

[mrbling]

Question: One of the beams of an interferometer, as seen in the figure below, passes through a small glass container containing a cavity D = 1.40 cm deep.

When a gas is allowed to slowly fill the container, a total of 230 dark fringes are counted to move past a reference line. The light used has a wavelength of 600 nm. Calculate the index of refraction of the gas, assuming that the interferometer is in vacuum.

I'm not even sure where to start with this question.. any help?

 

[swansont]

Question: One of the beams of an interferometer, as seen in the figure below, passes through a small glass container containing a cavity D = 1.40 cm deep.

When a gas is allowed to slowly fill the container, a total of 230 dark fringes are counted to move past a reference line. The light used has a wavelength of 600 nm. Calculate the index of refraction of the gas, assuming that the interferometer is in vacuum.

I'm not even sure where to start with this question.. any help?

The wavelength of light in a medium is lambda/n, so the total number of wavelengths contained in the region with the gas gets bigger. IOW, its optical path length gets longer. This is what causes the fringes.

 

[mrbling]

swansont,

thanks for your response..

from the information given, I know that:

dark bands occur at: 2t=m lambda (where t is thickness)
m(order) =230
lambda(in vacuum) = 600nm

from what you mentioned, the light going through the gas chamber will have change wavelengths.. and that wavelength is proportional to the index of refraction (via: lambda(in gas) = lambda(in air)/n)

so my question is: how does the 1.4cm cavity figure into things?
Thanks

 

[swansont]

swansont,
so my question is: how does the 1.4cm cavity figure into things?
Thanks

How many wavelengths fit into the cavity before and after the gas is introduced? That's the reason for the fringes, so the length has to be known.