How Many Fringes Shift in a Michelson Interferometer When Air is Evacuated?

[Skye77]

Here's my problem:
One of the arms of Michelson interferometer without the compensator consists of a cell 7.5 cm long. How many fringes would be observed shifting if all the air were evacuated from the cell? Use n_vac= 1, n_air= 1.00029 and the source wavelength 5,000 Angstoms (1 Angstrom = 10^-10m) *Please see attached diagram for clarification*

I'm so confused as to what my teacher is asking. He hasn't covered Michelson's experiment and it's not in my textbook.

Also, I was given a hint: The light will pass twice through the cell due to the round trip. The answer will be obtained by counting the number of wavelengths to fit into the path difference between with and without air paths.

This was all he gave me. English is not his first language, so I'm not sure if I'm missing something in the translation, but any direction in how to solve this would be greatly appreciated.

Thanks.

 

[Delphi51]

A certain number of wavelengths will fit in the 15 cm when n = 1.00029.
A different number will fit in the 15 cm when n = 1 (no air).
You are asked for the difference between these two numbers.

The wavelength is given for air. Look for a refraction formula that you can use to calculate the wavelength when n = 1. The formula will have two n's and two λ's in it.

 

[Skye77]

I'm still confused. :(

 

[Delphi51]

The wavelength is 5,000 Angstoms. How many of these fit in 15 cm?

It is like saying you have bricks 5 cm wide. How many will fit across your driveway which is 3 m wide?

 

[Skye77]

Got it! Thanks so much for the help. I finally figured out what I was doing wrong. The difference is 87. I was forgetting to multiply the results by 1.00029 for the cell. Once I did that I got 300,087 for number of wavelengths through the cell versus 300,000 for the number of wavelengths through air where n= 1.0.