Calculating Refractive Index Using Fabry-Perot Interferometer

AI Thread Summary
The discussion focuses on calculating the refractive index of air using a Fabry-Perot interferometer with a spacing of 2 cm and a wavelength of 5 × 10^−7 m. As air is pumped out, the beams undergo 23 cycles of interference, indicating a change in the integer m by 23. The equation used is 2d = mλ/n, leading to the conclusion that n = 1 + 23λ/2d. The numerical result for n-1 is approximately 0.0002875, which aligns closely with standard values for the refractive index of air at STP. The calculations confirm the theoretical understanding of how refractive index changes with air density.
HotMintea
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Homework Statement



A Fabry-Perot interferometer has spacing d = 2 cm between the glass plates, causing the direct and doubly reflected beams to interfere. As air is pumped out of the gap between the plates, the beams go through 23 cycles of constructive-destructive-constructive interference. If the wavelength of the light in the interfering beams is 5 × 10^−7 m, determine the index of refraction of the air initially in the interferometer.

Homework Equations



(I think) If constructive interference is occurring, 2d = mλ/n, for wave length λ, refractive index n and some integer m.

The Attempt at a Solution



I cannot figure out how to incorporate the following part in the equation: "As air is pumped out of the gap between the plates, the beams go through 23 cycles of constructive-destructive-constructive interference."
 
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HotMintea said:
...
I cannot figure out how to incorporate the following part in the equation: "As air is pumped out of the gap between the plates, the beams go through 23 cycles of constructive-destructive-constructive interference."

It means that m changes by 23.

Can you say which it did? Did m increase? or Did m decrease ??

Also, assuming that virtually all of the air was pumped out, what is the index of refraction for a vacuum ?
 
SammyS said:
It means that m changes by 23.

Can you say which it did? Did m increase? or Did m decrease ??

Since wavelength gets longer as the air gets thinner, I think the number of wavelengths in the distance 2d will decrease as the air gets pumped out.

SammyS said:
Also, assuming that virtually all of the air was pumped out, what is the index of refraction for a vacuum ?

The index of refraction in vacuum is defined to be 1.

I have: m λ/n = (m - 23) λ/1 = 2d. The left side says integer m times the contracted wavelength λ/n. The middle says (m - 23) times full length λ/1. The right side is twice the gap.

Solving the left and middle, I get m = (2d + 23λ)/λ. Substitute it for m in the left and solving the left and right, I get n = 1 + 23λ/2d.

Thanks for your help!
 
Last edited:
What do you get for a numerical answer for n-1? (Just out of curiosity .)
 
Excellent !
 

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