Michelson Interferometer problem

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SUMMARY

The discussion centers on calculating the number of bright-dark-bright fringe shifts observed in a Michelson interferometer as air is admitted into a 2.00 cm glass cell. Operating at a wavelength of 600 nm, the initial number of wavelengths in the evacuated cell is determined to be 33,333. When filled with air, the number of wavelengths increases to 33,344. The correct calculation for fringe shifts is 2(33,344 - 33,333) = 22, although participants caution about potential precision errors in the subtraction process.

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bcjochim07
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Homework Statement


A Michelson interferometer operating at a 600nm wavelength has 2.00 cm glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot ant the center of the interference pattern. Then a valve is opened and air is slowly admitted to the cell. The index of refraction of air at 1 atm of pressure is 1.00028. How many bright-dark-bright fringe shifts are observed as the cell fills with air?


Homework Equations





The Attempt at a Solution



To start out the problem, in order to form a bright spot at the center, I think that there must be an equal number of wavelengths going through both arms.

The number of wavelengths in the evacuted glass tube is (.02/6.0e-7)= 33,333

The wavelength of the light in the air is: (6.0e-7)/1.00028 = 5.998e-7

The length of the other arm must then be adjusted to be:

33333= L2/(5.998e-7) L2= 0.01999 m

When the glass tube is filled w/ air, the number of wavelengths will be (.02m)/(5.998e-7) = 33344 wavelengths.

But I'm not sure how to put this into equation form to calculate the number of shifts I will see. Can someone explain this to me? I'm really having a tough time visualizing
 
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Ok... I think I have it... first I need to multiply both 33333 and 33344 by 2, then I will subtract them.


2(33344-33333) = 22. Is this right?
 
bcjochim07 said:
Ok... I think I have it... first I need to multiply both 33333 and 33344 by 2, then I will subtract them.


2(33344-33333) = 22. Is this right?

The procedure you are following looks right to me; everytime you "fit" in an extra half-wavelength into the air cell you get a transition from bright to dark fringe (or vice versa).

However, numerically I believe you might need to be more careful. Subtracting two numbers that are close together causes a loss of precision. If you let your calculator keep all the digits it can (intead of rounding off during the problem) I think you'll find that the answer 22 has a large percent error.
 
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