Michelson Interferometer problem

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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.