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Michelson interferometer

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data

    In one arm of a Michelson interferometer the light passes through a glass tube of length L which contains air. The air is pumped slowly out of the tube and during this process it is found that the intensity of light reaching the eye has increased and decreased n times. What is the velocity v of light in air? Express your answer in terms of ,L n, c, λ(the wavelength of light). You may assume (correctly) that the difference between v and c is small.

    2. Relevant equations



    3. The attempt at a solution
    Here is the solution
    [img=http://s24.postimg.org/4elfpk629/Untitled.jpg]


    For the delta t, the decrease in time when the light Is pumped out, I am having trouble understanding why the equation Is 2(L/v – L/c). If we just looked at when the light traveled up there would be moments where the light would be traveling at a speed of c and other moment where the light would be traveling at a speed v. Wouldn’t the time difference be x¬1/v – x2/c where x1+ x2 = L and x1is the distance the light travels at a speed of v when the gas is still in the tube, and x2 the distance that the light travels going up the leg at a speed c when the gas is pumped out. Why is it not the time difference, and why is it so simplified. And the solution is correct because it is a solution to my physics problem set from 2 weeks ago, but the professor never explained it, and it is spring break now so I can’t ask him.
     
  2. jcsd
  3. Mar 26, 2013 #2

    mfb

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    Staff: Mentor

    You are not interested in the total time the light needs - you always have the other arm of the interferometer as reference. The relevant quantity is the additional time the light needs in air: It needs L/v if air is present, and L/c if there is no air. Therefore, the difference is L/v-L/c.
     
  4. Mar 26, 2013 #3
    Thank you very much for your response.
     
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