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Michaelson Interferometer and changing length of glass rod

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data
    The index of refraction of a glass rod is 1.48 at T=20°C and varies linearly with temperature, with a coefficient of 2.5 E -5 °C-1. The coefficient of linear expansion of the glass is 5 E -6 °C-1. At 20.0°C the length of the rod is 3.00 cm. A Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 °C/min. The light source has wavelength λ = 599nm, and the rod initially is at 20°C.

    How many fringes cross the field of view each minute?

    2. Relevant equations

    y = [itex]\frac{mλ}{2}[/itex]
    Where y is the change in distance.
    m is the number of fringes.
    λ is the wavelength
    (just for simplicity)
    α = 5 E -6 °C-1
    β = 2.5 E -5 °C-1

    3. The attempt at a solution
    I've attempted this problem several times. I had gotten it correct once before, but can't seem to do it this time (practicing it for my upcoming final).
    First, I find the change in distance. 5α since the temperture rises by 5°C in one minute.
    Now, the change in the index of refraction: 5β.
    Since the equation deals with the change in distance rather than the total distance, we plug in to the original equation.

    (5α) = .5m(λ/(n*5β) or, solving for m
    m = [itex]\frac{2*5α cm*1.48*5β}{599nm}[/itex]
    Make the units equal, and the answer is wrong.

    So I repeated my steps, this time setting the temperture to 25°C. No dice.
    Tried multiplying the length to 5α and both methods (°C = 5 and 25). But Nope.avi

    The answer should be around 14 I believe. Some guidance and explanation would be much appreciated.
     
  2. jcsd
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