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Michelson Interferometer - calculating errors

  1. Apr 20, 2014 #1
    1. The problem statement, all variables and given/known data

    2dng9y.png

    Part (a): Find an expression for intensity, interference phase and total change of interference phase.

    Part (b): Find an expression for <L1>/L0 and its error.

    Part (c): Find an expression for Δø1 and proportionality constant.

    Part (d): Find an expression for n(0) and α.

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    [tex]v_t = (<v> - \frac{\Delta v}{2}) + \Delta v(\frac{t}{T})[/tex]

    The phase difference is given by ##\Phi_0 = \frac{2\pi}{\lambda} (2L_0) = 4\pi v_t L_0##.

    [tex]\Phi_0 = 4\pi L_0 \left[ <v> + \frac{\Delta v}{2}(\frac{2t}{T} - 1)\right][/tex]

    Change in phase:
    [tex]\Delta \Phi_0 = \Phi_{0(T)} - \Phi_{0(0)} = 4\pi L_0 (\Delta v)[/tex]

    Part (b)
    [tex]\Delta \Phi_0 = 4\pi L_0 (\Delta v)[/tex]
    [tex]\Delta \Phi_1 = 4\pi L_1 (\Delta v)[/tex]

    Therefore,##\frac{<L_1>}{L_0} = \frac{\Delta \Phi_1}{\Delta \Phi_0} = \frac{\sigma_{(\Delta \Phi_1})}{\sigma_{(\Delta \Phi_0)}}##

    To find the change of this, consider ##x = \frac{\Delta y}{\Delta z}##, ##dx = (\frac{\partial x}{\partial y})dy + (\frac{\partial x}{\partial z})dz##. Therefore ##\sigma_{(\frac{<L_1>}{L_0})} = 2 \frac{\sigma_{(\Delta \Phi_1)}}{\sigma_{(\Delta \Phi_0)}}##.

    Part (c)

    Now the length is replaced by:
    [tex]L_{1(t)} = <L_1> + \frac{\Delta L_1}{2}\left[ 2\frac{t}{T} - 1\right][/tex]
    [tex] = 4\pi(<L_1> + \frac{\Delta L_1}{2})(<v> + \frac{\Delta v}{2}) - 4\pi(<L_1> - \frac{\Delta L_1}{2})(<v> - \frac{\Delta v}{2})[/tex]
    [tex] = 4\pi\left[\Delta v <L_1> + \Delta L_1 <v>\right][/tex]

    The proportionality constant ##C = \frac{\sigma_{\Delta L_1}}{\Delta L_1}## I have no idea how to find it.

    Part (d)

    Using ##n_i = n(<v> - \frac{\Delta v}{2})## and ##n_f = n(<v> + \frac{\Delta v}{2})##, and since ##v_t## grows linearly with time, therefore ##n_{(\gamma)}## grows linearly with time.

    Thus ##n_{(\gamma)} = n_i + \frac{(n_f - n_i)}{T}t = n_i + \Delta v(\frac{t}{T})##

    Equating to the expression given: ##n_{(\gamma)} = n_{(0)} + \alpha \gamma = n_0 + \alpha \left[<v> + \frac{\Delta v}{2}(\frac{2t}{T} - 1)\right]##:

    [tex] \alpha = \frac{n_i - n_0}{(<v> - \frac{\Delta v}{2})}[/tex]

    To solve for ##n_{(0)}##, we need path difference to balance out:
    [tex] \sigma_{L_1}= C\Delta L_1 = \frac{L_b}{ n_{(0)} + \alpha \gamma = n_0 + \alpha \left[<v> + \frac{\Delta v}{2}(\frac{2t}{T} - 1)\right]}[/tex]

    But this seems strange as there is a time dependence in the denominator.. Also, how do I solve for proportionality constant C in part (c)?
     
  2. jcsd
  3. Apr 21, 2014 #2
  4. Apr 22, 2014 #3
    Part (c) - how do i find the proportionality constant?
    Part (d) - it is weird that I end up with a time dependence in the error of L1
     
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