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modphysnoob

  • one year ago

2. In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the measurable fringe shift was 0.005 fringes. If sodium light was used (λ = 589nm), what upper limit did the null experiment place on the speed of the Earth through the expected ether?

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  1. modphysnoob
    • one year ago
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    http://physicsweb.phy.uic.edu/244/Phys244notes_wk10.pdf

  2. modphysnoob
    • one year ago
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    t1= l/ (c-v )+ l/ (c+v)

  3. Jemurray3
    • one year ago
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    Right, so you need to find the phase difference as a function of velocity. If a river flows downstream at speed u, and I can paddle my boat and a speed c relative to the river, how long does it take me to go a distance L and back?

  4. Jemurray3
    • one year ago
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    at a speed c*

  5. modphysnoob
    • one year ago
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    l/c+v

  6. Jemurray3
    • one year ago
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    To summarize, the guy who has to go upstream and downstream takes a time \[ t = \frac{2L c}{c^2-u^2} \] and goes through a number of cycles \[ \frac{t}{T} = \frac{2L}{\lambda} \cdot \frac{c^2}{c^2 - u^2} \] The guy who is unaffected takes time \[ t' = \frac{2L}{c} \] And goes through a number of cycles \[\frac{t'}{T} = \frac{2L}{\lambda} \] Everything else was a typo -- I got my c's and u's mixed up, sorry.

  7. modphysnoob
    • one year ago
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    so we are trying to find number of oscillation?

  8. Jemurray3
    • one year ago
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    We're more interested in the difference between the two.

  9. Jemurray3
    • one year ago
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    The phase difference is the difference between those two quantities times 2 pi (if the difference is zero, that's fine. If it's one, that means they are 1 whole wavelength or 2pi radians out of phase). So, the phase difference is \[ \Delta \phi = 2\pi \frac{2L}{\lambda} \frac{u^2}{c^2-u^2} \]

  10. Jemurray3
    • one year ago
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    But the main point of your question is this: The "fringe shift" is just the difference in number of cycles between the two. Therefore, \[ N = \frac{2L}{\lambda} \frac{u^2}{c^2-u^2} \] if we assume that the speed of light is immensely greater than the speed with which we move through the aether, \[N \approx \frac{2L}{\lambda} \frac{u^2}{c^2} \] given N, which for you is .005, we can find u: \[ u = c \sqrt{ \frac{N\lambda}{2L} } \] Which is a few hundred meters per second, if I'm estimating that correctly.

  11. modphysnoob
    • one year ago
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    Thanks you for you help, we just started relativity this week and it's very tricky

  12. Jemurray3
    • one year ago
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    No problem. Just as a caveat -- I lied just a little bit during that derivation (in particular, the part where I calculated the travel time for the guy traveling across, rather than up or downstream) but in the end it doesn't meaningfully affect the calculation and you'd get the same answer.

  13. modphysnoob
    • one year ago
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    but wouldn't guy going across be affected by river current as well since he have to go diagnally? since|dw:1365735827635:dw|

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