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Michelson-morley interforometer.

  1. Aug 5, 2007 #1
    I had a question in the exam that asked how much percantege of the photons are detected when the beam splitter, is a symmetrical beam splitter, i.e: |In>->(1/sqrt(2))(|Transmitted>+i|Reflected>)

    well, at first 50 percentage is reflected and 25 goes to the right hand mirror, and quarter goes to the up mirror, afte the mirrors refelct the waves back, 50 percent form each wave is refelceted, and 50 percent goes through, (the detector is placed parallel to the up mirror), 50 percent from the 25 percent goes through which goes downside is getting detected, which mean 1/8 of the photons are being detected, is my reasoning correct?

    if this question is suitable to HW then move it there, im just not sure it's suitable for Advanced physics cause it's in a first year course, but then agian it's not for high school mechanics.

    thanks in advance.
     
  2. jcsd
  3. Aug 5, 2007 #2

    Cthugha

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    No. This is wrong.
    A 50/50 BS splits a beam in one half, which goes to mirror one (50%) and one half, which is reflected and goes to mirror two (50%). It does not split the beam in three parts as you suppose. Where should a reflection back to the source come from?

    So during the second reflection 50% of each beam go to the detector and 50% are directed back to the source. This adds up to a total of 50% being detected.
     
  4. Aug 5, 2007 #3
    Remember that the interference of the recombined beams can be distructive or constructive toward the detector. So the fraction reaching the detector can be 100% or zero or anything in between, depending on the path difference between the two arms of the interferometer. For white light and an arbitrary path difference, the fraction is generally close to 50%.
     
  5. Aug 6, 2007 #4
    now as i look at it, it should be 100 percent.
    here's my calculation:
    |R>->1/(sqrt2)*(|R>+i|U>)->1/sqrt(2)*(|L>+i|D>)->1/sqrt2*(1/sqrt2(|L>+i|D>)+(1/sqrt2)i(|D>+i|L>))=|D>
    which mean it's 100 precent, or am i wrong in my calculations?
     
  6. Aug 6, 2007 #5
    anyway my english perhaps isn't that good but i know that reflected means returned, goes back, and refracted means like in light refraction.

    in the exam, it was written (in english!) reflected!
    how on hell should i know, he means refracted, the TA and the Lecturer who also visited in the exam to answer to questions didn't see this mistake, i guess the other students understood as refracted.
     
  7. Aug 6, 2007 #6

    Cthugha

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    Ok, as there was no path difference given, I assumed that.

    Reflection does not mean going back necessarily. Reflection means that the angle of incidence equals the angle of reflection. Both are 45° here.
    If you shine light on a mirror, which is turned around 45° from your point of view, the reflected light won't return to you as well.
     
    Last edited: Aug 6, 2007
  8. Aug 6, 2007 #7
    what about my post 4 in this thread is my calculation wrong there?
     
  9. Aug 6, 2007 #8

    Cthugha

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    In case, that the paths to both of the mirrors have equal length and the beam splitter is an ideal 50:50 BS, this is true, because there is a phase difference of pi/2 between the split light beams. This is quite remarkable as usual mirrors give a phase difference of pi.
     
  10. Aug 6, 2007 #9
    ****, i could have got a good grade, i hate this when i know the material but get ****ed by trivialities.
    btw, it wasn't given that the mirrors are placed with the same length from the splitter, although the picture does suggest that's the case, i thought you can deduce this solely by the algebra that iv'e done.
     
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