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Michelson Interferometer

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  1. Sep 26, 2014 #1
    1. The problem statement, all variables and given/known data

    My personal question:

    What does a Michelson interferometer tell us?

    The actual problem:

    A Michelson interferometer is used to precisely measure distances of the order of 1 mm using a laser of wavelength 632.8 nm.

    a) If the motion of one mirror is 1 mm, what is the corresponding optical path length?
    How many fringes would have to be counted?

    b) A more convenient method may be to add a second wavelength which provides another set of independent fringes. Calculate the value of the second wavelength required to produce a maximum fringe coincident with the maximum fringes of the first wavelength at intervals of 0.1 mm and 1 mm. Comment on the practicality of this approach, and on which choice of wavelength would be preferable.

    2. Relevant equations

    d*sin##\theta## = m*##\lambda## for constructive interference, m = 0, 1, 2, ... (?)

    3. The attempt at a solution

    To my own question: I have read about the set-up of a Michelson interferometer. It consists of two mirrors, one fixed and the other movable. There is a third beam splitter mirror positioned in the middle. Incident light is split by the beam splitter, travels to the other two mirrors and back to the beam splitter where they recombine. Depending on the relative distances to each mirror from the beamsplitter, the light can constructively or destructively interfere when recombined, and this can be determined when viewed at a detector.

    But what does it actually do? From the intro of the actual problem, I guess it has something to do with length measurement. How? My lecturer spent just about zero time explaining anything about this device (and a lot of other things as well... it was basically 'you'll need to know this, now let's move on'). I have read up on a few websites (Wikipedia for example) but I still have no idea as to what sort of useful information it tells us.

    To the actual problem:

    a) If motion is 1 mm then a ray of light going to and from the mirror travels an extra 2 mm so OPL = 2 mm. (Maybe? From what I read on Wikipedia, anyway, since I can't remember the lecturer mentioning this either...)
    As to how many fringes are counted -- well, I still have no idea how fringes are done except that tilting the movable mirror produces these fringes.

    b) I really am quite lost. It would probably help if I understood the device first before trying to come up with... stuff...

    Please help.
     
  2. jcsd
  3. Sep 26, 2014 #2
    Alright, so I have spent a bit of time looking for stuff and I think I might have got it - if someone can confirm my responses that would be appreciated.

    So firstly, OPL is 2 mm, and there is an equation that makes part a) trivial, which is OPL = n*##lambda## where n is the number of fringes. Funny enough, this equation is nowhere in the lecture notes nor mentioned in lectures...:rolleyes:

    So solving gives n = 3160 fringes that need to be counted.

    Part b) is still a little iffy: firstly there is a fringe every 0.1 mm and *at* 1 mm (accidentally left the word out of the original post). So there is a fringe at 0, 0.1, 0.2, ..., 1 mm i.e. 11 fringes in total...?

    But we don't need 11 fringes (that is only the minimum) because we can have fringes every, say, 0.05 mm and still satisfy the condition. So number of fringes can be n = 11*x where x = 1, 2, 3, ...

    If so, then subbing n = 11*x and OPL = 2 mm gives ##lambda## = ##\frac{0.00018}{x}## mm

    Then I subbed in x and found for x = 300, ##lambda## = 610 nm which is around visible yellow/orange. So I said that this choice is preferable so that fringes are easily visible.

    So how did I do so far? I'm still not sure how I am supposed to comment on the practicality of this, though. Any insight will be appreciated.
     
  4. Sep 27, 2014 #3

    Simon Bridge

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    ... it tells us the phase difference between two light beams as a result of travelling along different paths. You have just described what it "actually" does in post #1. That's it .

    For part (b) they are sort-of talking about making a vernier out of the interference fringes. Your reasoning looks good. YOu just need to discuss the idea next: what would be the pros and cons? How practical would it be?

    The way the question is phrased, it isn't one of those where you are expected to regurgitate stuff from notes, you are supposed to figure it out, so I am limited in how I can help you there.

    It does help to understand the device, and you won't really get a feel for it without using one.
    If you have enrolled in a laboratory paper, they may have one you can use: it is well worth it.
     
  5. Sep 1, 2015 #4
    what is the difference between michelson interferometer's circular fringes and newton's rings?
     
  6. Sep 3, 2015 #5

    Simon Bridge

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    The way they are formed.
    I should say: the apparatus used to produce them.
     
  7. Sep 3, 2015 #6
    How circular fringes are formed in michelson interferometer?
     
  8. Sep 10, 2015 #7

    Simon Bridge

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