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Homework Help: Interference fringes in Young's Double Slit Experiment

  1. Dec 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Light of two wavelenghts L1 and L2 (L1,L2 = lambda)creates interference fringes in the Young's double slit experiment. To calculate the least distance from the central maximum where the bright fringes for both L1 and L2 coincide. Given: the distance 'd' between the two slits, and 'D' between the plane of the slits and the screen.

    2. Relevant equations

    Distance of the nth bright fringe from the central maximum = nLD/d
    For the two bright fringes to overlap,

    n1L1 = n2L2 (since D/d is common for both wavelengths)
    n1 corresponds to the first wavelength
    n2 corresponds to the second wavelength

    3. The attempt at a solution

    By the above I am able to get the ratio n1/n2. However, I am unable to proceed any further to determine the required distance to the first point of overlap of the bright fringes.
     
  2. jcsd
  3. Dec 22, 2009 #2
    It is not clear to me here what you are asking to calculate. The least distance from the central maximum to what?
     
  4. Dec 22, 2009 #3
    For both the wavelengths there is a maximum of brightness at the centre. However, the distance of the respective bright fringes from the central maximum is different for the two wavelengths. At some distance (i.e. a minimum distance from the centre) two bright fringes (one for each wavelength) coincide. It is this minimum distance that has to be calculated. I hope the query is clear now.
     
  5. Dec 22, 2009 #4
    You have n1/n2. You need n1 and n2.

    First, you need to be able to express n1/n2 as a ratio of whole numbers. (why can't n1 and n2 be fractions?) For instance, suppose it is 7/2. Then n1 would be 7 and n2 would be 2. (Why not n1=14 and n2=4, since 14/4 = 7/2? Hint: n1L1=n2L2 at every coincidence of the fringes, not just the first one closest to the central peak. )

    If you get as far as getting n1 and n2, and are still stuck, let us know. If you don't see the answers to the why-questions I pose, i can explain further.
     
    Last edited: Dec 22, 2009
  6. Dec 22, 2009 #5
    I don't quite get your point. In this case n1/n2 is a given, viz. L2/L1. What we really need is another relation between n1 and n2 so that we xould solve for n1 and n2 individually. I agree that n1L1 = n2L2 at every point of coincidence of the bright fringes, but the question relates to the one that is closest to the central maximum.
     
  7. Dec 22, 2009 #6
    Why don't you share the actual value of n1/n2 and I can make the explanation more concrete.
     
  8. Dec 22, 2009 #7
    OK. Let us take L1 = 400 nm and L2 = 490nm. n1/n2 could be calculated from this.
     
  9. Dec 23, 2009 #8
    ok.

    L1n1=L2n2
    400n1=490n2
    n1/n2=490/400

    Both n1 and n2 have to be whole numbers. They represent the number of fringes we are away from the center maximum. We are not interested in two-thirds of a fringe. We want to know when two fringes coincide.

    Now n1=490 and n2=400 is one such place where they coincide (that's what L1n1=L2n2 means). That is, at fringe 490 for L1=400 and fringe 400 for L2=490. But that is not the first coincidence of the fringes, the one closest to the central maximum. For that we want the smallest (whole number) values of n1 and n2 such that n1/n2=490/400. (Hint: reduce the fraction 490/400. )

    Once you have n1 (or n2) just plug it into nLD/d.
     
  10. Dec 23, 2009 #9
    I have now got the clue. Many thanks for your guidance.
     
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