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Number of interference fringes visible in the central diffraction peak

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data

    How many interference fringes are visible in the central diffraction peak if the ratio of slit separation to slit width is 3.125?

    2. Relevant equations

    d*sin(θ) = m*λ for interference maxima with slit separation d, maxima are at m = 0, 1, 2, 3 ...

    a*sin(θ) = m*λ for diffraction minima with slit width a, minima are at m = 1, 2, 3 ...

    3. The attempt at a solution
    md = m for diffraction minima
    ma = m for interference maxima
    ma/md = 3.125

    d*sin(θ) = md*λ
    a*sin(θ) = ma*λ
    -> d/a = (md/ma)*λ

    When md = 1, ma = 3.125, so the central diffraction peak contains the 3rd order interference maxima. This means that there are 3 interference maxima on the right of the central interference fringe, and 3 on the left, giving a total of 7.

    My instructor gave a formula to calculate the number of interference fringes visible in the central diffraction peak: 2(d/a) - 1
    Using this, I get 2(3.125) - 1 = 5.25, or after rounding, 5 peaks. What am I missing?

    Also, this formula allows for an even number of interference fringes in the central peak (consider d/a = 2.5, 2(2.5 - 1) = 4), is that possible?
     
  2. jcsd
  3. Nov 26, 2011 #2
    Does anyone have any insight into this problem? I still haven't figured it out. Also, is it possible to have an even number of interference fringes in a central diffraction peak?
     
  4. Nov 26, 2011 #3
    I think it is safe to assume that angles are small and therefore it is safe to assume
    that Sin∅ = ∅
    For the interference maxima the angular separation is therefore λ/d
    The width of the central diffraction max is λ/a
    Can you take it from there?
    Also I think this gives the n=3 interference max at the min of the diffraction pattern so the interference max will not be observed. Therefor 2 each side of the central max (n=0 max)
     
  5. Nov 26, 2011 #4
    I have that, I would just like to verify if my working and answer is correct. :smile:

    And, if I could find out if it is possible for there to be an even number of interference fringes in a central diffraction peak.
     
  6. Nov 26, 2011 #5

    ehild

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    Look at these patterns.http://www.sciencephoto.com/media/157199/enlarge
    The width of the slits is 25 μm, and their separation is 75 μm in the upper pattern and 58 nm in the bottom one.
    If the last interference maximum coincides with a diffraction minimum, you can not see it. That is seen in the upper picture when d/a=3: 5 maxima are seen. In your case, the 3rd maximum is inside the first diffraction maximum, so it is visible, although faint. You will see 7 maxima. In the bottom pattern, d/a =2.3. The second maxima are inside the main diffraction maximum, so you see 5 maxima again.
    The number of maxima is integer. You need to take the integer part of d/a. The number of maxima is 2 int(d/a)+1, if d/a is not integer. In case it is an integer, the last interference maximum coincides with the diffraction maximum, and the number of maxima is 2int(d/a)-1.

    There is no such thing that 2and a half maxima. If d/a=2.5, 2 maxima are seen on both sides (as int 2.5=2) and one in the middle, so 5 altogether.

    ehild
     
  7. Nov 26, 2011 #6
    Great, thank you for clarifying. That was very helpful.
     
  8. Nov 26, 2011 #7

    ehild

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    Homework Helper
    Gold Member

    The pleasure was mine:smile:

    ehild
     
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