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Young's Experiment (destructive interference)

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm having some trouble understanding the following solved problem:

    In Young’s experiment, narrow double slits 0.20 mm apart diffract monochromatic light onto a screen 1.5 m away. The distance between the 5th minima on either side of the zeroth-order maximum is measured to be 34.73 mm. Determine the wavelength of the light.

    Solution:

    I don't understand why they have used ##(m-\frac{1}{2}) \lambda## as the condition for destructive interference, instead of ##(m+\frac{1}{2}) \lambda##?

    2. Relevant equations

    For destructive interference my textbook uses:

    ##d \ sin \theta_{dark} = (m+\frac{1}{2}) \lambda##

    (this is an approximation)

    3. The attempt at a solution

    If I use -1/2, then 5th minima turn out to be are 9λ apart, just like the model answer.

    However when I use +1/2 I get a totally different solution:

    ##\therefore \ \lambda = \frac{\Delta y a}{(5.5-(-5.5)) L}= \frac{\Delta y a}{11 L} =421 \ nm##

    So, why do they use -1/2? And which method is correct? :confused:

    Any help is greatly appreciated.
     
  2. jcsd
  3. Nov 7, 2013 #2

    Zondrina

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    I think both you and your book might be wrong on this one. I'm getting ##1029nm## as the answer.

    I used: ##\frac{wx_n}{L} = (n - \frac{1}{2}) \lambda##

    Where ##n## represents a nodal line number.
     
    Last edited: Nov 7, 2013
  4. Nov 8, 2013 #3
    No, that's wrong because you are not given y, you are given ##\Delta y##.
     
  5. Nov 8, 2013 #4
    My textbook ("Optics" by Hecht) uses "m+1/2", while my other book ("Introduction to Optics" by Pedrotti) uses "m-1/2" for destructive interference. The two methods clearly produce different results! :confused:

    Why in this particular situation, do we need "-1/2" and not "+1/2"?
     
  6. Nov 8, 2013 #5

    Zondrina

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    No, you are not given ##\Delta x##.

    ##\Delta x## represents the distance between two consecutive nodal or antinodal fringes.

    ##x_n## represents the distance from a minimum to the right bisector (central maximum).

    They are not talking about the distance between two consecutive node/antinodal fringes, rather the distance from the 5th minima on either side of the fringe pattern to the right bisector.

    Also, because the fringe pattern is symmetric, I beleive it wouldn't matter whether you used ##(n - 1/2)## or ##(n + 1/2)##.

    It depends on which direction the author likes to measure I believe.
     
    Last edited: Nov 8, 2013
  7. Nov 8, 2013 #6
    You are using a different notation. I never defined ##\Delta y## as the distance between consecutive fringes! It is the distance between the position of 5th minima on one side, to the 5th minima on the other.

    ##y## itself is the linear positions of fringes measured along the screen from the bisector you mentioned.
     
    Last edited: Nov 8, 2013
  8. Nov 8, 2013 #7

    Zondrina

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    Indeed I use a different notation myself, but that really doesn't change the logic too much.

    Just use ##\frac{wy}{L} = (n - \frac{1}{2}) \lambda## then.
     
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