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A problem using Young's equation from the double-slit experiment

  1. Nov 24, 2006 #1

    Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by 0.200 mm and the interference pattern is observed on a screen 4.00 m from the slits.


    What is the width (in mm) of the central interference maximum?

    [tex]y_m = R \cdot \frac{m \cdot \lambda}{d}[/tex]

    I thought I would get the width of the central interference maximum by doubling [tex]y_m = 4.00 \cdot \frac{4.00 \cdot 1 \cdot 4.00 \cdot 10^(-9)}{.00200} = .016[/tex].

    That's not correct, though.

    Why was my approach incorrect?
  2. jcsd
  3. Nov 24, 2006 #2
    for central maximum doesnt m=0?
  4. Nov 24, 2006 #3
    But, does [tex]y_0 = 0[/tex] make sense physically?
  5. Nov 24, 2006 #4


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    Your substitution doesn't look right to me

    YOu appear to have one too many 4.0s in there.

    400 nm is either 400 x 10^-9 or 4 x 10^-7 - you appear to have used 4.0 x 10^-9

    2mm is 0.0002m, not 0.002

    [Edited to correct should read 0.2 = 0.0002 not 0.002]

    Incidentally, I don't know whether that's why you've gone wrong, because either using the extra 4 or not, I can see the answer being 0.016. 4x4 is 16 so divided by 2 should give you an 8 - using an extra 4 would give you a 32 in your answer.
    Last edited: Nov 24, 2006
  6. Nov 24, 2006 #5


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    Homework Helper

    2 [mm] = 0.002 [m].
  7. Nov 24, 2006 #6


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    My typing mistake - he's got 0.2 mm in the original, which gives 0.0002
  8. Nov 24, 2006 #7

    Doc Al

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    Staff: Mentor

    It looks like you found the distance of the first maximum (m = 1) from the center and then doubled it. (There are typos in your expression.) That gives you the distance between the two first non-central maxima, which is not the same thing as the width of the central maximum. A good estimate of the width of the central maximum would be the distance between the first minima. (For double slit interference, the minima are centered between adjacent maxima.)
  9. Nov 25, 2006 #8
    Thanks for all the responses. Indeed, my tex equation had numerous typos, I apologize for those.

    I was finally able to do the problem by using [tex]d \cdot \sin(\theta) = (m + \frac{1}{2}) \cdot \lambda[/tex] and its corresponding constructive counterpart.

    I used the equation [tex]y_m = R \cdot \sin(\theta_m)[/tex], and I got my answers for part A and part B (not shown).
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