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Homework Help: Optics (double slit experiment)

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data

    I have some difficulty understanding a part of the following 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.

    2. Relevant equations

    Here is a diagram of the double slit experiment:

    http://imageshack.us/scaled/landing/850/doubleslit.jpg [Broken]

    Condition for minima: ##d \ \sin \theta = (m+\frac{1}{2}) \lambda##

    Linear positions measured along the screen: ##\tan \theta = \frac{y}{L}##

    ##\therefore \ y_{dark} = \frac{L (m+\frac{1}{2})\lambda}{d}##

    3. The attempt at a solution

    So, in the question what is meant by "distance between the 5th minima on either side of the zeroth-order maximum"? I'm not sure if I understand this. :confused:

    Does this mean the distance from the 5th minima on one side of the maxima to the 5th minima on the other side like this:

    http://imageshack.us/scaled/landing/706/52496318.jpg [Broken]

    Did I understand the question correctly? If this is correct, then

    ##34.73=2y \implies y=17.37 \ mm##

    And I can find the λ by rearranging the above equation:

    ##\lambda=\frac{y \ d}{L(m+\frac{1}{2})} = \frac{(17.365\times 10^{-3})\times (0.2 \times 10^{-3})}{1.5(5+\frac{1}{2})} = 4.2096 \times 10^{-7}##

    Is this correct? The answer looks reasonable (about 420.96 nm in the blue/violet region of spectrum), but I doubt it is correct. Any help would be greatly appreciated.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 26, 2012 #2


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    Homework Helper

    Yes, it is meant the distance between the fifth minimum at one side of the central maximum (at O) to the fifth minimum at the other side.

    You approximated sinθ by tanθ, it is all right in this case.

    But reconsider what number you should use for m. The first minimum is when the dsinθ=λ/2. m starts from zero.

  4. Dec 29, 2012 #3
    Thank you for the reply. I used m=5 because we want the fifth-order dark fringe. Personally I think that in a double-slit experiment there should be no zeroth-order minimum (i.e., m=0), but my book also says ##m=0, \pm1, \pm2, ...##. Does this mean I have to use m=4 instead (pretending m=0 is m=1)? :confused:

    Also I am wondering when do we use the negative values of m?
  5. Dec 29, 2012 #4


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    Homework Helper

    You get destructive interference, that is a dark fringe when the path-length difference δ=r2-r1 is odd multiple of the half wavelength: dsinθ=(2m+1)λ/2. If m is 0 or positive integer you get the dark fringes above O, with δ= λ/2, 3λ/2, 5λ/2,7λ/2, 9λ/2. For the dark fringes on the other side, m is negative (-1,-2,..) and δ= -λ/2, -3λ/2, -5λ/2, -7λ/2, -9λ/2. The fifth dark fringe corresponds to δ=9 λ/2.

  6. Jan 2, 2013 #5
    Thank you very much, that makes perfect sense now.
    Last edited: Jan 2, 2013
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