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Homework Help: Diffraction Minima confusion- please help

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data
    I did a physics lab the other day that dealt with two slit diffraction. In my setup, I also saw two faint diffraction minima, and it is my understanding that there is also some single slit diffraction effects.

    I measured the position a diffraction minimum to be y= 8.775 cm. I am asked to calculate the slit width a using equation sin(theta)=p*lambda/a, where a is slit width.

    2. Relevant equations

    3. The attempt at a solution
    My question is do I use a= (p)*lambda/(sin(theta)) or a=(p+.5)*lambda/sin(theta)

    I know that in single slit diffraction you don't add a .5 to the value for p, but I was just wondering if it is any different for this case.
  2. jcsd
  3. Sep 17, 2008 #2

    Chi Meson

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    It is no different. You are seeing the two interference patterns overlaid on each other.

    The single slit interference pattern creates the first minimum at the point where rays from either side of the gap meet (ironically) with constructive interference; where one has traveled an extra distance equal to one wavelength. Therefore, you do not use the ".5" .

    You will notice that the central fringe is twice as wide as the higher order fringes. This is indicative of the "single slit" pattern, and the "full integer multiple."

    The fringes that are due to the distance between the two gaps, you will notice, are all the same width, including the central fringe. This indicates the necessity of the "integer +.5" multiples when determining the minimums.
  4. Sep 17, 2008 #3
    Also, in this case I would use p=1, because these are the first pair of diffraction minima. The wavelength of the laser was 650 nm and the distance to the screen was 5.52 m.

    so tan(theta)= y/L

    tan(theta)=(8.775cm/552cm) theta= .911 degrees.

    a= (1)(6.50*10e-7m)/(sin(.911)) a= 0.0409mm. Is this correct?
  5. Sep 17, 2008 #4
    Thanks.. so why do I see single-slit diffraction-behavior with this double slit experiment
  6. Sep 17, 2008 #5

    Chi Meson

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    Because each slit has a width!

    If the two slits had narrower widths, then the overlaid single-slit interference pattern would be wider apart, but the two-slit pattern would stay the same.

    If the space between the slits narrowed, while the widths stayed the same, then the two-slit pattern would change, and the single slit pattern would be the same.
  7. Sep 17, 2008 #6
    So... just to get this straight, on my screen the diffraction minima will not be dark spots, but rather sort of faded-out looking maxima, where constructive interference occurs. Calling them diffraction "minima" is kind of misleading because of this fact. The reason they are faded out is because the two slits have widths, and therefore, the wavelets in each aperture have the opportunity to interfere with each other, so that the amplitude of the waves is not uniform when meet. This is why the intensity of the fringes decreases as y increases. Right?
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