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Diffraction Question

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    The distance between the first and fifth minima of a single-slit diffraction pattern is 0.350 mm with the screen 45.0 cm away from the slit, when light of wavelength 550 nm is used. (a) Find the slit width. (b) Calculate the angle θ of the first diffraction minimum

    2. Relevant equations
    a(sin θ) = m(λ)
    a = slit width
    λ = wavelength
    θ = angle between ray and central axis
    m = which minimum

    3. The attempt at a solution
    (a)
    sin of angle θ to fifth minimum:
    (5)(550 * 10[-9] m)/(a) = 0.00000275/a

    By the small angle approximation, 0.00000275/a also gives the angle to the fifth minimum and the tangent of that angle.

    Distance from center of diffraction pattern to fifth minimum:
    [(0.00000275)(0.45 m)]/a = 0.0000012375/a

    Distance from center of diffraction pattern to first minimum:
    [(550*10[-9])(0.45 m)] = 0.0000002475/a

    Solving for a:
    0.000350 m = 0.0000012375/a - 0.0000002475/a
    a = 0.002828 m

    (b)
    (0.002828 m)(sin θ) = 550 * 10[-9] m
    sin θ = 0.000194 rad
    θ = 0.000194 rad

    When I checked, my answer for the slit width was correct, but the angle wasn't.
    Can anyone please find what went wrong with the angle in part b?
    Thank you!
     
  2. jcsd
  3. Nov 25, 2014 #2
    I think you may have used an incorrect equation. For destructive interference, the equation should be
    dsinΘ = (m + 1/2)λ​
    this will yield a different value for the angle.

    If I am correct, then I think that the anomaly of you having the correct answer for one component and yet not the other arises because conceptually the difference in slit produces both maximum and minimum fringes and their angles for the same value of d; so by calculating the maximum constructive interference angle you have the same distance as the minimum destructive interference angle.
     
    Last edited: Nov 25, 2014
  4. Nov 25, 2014 #3
    I still couldn't get it to work, but thank you!
     
  5. Nov 26, 2014 #4


    This equation is actually for double-slit destructive interference. ##a\sin\theta = m\lambda## is indeed the correct equation for single-slit minima. Since your slit width is correct, I'm not sure where your error lies since it looks like you calculated the angle correctly...​
     
  6. Nov 26, 2014 #5
    Thank you for taking the time to look at this!
     
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