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Homework Help: Multiple Slit Interference and Driffraction Grating

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    A five-slit system with 8.4*10-6 slit spacing is illuminated with 610*10-9 light.

    Find the angular position of the first two maxima, and the 3rd and 6th minima.
    2. Relevant equations
    For Maxima:
    d*sin(θ) = m*λ
    θ = sin-1((m*λ)/d)

    For Minima:
    d*sin(θ) = (m/N)λ
    θ = sin-1((m/(N*d))λ)

    3. The attempt at a solution
    λ = 610*10-1
    d = 8.4*10-6
    N = 5

    Subbing my values in to find the first and second maxima I solve for θ = 4.16, 8.35 respectively. These answers are correct according to my online module.

    However, I cannot find the 3rd and 6th minima.
    My attempt:

    for m = 3
    θ = sin-1((m/N*d)λ)
    θ = sin-1((3/(5*8.4*10-6))610*10-9)
    θ = 2.49°

    for m = 6
    θ = sin-1((6/(5*8.4*10-6))610*10-9)
    θ = 4.99°

    These answers are given as incorrect using the online module.

    I read further into the problem and discovered that a system with N = 5 slits should have N-1 = 4 minima between each maxima. From the equation we can see that the 3rd minima will occur at 3/5 of one wavelength. Maybe I should try a trigonometric approach?
    I am really stuck and any input is greatly appreciated.

  2. jcsd
  3. May 29, 2013 #2
    I guess it depends on how you interpret:
    I'd argue that the first two maxima are at θ = asin(0*λ/d),asin(1*λ/d) and the 3rd and 6th minima are at θ = asin(3/5*λ/d) and θ = asin(7/5*λ/d), respectively.

    To illustrate:
    asin(0/5*λ/d) → maxima
    asin(1/5*λ/d) → minima
    asin(2/5*λ/d) → minima
    asin(3/5*λ/d) → minima
    asin(4/5*λ/d) → minima
    asin(5/5*λ/d) → maxima
    asin(6/5*λ/d) → minima
    asin(7/5*λ/d) → minima


    Second edit:
    I think arcsin, but I write asin. MATLAB habit.
    Last edited: May 29, 2013
  4. May 29, 2013 #3
    Thanks, I was using 6/5 instead of 7/5

    This still leaves me with the values 2.49, and 5.84 as the values for the 3rd and 6th minima.

    The online module gives three attempts, and I've already used 2 (my friend told me to treat it as a double slit system, not surprisingly it was incorrect). Is there any other errors I could have made before I try this solution?
  5. May 29, 2013 #4
    If whoever wrote the problem statement agrees with my interpretation, then I can't see any problem with your solution for the 3rd (round to nearest integer perhaps, 2.50) and 6th minima. You're sure the result is supposed to be given in degrees?
  6. May 29, 2013 #5
    Yeah the answer is required in degrees, appreciate the help. Thanks
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