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Double Slit Interference (nonzero slit width)

  1. Apr 1, 2009 #1

    TOD

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    1. The problem statement, all variables and given/known data
    A mask with 2 slits is illuminated by a light of 589nm wavelength. Slits each have a 0.1mm width separated (centre to centre) by 3.0mm. Calculate the distance between adjacent bright fringes on the screen if a screen is placed 5.0m away from the slits.


    2. Relevant equations
    Where

    [tex]\beta[/tex]=2[tex]\pi[/tex]dsin[tex]\theta[/tex]/[tex]\lambda[/tex]

    [tex]\phi[/tex]=2[tex]\pi[/tex]asin[tex]\theta[/tex]/[tex]\lambda[/tex]

    d=slit separation
    a=slit width

    E=E(max)cos[tex]\frac{\phi}{2}[/tex]*([tex]\frac{sin(\beta/2)}{\beta/2}[/tex])

    (sorry, no good with this latex stuff, some stuff not showing right)

    3. The attempt at a solution
    First, I assume that if given a slit width of some size then we cannot assume that the interference comes from points sources. So to calculate the distance between adjacent bright fringes, I took on the formula for intensity and turned it into electric field of some form (just square rooted) and tried to differentiate the electric field and then find the roots of the equation. However I have found this extremly lengthy and most probably unnecessary. Can anyone get me started in the correct direction?
     
    Last edited: Apr 1, 2009
  2. jcsd
  3. Apr 1, 2009 #2
    Treat the problem as the superposition of two diffraction patterns, one from each 0.1 mm slit. Where a bright fringe from one source occupies the same position as the other source's bright fringe this is a bright fringe of the double slit system.
     
  4. Apr 1, 2009 #3

    Redbelly98

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    Looks like you have φ and β switched around here, given your expression for E below.

    Okay. And what must
    cos(φ/2) = cos[π d sin(θ) / λ ]​
    be equal to, for a bright intensity fringe to occur?
     
  5. Apr 2, 2009 #4

    TOD

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    Ah yes, my bad.


    Plus or minus one? So I could in theory get that thing to equal 1, but that doesnt mean the product of this and the sine of a sine function is maximum does it?
     
  6. Apr 2, 2009 #5

    Redbelly98

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    Yes, exactly.

    That's a good question. The answer, in practice, is yes it does. Since measurements of fringe spacing are typically done near the central axis, where
    sin(β/2) / (β/2) ≈ 1,​
    the effect of the sin(β/2) term can be ignored.

    Or put another way, the cos(φ/2) term oscillates many many times for each oscillation of the sin(β/2) term. This is because d and φ are much larger than a and β, respectively. (In the present case they are 30 times larger.)

    Hope that helps.
     
  7. Apr 2, 2009 #6

    TOD

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    Very interesting. Thanks for the help.
     
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