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HW help (Double Slit, Resolving power etc)

  1. Oct 25, 2006 #1
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    (a) A double-slit experiment is set up using red light (l = 708 nm). A first order bright fringe is seen at a given location on a screen. What wavelength of visible light (between 380 nm and 750 nm) would produce a dark fringe at the identical location on the screen?
    l = nm

    All right so i know
    d * sin theta = m * Wavelenght

    How do i find the Distance and the angle?
     
  2. jcsd
  3. Oct 25, 2006 #2
    The condition for constructive interference is what you have written down (i.e. peak on peak), but destructive interference happens when you have a trough meet a peak, i.e. there is a half wavelength difference so for destructive interference (dark fringe) [tex]dsin (\theta) = (m + 0.5)\lambda[/tex]. You should be able to go from there.
     
  4. Oct 28, 2006 #3
    (b) A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.08 mm). What is the distance between the second order bright fringe of light with l = 691 nm and the third order bright fringe of light with l = 414 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)







    Ok so
    D sin theta = m * Wavelenght

    Distance is 2.2?
    I am going to find theta
    m = 1
    Wavelenght is 698 in one part, and then i do the quation again for 413 right?
     
  5. Oct 28, 2006 #4

    OlderDan

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    Don't lose track of the statement about the different orders of the finges. What is m for second order, and for third order?
     
  6. Oct 28, 2006 #5
    2 for second order, and 3 for third order
     
  7. Oct 28, 2006 #6
    got the answer, thanks
     
    Last edited: Oct 28, 2006
  8. Oct 28, 2006 #7
    You don't actually need to find the distance at all. You will have two equations equal to [tex]dsin (\theta) [/tex], and you just have to solve for the 1 unknown (wavelength).
     
  9. Oct 28, 2006 #8
    and that would be?
     
  10. Oct 28, 2006 #9
    .00315 m
    ................
     
  11. Oct 28, 2006 #10
    Yes that's correct for (b). What answer did you get for (a).
     
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