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Wave Optics of laser

  1. Aug 6, 2009 #1
    1. The problem statement, all variables and given/known data
    I have a few questions because of the fact I just don't understand any of this.......

    1.A helium-neon laser (l=656 nm) illuminates a single slit and is observed on a screen 1.9 m behind the slit. The distance between the first and second minima in the diffraction pattern is 5.45 mm. What is the width (in mm) of the slit?


    wavelength= 656 x 10^-9 m
    L=1.9 m
    a=5.45 x 10^-3 m

    3. The attempt at a solution

    I plugged and chugged and got 4.574 x 10^-4.....what did i do wrong?
  2. jcsd
  3. Aug 6, 2009 #2

    Chi Meson

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    5.45 mm is the distance between the first and second minima. How far would it be between the central maximum and the second minimum? That's the distance you need in the equation (as you have it).
  4. Aug 6, 2009 #3
    how do you find that, or how would you appoarch this?
  5. Aug 6, 2009 #4

    Chi Meson

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    The distance from the central max to the first min is the same as the distance from the first min to the second min. You can do the math from here.
  6. Aug 6, 2009 #5
    so is it 22.87 mm?
  7. Aug 6, 2009 #6
    opps I moved the decimal the wrong place .2287 mm
  8. Aug 6, 2009 #7
    Instead of starting a new post I thought that you could just help me on this forum......

    A 578 line/mm diffraction grating is illuminated by light of wavelength 599 nm. How many bright fringes are seen on a 2.73-m-wide screen located 3 m behind the grating

    L=3 m
    Lamda= 599 x 10^-9
    solving for m?

    How do I do this with out theata?
  9. Aug 6, 2009 #8

    Chi Meson

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    You have a 2.73 m wide viewing screen. The fringes diffract in both directions from the central "zeroth" order fringe ("bright spot"). There will be two first order fringes, one on each side of the zeroth, two 2nd order fringes, etc.

    You don't need theta just as you didn't need theta in the earlier problem. It's the same dang formula used for single slit, double slit, and grating solutions. WHere you had "w" for width of the single slit, you now use the distance between the lines (inverse of # of lines per meter).

    We assume that you aim the central "zeroth" max at the center of the viewing screen. Which order will be on the edge(s)?
  10. Aug 7, 2009 #9
    Is it second order then?

    This is how I tried solving this......

    d= 1/(578x10^-3)
    m=? what we are solving for

  11. Aug 7, 2009 #10

    Chi Meson

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    Except that the full screen is 2.73m wide. The distance from the central maximum to the "mth" order fringe will be half of that, since there is an mth order fringe to both sides of the center.
  12. Aug 8, 2009 #11
    Oh got it, it is starting to make since. Do you always have to divide the distance?
  13. Aug 9, 2009 #12

    Chi Meson

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    Not necessarily. You need to remember that the distance involved is from the center of the pattern to the "mth" fringe to either side of the center.
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