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Interference/Diffraction formulas

  1. Dec 3, 2017 #1
    1. The problem statement, all variables and given/known data
    So I'm trying to determine the slit separation of a 2-D grid (as I understanding it, basically a grid with a lot of parallel, equally spaced slits in the x and y directions). I know I have to use some form of the formula d*sin(theta) = m*lambda, and I'm having some trouble deciding which one. I'd appreciate it if someone could clarify if these are correct:

    d*sin(theta) = m*lambda for constructive interference, double slit
    d*sin(theta) = (m+0.5)*lambda for destructive interference, double slit
    b*sin(theta) = m*lambda for destructive interference, single slit
    b*sin(theta) = (m+0.5)*lambda for destructive interference, single slit
    where d=slit spacing and b=slit width

    2. Relevant equations


    3. The attempt at a solution
    If my equations are right, then I would use the one for constructive interference of a double slit?
     
  2. jcsd
  3. Dec 3, 2017 #2
    I don't think anyone can help you unless you give some more context. Either you can give the original question text or explain whether you have a diffraction grating, single slit, double slit and what exactly you want to find.

    I don't know whether you have a single slit, double slit or diffraction grating and I also don't know whether you want the width of a fringe or the spacing between the fringes.

    Could you help out by explaining a little bit more?
     
  4. Dec 3, 2017 #3
    You can view the 2-D grid as a multiple-slit system - it's 2-D because the slits go in two directions and are perpendicular to one another, so it forms a cross-shape pattern. I'm trying to find the slit spacings (given the position of the maxima projected on a screen a given distance away), not the width.
     
  5. Dec 3, 2017 #4
    Ahh, I think I understand. It is a diffraction grating you are looking at. The equation will be: d sin (theta) = m (lambda). This is assuming that d will be the space between each slit. You can find the theta value of different m-values, and then you should be able to find the distance between fringes and anything you wish.
     
  6. Dec 3, 2017 #5
    Yep, d is the distance between the slits. Thanks for your help - do you think you'd also be able to confirm the four equations I wrote above? I know they vary from convention to convention, but is the relationship between the equations correct? This is more for my understanding, rather than for the problem.
     
  7. Dec 3, 2017 #6
    All of the equations for finding fringes from slit separation and wavelength are all essentially the same. The only thing really changing is what "d" means. In single slit diffraction, "d" is the width of the slit. In double slit diffraction "d" is the distance between slits. In diffraction gratings, "d" is the distance between each slit. The only other difference is that you have to add a half to the m-value to find the destructive interference.

    When I studied diffraction, it struck me as odd that they are all the same. But it makes sense they are related since they all correspond to the same effects (the wave nature of light). The equations you wrote above are correct and so are the relations.
     
  8. Dec 3, 2017 #7
    What I get confused about is why it's m for destructive, single slit and m+0.5 for constructive, single slit. My teacher told me that it's m for constructive, double and m+0.5 for destructive, double, and that these two switch when talking about single slit.
     
  9. Dec 3, 2017 #8
    In single slit b is width of slit when light pass through this slit at an angle theta from perpendicular bisector the optical path difference between waves sent by edage and waves sent by centre of the slit is b/2sinθ and for distructive interference b/2sinθ=λ/2
    For other minima we can write bsinθ=mλ here interference take place due to single silts of differected and non differected wave but in double slit there is two long parallel slit and d is distance between them and here path difference travel by two waves are dsinθ
     
    Last edited: Dec 3, 2017
  10. Dec 3, 2017 #9
    Your teacher is correct, but let me explain why.

    The values of theta when you have a single slit are always all the points of destructive interference. These values are found by using whole number m-values. In order to find the point of the constructive interference, you must find the point between two destructive fringes. In order to find this midway point, you must use a value of m+1/2. This is why it is swapped.

    In the case of the double slit experiment, the values of theta with whole number m-values are the constructive fringes. In order to find the destructive fringes, you must find the point between constructive fringes. Hence, m+1/2, for destructive fringes.

    This is why it swapped around. The two equation give you different values.

    If you need any more help, this is the textbook I am referring to: http://fcis.aisdhaka.org/personal/chendricks/IB/Giancoli/Giancoli Ch 24 Wave Nature.pdf
     
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