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Young's Double Slit Experiment

  1. May 12, 2011 #1
    1. The problem statement, all variables and given/known data

    [​IMG]


    so here i have shown the distances of two dark fringes ...

    from here ... fringe width is [tex]\frac{\lambda L}{2d}[/tex]
    L is distance of screen ... d is distance b/w 2 fringes.

    but fringe width is [tex]\frac{\lambda L}{d}[/tex] , right ???

    What mistake am i making?
     

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    Last edited: May 12, 2011
  2. jcsd
  3. May 12, 2011 #2
    What's the question?
     
  4. May 12, 2011 #3
    i wrote it above

    by using formula of distance of dark fringe ...

    fringe width comes out to be ... [tex]
    \frac{\lambda L}{2d}
    [/tex]

    but fringe width is [tex]
    \frac{\lambda L}{d}
    [/tex]

    so why am i getting wrong answer
     
  5. May 13, 2011 #4
    hello ... someone?
     
  6. May 13, 2011 #5
    The fringe width is the distance between the centres of alternate dark(or bright) fringes.
    Lamda D/d as marked on your diagram goes from the centre of a bright fringe to the centre of an adjacent dark fringe in other words you have marked in half the fringe width.
     
  7. May 13, 2011 #6
    So you mean this is fringe width ??
    [​IMG]
    well this looks satisfying
     

    Attached Files:

  8. May 13, 2011 #7
    You have now marked the fringe width correctly.
     
  9. May 14, 2011 #8
    are you sure?
    because fringe width means "width of 1 fringe"
    but (as you said) i have marked width of 2 fringes
     
  10. May 14, 2011 #9
    I did not say you marked in the width of two fringes I said you marked in the width of half a fringe.The fringe width as marked above in red is correct since it goes from the centre of a bright fringe to the centre of an adjacent bright fringe.
    Look at a graph of how the intensity varies smoothly across an interference pattern and this should clarify things for you.
     
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