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Double slit equations

  1. Nov 20, 2007 #1
    [SOLVED] Double slit equations

    In an experiment, blue light with a wavelength of 645 nm is shone through a double-slit and lands on a screen that is located 1.35 m from the slits. If the distance from the centre maximum to the 8th order bright fringe is 2.6 cm, calculate the distance between the two slits.

    Here is what I did:
    wavelength=6.45*10^-7 m
    L=1.35 m
    m=8
    X=0.026m

    I used X/L = m*wavelength/d to find d.
    Is this right?
     
  2. jcsd
  3. Nov 20, 2007 #2
    got another question:

    Two slits are separated by a distance of 2.00 10-5 m. They are illuminated by light of wavelength 5.60 10-7 m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?

    Have no idea to solve it since it involves both a bright and a dark fringe..
     
  4. Nov 20, 2007 #3
    Anyone??
     
  5. Nov 20, 2007 #4
    i would really appreciate it if someone could help me with this..
     
  6. Nov 20, 2007 #5

    rl.bhat

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    I used X/L = m*wavelength/d to find d. This formula is correct.
    The position of the dark fringe is given by X/L = (2m + 1)*wavelength/2d .
    m = 0 gives the first dark fringe
     
  7. Nov 20, 2007 #6
    ok but I am asked to find the distane between the middle bright fringe and the 3rd dark fringe.
     
  8. Nov 20, 2007 #7
    Oh wait, so you are saying I sub in m=2, and the given values to find X?
     
  9. Nov 20, 2007 #8

    rl.bhat

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    All the distances are measured from the central bright fringe. You want to find the distance of third dark fringe. So put m = 3
     
  10. Nov 20, 2007 #9
    umm..ok..got it..!..thnkz..got another question..this one is on single slit phenonmenon:
    Light of wavelength 625 nm shines through a single slit of width 0.32 mm and forms a diffraction pattern on a flat screen located 8.0 m away. Determine the distance between the middle of the central bright fringe and the first dark fringe?

    So what i did was using wavelength/width= X/L, i found X. this is the distance from one dark to another dark fringe (on the other side of the middle light fringe) then i divided tht by 2. tht right?
     
  11. Nov 20, 2007 #10

    rl.bhat

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    So what i did was using wavelength/width= X/L, i found X. this is the distance of the first dark fringe from the central bright fringe.
     
    Last edited: Nov 20, 2007
  12. Nov 20, 2007 #11
    um..ok..so this is the dist. of the 1st dark fringe to the dark fringe after the central bright fringe yea?
     
  13. Nov 20, 2007 #12
    i am still unclear on how to find the distance from the 1st dark fringe to the MIDDLE of the central bright fringe..
     
  14. Nov 20, 2007 #13

    rl.bhat

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    The distance of the first dark fringe is half the width of the central bright fringe.
     
  15. Nov 21, 2007 #14
    so my answer is write yea..?..i mean the method..find findind the distance, and then divinding it by 2 gives the right answer..
     
  16. Nov 21, 2007 #15

    rl.bhat

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    No need to divid by 2. X itself is the distance of the first dark fringe from the central maximum.
     
  17. Nov 21, 2007 #16
    Oh..ok..thankz..:)..One more question..Light of wavelength 600 nm is incident upon a single slit with width 4.0 × 10-4 m. The figure shows the pattern observed on a screen positioned 2.0 m from the slits.Determine the distance s. see attached pic..
    So here we have the wavelenght, w and L. Now since both the bright and the dark fringe is invloved do we have to cobine their equations??
    like: (m+0.5)wave/w= X/L...Dont know how to proceed with this problem..
     

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  18. Nov 21, 2007 #17

    rl.bhat

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    I can't see the picture. Any way, the distance of the nth secondary maximum from the mid point of the central maximum = (2n + 1)lambda*L/2d and the distance of the nth secondary minimum from the mid point of the central maximum = n*lambda*L/d
     
  19. Nov 22, 2007 #18
    umm..but in the pic..we are supposed to find the distance between the middle of the central max...to the middle of the 2nd dark fringe from the central max..
     
  20. Nov 22, 2007 #19

    rl.bhat

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    All the fringes are of finite width. In the case of diffraction the fringe widths are not same. As you move away from the central bright fringe the width decreases. So always you have to take the reading from center to center of any fringe.
     
  21. Nov 22, 2007 #20
    so then, for solving this problem we would use X=Lnlamda/w. I have attached the pic again for you to see.
     

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