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Max # of bright fringes

  1. Apr 30, 2011 #1
    1. The problem statement, all variables and given/known data
    A diffraction grating with 620 lines per mm is illuminated with light of wavelength 520 nm. A very wide viewing screen is 2.0 m behind the grating. (Note: 1 mm = 10-3 m; 1 nm =10-9 m)
    a) What is the distance between the two m=1 bright fringes? (Express your answer in meters)
    b) How many bright fringes can be seen on the screen


    2. Relevant equations
    m = d(sinθ)/λ
    y=Ltanθ

    3. The attempt at a solution

    A)
    d(sinθ)=mλ
    =sin-1 (1x(520x10-9)/(1/620 x10-3))
    =18.81

    y=Ltanθ
    tan(18.81)x2
    =.34
    .34 x 2 x 2 = 1.36 meters

    B) ?
    m = d(sinθ)/λ
    but when I use this equation I do not get the correct answer. I am at a loss if someone can walk me through this part.

    Thanks!
     
  2. jcsd
  3. Apr 30, 2011 #2

    Doc Al

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    What did you use for θ?
     
  4. Apr 30, 2011 #3
    I tried the theta I solved for, 18.81, and 90. I wasn't sure about that part of the equation
     
  5. Apr 30, 2011 #4

    Doc Al

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    Trying θ = 18.81 will just give you back m = 1, which you already know. θ = 90 is what you want.
     
  6. Apr 30, 2011 #5
    Alright. When I do that I get 3.1, which is 3.1 x 2 = 6.2 but the answer is 7. What I am doing wrong?

    Thanks for your help help!
     
  7. Apr 30, 2011 #6

    Doc Al

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    Getting m = 3.1 tells you that the highest order of fringe that you could see would be m = 3. (You won't see m = 4.)

    Don't forget the center fringe, where m = 0.
     
  8. Apr 30, 2011 #7
    Ok, that's what I was thinking...so it's 3 above the central max and 3 below plus 1 b/c you include m=0 ?
     
  9. Apr 30, 2011 #8

    Doc Al

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    You got it.
     
  10. Apr 30, 2011 #9
    Great! thanks for your help.
     
  11. Apr 30, 2011 #10
    I have one more question:

    To get the value of m=0 do you always add 1?
     
  12. Apr 30, 2011 #11

    Doc Al

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    I don't understand the question. m=0 is the central maximum. (Add 1 to what?)

    If you mean: To find the total number of fringes given a maximum value of m, then yes add 1 to 2*mmax.
     
  13. Apr 30, 2011 #12
    after solving for the no. of fringes per side of the central max you get 6.2 but must account for m=0, which is why 1 was added to 6.2 and you get 7?
     
  14. Apr 30, 2011 #13

    Doc Al

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    Right. Just think in terms of counting them up. 3 fringes on each side of the center plus one in the middle.
     
  15. Apr 30, 2011 #14
    great, thanks Doc Al!
     
  16. May 2, 2011 #15
    Doc Al,

    I want to run this by you if you get this post...
    A 3600 line/cm diffraction grating produces a third-order bright fringe at a 31.0 degree angle.
    (Recall: 1 cm = 10-2 m; 1 mm = 10-3 m; 1 nm =10-9 m)
    A) What wavelength (in nm) of light is being used?
    a. 477 nm b. 1430 nm c. 233 nm
    B) What is the distance (in meters) between this third-order bright fringe and the central maximum on the screen 2.0 meters away?
    a. 2.4 m b. 1.2 m c. 2.2 m
    C) How many total bright fringes can be seen on the screen?
    a. 3 b. 5 c. 11

    For C) would you do the following:
    (1/3600) x 10-2 sin(90)/(477 x 10-9)
    = 5.8
    so 5 fringes up and 5 fringes down for a total of 10 fringes but to account for the central max you add 1 to get 11, choice C?
    And is 90 degrees what you use in every scenario?
     
  17. May 2, 2011 #16

    Doc Al

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    Yes to all of that. Good!
     
  18. May 2, 2011 #17
    Thanks for your help!
     
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