Max # of bright fringes

  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

    3. The attempt at a solution

    =sin-1 (1x(520x10-9)/(1/620 x10-3))

    .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.

  2. jcsd
  3. Doc Al

    Staff: Mentor

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

    Staff: Mentor

    Trying θ = 18.81 will just give you back m = 1, which you already know. θ = 90 is what you want.
  6. 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. Doc Al

    Staff: Mentor

    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. Ok, that's what I was it's 3 above the central max and 3 below plus 1 b/c you include m=0 ?
  9. Doc Al

    Staff: Mentor

    You got it.
  10. Great! thanks for your help.
  11. I have one more question:

    To get the value of m=0 do you always add 1?
  12. Doc Al

    Staff: Mentor

    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. 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. Doc Al

    Staff: Mentor

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

    Staff: Mentor

    Yes to all of that. Good!
  18. Thanks for your help!
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?