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Two Slit Interference Question =] Thanks!

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Monochromatic light of wavelength 656 nm is normally incident on a vertical plate with two slits at positions +-d/2 where d=.25mm, the resultant interference pattern is observed on a screen at distance D= 3.4m away, let y denote vertical position on the screen with y=0 at the same height as the midpoint between the two slits. Picture attached
    [​IMG]

    a) What is the location y1 of the first intensity minimum on the screen, measured up from the central maximum?

    b) what is the location y3 of the third intensity minimum on the screen?

    c) A Second monochromatic beam is normally incident on the same slits. The second maximum pof the second beam occurs at y3, the third minimum of hte first beam. What is the wavelength of the second beam?


    2. Relevant equations
    y=L tan theta
    d*sin theta = m lambda
    d*sin theta = (m-1/2) lambda

    3. The attempt at a solution

    Part a) It's a dark fringe so dsin theta = (m-1/2) lambda and m=1 because it's the first one, and solve for theta, use y= L tan theta, take the difference which is y-0 and you get y1, whihc in my case is 4.4608 mm

    b) Same thing except m is now 3, ends up being 22.3045 mm

    c) here's where I got stuck, It said the second max of the second beam occurs at y3, the third minimum of the first beam, here's what I thought I Could do,

    d*sin theta = (3-1/2)lambda1 This is for the y3, third minimum
    d*sin theta = 2lambda2 This is for the second beam, set them equal to eachother
    (3-1/2)lambda1 = 2 lambda2
    And solve for lambda2, but this didn't work, i got a value of 8.2e-7 and it's not correct :(, I also tried going backwards with my location y3, solving for the angle and doing (d*sin theta)/2 but i go the same answer.
     
  2. jcsd
  3. Nov 6, 2008 #2
    Second maximum is for m=1 (first maximum is in the center, at m=0)
     
  4. Nov 6, 2008 #3
    So it will be like

    (3-1/2)lambda1 = lambda2

    ?
     
  5. Nov 7, 2008 #4
    +1 Bump
     
  6. Nov 7, 2008 #5
    Bump, that still didn't work and I'm kind of out of ideas >.<
     
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