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Find wavelength in double slit experiment

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Two narrow slits are 0.4 mm apart. The dark fringe of order 5 is 1.2 degrees from the central bright fringe. What is the wavelength of the light in nm?


    2. Relevant equations

    dsin(theta) = (m +1/2)lambda
    where d = distance between silts
    m=order
    lambda = wavelength

    3. The attempt at a solution

    It seems simple enough right? I converted the 0.4mm into meters and plugged all the values into the equation which gives me:

    (4x10^-4)(sin1.2) = (5 + 1/2)lambda

    then lambda = (4x10^-4)(sin1.2)/5.5

    Since the answer i get is in meters i have to multiply it by 10^9 right? but that answer is way too big. What am I doing wrong?
     
  2. jcsd
  3. Mar 30, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    1523 nm is in the infrared range of the optical spectrum. Why do you think that it is too big?

    ehild
     
  4. Mar 30, 2010 #3
    Because according to my homework set that answer's not right. The wavelengths we're using in class is visible light between 400nm and 700nm and the answer I'm getting from my calculations is in the thousands.
     
  5. Mar 30, 2010 #4
    I find that your solution is correct. I obtained the same result, 1523 nm for the wavelength. You never know, sometimes homework sets/ exercises in books are not 100% fine.
     
  6. Mar 30, 2010 #5
    Well ok but i didnt get 1523 nm as an answer either. Did I use a wrong equation or wrong values somewhere?
     
  7. Mar 30, 2010 #6
    No, I don't think so. the distance between slits (a) multiplied by the sin of the angle given (sinQm) equals wavelength (landa) x (order). In this case, you are refering to dark fringes and the order gets a 1/2 added to it.
    So a.sinqm=landa.(5+0.5)
    comes out to be 1523 x 10^-9 m.
    The equation is okay, but whether or not the values are wrong I can't know, you have the exercise.
     
  8. Mar 30, 2010 #7
    Yes all the values I posted are exactly the ones from the exercise. I set up the equation exactly like that too. There must be something really obvious that I'm not getting. Let me do it step by step to see exactly where I'm messsing up.

    dsin(angle) = lambda(m+0.5)
    lambda = [dsin(angle)]/(m+0.5)
    lambda = [(4*10-4)(sin1.2)]/5.5
    lambda = 3.73*10^-4/5.5

    This gives me lambda = 6.78 * 10^-4 m and so when I convert it to nm its definitely not 1523 nm. So exactly what's my mistake?
     
  9. Mar 30, 2010 #8
    (4*10-4)(sin1.2) actually comes up to ~ 8.377 x 10^-6 which divided by 5.5 gives ~ 1.523 x 10 ^ - 6m or 1523 nm. Remember, you are working with degrees, not radians.
     
  10. Mar 30, 2010 #9
    Oh wow i get it. Ya it really was something obvious after all
     
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