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Youngs slits

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A screen is illuminated by monochromatic
    light.
    The distance from the slits to the screen is
    7.4 m.
    What is the wave length if the distance from the central bright region to the fourth
    dark fringe is 2.6 cm.
    Answer in units of nm.


    2. Relevant equations
    x=L*tan(theta)

    sin(theta)=(m+.5) lambda/d


    3. The attempt at a solution

    theta= inverse tan( 2.6e-2 / 7.4) = .201309

    lambda = (.55e-3) sin(.201309) / 4.5 = 429.427 nm
     
  2. jcsd
  3. Nov 23, 2008 #2

    Chi Meson

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    correct, but way too many digits. 430 nm is the proper way to give the answer, since only two digits are given for L and x.

    By the way, when the angle is very small (such as is the case here) sin theta = tan theta.

    That way, you can save some time and see that x/L=(m lambda)/d

    It'll be like that whenever L is much greater than x
     
  4. Nov 23, 2008 #3

    alphysicist

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    I don't believe this is correct. The fourth dark fringe does not correspond to m=4 in that equation.
     
  5. Nov 24, 2008 #4

    Chi Meson

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    Alphysicist is right. Fourth dark fringe is halfway between the 3rd & 4th bright fringe, so the "m-factor" should be 3.5.
     
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