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Youngs double split experiment

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Light of wavelength 473 nm falls on two
    slits spaced 0.389 mm apart.
    What is the required distance from the slit
    to a screen if the spacing between the first and
    second dark fringes is to be 6.71 mm?
    Answer in units of m.



    2. Relevant equations

    sin[tex]\theta[/tex]= m[tex]\lambda[/tex]/d

    x=tan[tex]\theta[/tex]*L

    3. The attempt at a solution

    [tex]\theta[/tex] = sin-1{2*473e-9m / .389e-3 m} = .139336 degrees

    L = 6.71e-3 m / tan .139336 = 2.75919
     
  2. jcsd
  3. Nov 17, 2008 #2

    alphysicist

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    Hi Kris1120,

    I don't believe this is the formula you want. They are referring to the dark fringes in this problem.


    Also, when you solve for the distance remember they want the distance between two fringes, not the distance from one fringe to the center point.
     
  4. Nov 17, 2008 #3
    Ok so I am using the equation sin(theta) = (m +.5) lambda / d ... Does that mean I do not use the equation to solve for x?
     
  5. Nov 17, 2008 #4

    alphysicist

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    You do still have to solve for x.

    However, I think here you should use the small angle approximation that combines both equations (because for small enough angles sin(theta)=tan(theta)). Does your book have that formula?
     
  6. Nov 17, 2008 #5
    No. I can't find it
     
  7. Nov 17, 2008 #6

    alphysicist

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    For constructive interference, you have the two equations that you had in your original post:

    [tex]
    \sin\theta= \frac{m\lambda}{d}
    [/tex]

    [tex]
    \tan\theta= \frac{x}{L}
    [/tex]

    For small angles,

    [tex]
    \frac{x}{L} =\frac{m\lambda}{d}
    [/tex]
    so you can find the distance along the screen x directly for each line order (again, when the angles are small). You can do the same thing with the destructive interference case, and get the same equation with m [itex]\to[/itex] (m+1/2).
     
  8. Nov 17, 2008 #7
    If I am not misunderstanding... I set the sin equation equal to the tan equation and solve for L... I am still not getting it correct though. I am trying to look the equation up on google.
     
  9. Nov 17, 2008 #8
    Maybe I am using the wrong m... m=2?
     
  10. Nov 17, 2008 #9

    alphysicist

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    The formula that I mentioned for the destructive interference:

    [tex]
    x = L \frac{(m+\frac{1}{2} )\lambda}{d}
    [/tex]

    gives the x value for each fringe. What they tell you in the problem is that the difference in the x values is 6.71mm. Do you see what to do?

    I just saw your last post. The first order dark fringe would have m=0, and the second order dark fringe would have m=1, based on how you wrote your destructive interference condition.
     
  11. Nov 17, 2008 #10
    6.71 mm = [L((1+.5) * 473 nm)/ .389 mm] - [L((0+.5)*473nm)/ .389 mm] and solve for L? I get L = 5.51809 m is that correct?
     
  12. Nov 17, 2008 #11

    alphysicist

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    That looks right to me.
     
  13. Nov 18, 2008 #12
    Great! Thank you for being so patient with me.
     
  14. Nov 18, 2008 #13

    alphysicist

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    I'm glad to help!
     
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