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Light: Fringes and Interference

  1. Apr 23, 2005 #1
    Coherent light with wavelength 597 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.

    For what wavelength of light in micrometers will the first-order dark fringe be observed at this same point on the screen?

    To solve this, I must use the formula for destructive interference:
    d*sin(theta) = (m + 1/2)*lambda, where m= 0 or -1 since it is the first dark fringe,

    but I am stuck on using small-angle approximations where sin(theta) = tan(theta) to find d.

    Could someone please explain this method to me clearly?

    After finding d, what formula must I use for the 2nd beam's interference pattern to find the 2nd wavelength?

    Thanks.
     
  2. jcsd
  3. Apr 23, 2005 #2

    Doc Al

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    Bright fringes are found where [itex]d \sin\theta = m\lambda[/itex]; dark fringes where [itex]d \sin\theta = (m + 1/2)\lambda[/itex]. m = 0, 1, 2, etc.

    The position (y) of the fringes is given by [itex]\tan\theta = y/L[/itex]. For small angles [itex]\tan\theta = \sin\theta = \theta[/itex]. So the bright fringes are found at [itex]y = m\lambda L /d[/itex]; the dark fringes at [itex]y = (m + 1/2)\lambda L /d[/itex].

    Note that d, L, and y stay the same for the two wavelengths. Hint: Solve for the new wavelength symbolically before plugging in any numbers.
     
  4. Apr 23, 2005 #3
    What value is y in the problem? 4.84 mm?
     
  5. Apr 23, 2005 #4

    Doc Al

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    That's correct. Note that the same value of y is used for both wavelengths:
    For the first, y = 4.84mm is the postion of the first bright fringe
    For the second, y = 4.84mm is the postion of the first dark fringe​

    Hint: You don't really need to plug in that y value! What you need to do is find the second wavelength in terms of the first.
     
  6. Apr 23, 2005 #5
    How do I do this? Do I use the formula for bright fringes for the second beam as a starting point?
     
  7. Apr 24, 2005 #6

    Doc Al

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    Here's what I suggest: Call the first wavelength [itex]\lambda_1[/itex] and the second [itex]\lambda_2[/itex]. Now write the appropriate equation for each and compare. I'll write the equation for the first wavelength. Since it gives the location of the first bright fringe, I'll set m = 1:
    [itex]y = \lambda_1 L /d[/itex]
    Rearranging that gives you: [itex]\lambda_1 = yd/L[/itex]

    Now you write the equation that applies to the second wavelength and compare.
     
  8. Apr 24, 2005 #7
    Well, for the second, is it this?

    y = [(m + 1/2)*lambda_2*L]/[d], where m = 0

    lambda_2 = (2*y*d)/L
     
  9. Apr 24, 2005 #8

    Doc Al

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    Exactly. So... how does [itex]\lambda_2[/itex] compare to [itex]\lambda_1[/itex]?
     
  10. Apr 24, 2005 #9
    It's twice the value of lambda_1.
     
  11. Apr 24, 2005 #10

    Doc Al

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    Staff: Mentor

    Correct. Now express the answer in micrometers and you're done.
     
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