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Easy wavelength problem

  1. Mar 15, 2005 #1
    Monochromatic light of wavelength 689 nm falls on a slit. If the angle between the first bright fringes on either side of the central maximum is 38 °, what is the slit width?

    a = λ / sin(theta)

    a = 689 nm / sin 38 °

    a = 1.12E-3mm

    is this all there is to it?
  2. jcsd
  3. Mar 15, 2005 #2


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    I believe that [tex]asin\theta=\lambda[/tex] where [itex]\theta[/itex] is the angle between the central line and the first bright fringe. So your theta here should be 19.

    a = 689 nm / sin 19 °

    EDIT: Is this a double slit or single-slit?
    For double slit:
    a = λ / sin(theta)

    For single slit:
    a = (3/2)λ / sin(theta)
    Last edited: Mar 15, 2005
  4. Mar 15, 2005 #3

    ahhh. thankyou :smile:
  5. Mar 15, 2005 #4


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    Please note my edit to my previous post. Be careful whether it is a double or single slit.
  6. Mar 15, 2005 #5

    It is a single slit...

    so then I do use the same theta of 19 degrees correct?

    a = (3/2)λ / sin(19°)
  7. Mar 15, 2005 #6


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    Yes, that's right. theta is 19. It comes from asin(theta)=(m+1/2)λ
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