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Wave Optics Problems

  1. Feb 22, 2006 #1

    G01

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    #1 A diffraction grating having 500lines/mm diffracts visible light at 30 degrees([tex]\pi/6[/tex]) What is its wavelength?

    the distance between slits is .002mm(500l^-1)

    [tex] d\sin\frac{\pi}{6} = \lambda = 1000nm[/tex], which is too big. The answer is 500nm. What am I doing wrong?

    #2

    The slit spacing in a diffration grating is .0002m. The screen is 2.0m behind the grating. The distance of the first bright fringe is .004m. What is the wavelength of light?

    y = L[tex]\tan\theta[/tex]
    [tex] \tan^{-1}.004/2 = \theta = .0019 [/tex]
    [tex] d\sin\theta = \lambda [/tex]
    [tex] .0002m\sin(.0019) = \lambda = 400nm [/tex]

    If this is right then I read from a graph wrong. If this is wrong, then I would appreciate anybody's input. Thank you for your time.
     
  2. jcsd
  3. Feb 22, 2006 #2

    nrqed

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    The correct equation is

    [tex] d\sin\frac{\pi}{6} = n \lambda [/tex]

    where is an integer. since n=1 gives you a result out of the visible spectrum (in the infrared), you try n=2, which gives something in the visible.


    Pat
     
  4. Feb 23, 2006 #3

    G01

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    AHHH icic, very simple now that I see it. Anybody for # 2
     
  5. Feb 26, 2006 #4

    G01

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    bumping the thread...
     
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