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Diffraction Grating and Visible Light

  1. Mar 25, 2017 #1
    1. The problem statement, all variables and given/known data
    "You're using a diffraction grating o view the 400 to 700 nm visible spectrum. Suppose you can see the spectrum through the third order. a) Show that the first and second order spectra never overlap regardless of grating spacing. b) show that the second and third order spectra always overlap.

    2. Relevant equations
    n λ = dsinΘ

    3. The attempt at a solution
    for violet light 400nm
    sinΘv = 1 × 400⋅10^-9 / 1/400/1000 = sin^-1(400⋅10^-9 / 1/400/1000) = 9.21°
    for red light 700nm
    sinΘv = 1 × 700⋅10^-9 / 1/700/1000 = sin^-1(700⋅10^-9 / 1/700/1000) = 29.34°

    At this point im completely lost. Any guidance on moving forward would be greatly appreciated. Thanks
     
  2. jcsd
  3. Mar 25, 2017 #2

    ehild

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    The grating spacing(d) is not given. You can not get the numerical values of theta.
     
  4. Mar 25, 2017 #3
    That is the problem in its entirety. I do not know how to determine the answer when so many variables are missing. I asked my instructor but they said. "What is the maximum angle that the light can be diffracted?" Super Confused!
     
  5. Mar 25, 2017 #4

    ehild

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    What does it mean that you have first, second and third order spectra? Which letter means the order?

    Express sin(theta) in terms of d, for the red and violet light for spectra of all the three orders,
     
  6. Mar 25, 2017 #5
    I am assuming that it means that I have the first, second and third bright fringe and that the first and second will never overlap and the 2nd and 3rd will.
     
  7. Mar 25, 2017 #6

    ehild

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    Y
    Those bright fringes are complete spectra.

    upload_2017-3-25_21-24-32.png

    What do the letters mean in the formula n λ = dsinΘ?
    Express sinΘv and sinΘr in terms of d.
     
  8. Mar 25, 2017 #7
    n = the bright fringe , λ = wavelength, d = space between each line, Θ = angle that the order is seen.

    min = 400nm
    max = 700nm
    For first order on each end of the spectrum it would be
    sinθmin = nλ / d = θ = arcsin(n(400) / d)
    sinθmax = nλ / d = θ = arcsin(n(700) / d)

    second order would be
    sinθmin = 2nλ / d = θ = arcsin(2(400) / d) = arcsin(800/ d)
    sinθmax = 2nλ / d = θ = arcsin(2(700) / d) = arcsin(1400 / d)

    Third order would be
    sinθmin = 3nλ / d = θ = arcsin(3(400) / d) = arcsin(1200n / d)
    sinθmax = 3nλ / d = θ = arcsin(3(700) / d) = arcsin(2100n / d)

    Because the second order maximum value is greater than the third order minimum, those would overlap. The first and second order do not overlap because the maximum of the 1st will never exceed the value for the 2nd order.
     
  9. Mar 25, 2017 #8

    ehild

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    See: http://www.physics.smu.edu/~scalise/emmanual/diffraction/lab.html
    n is a number, not a fringe. It is the order of the diffraction spectrum, produced by the diffraction grating.
    What do you mean on "line"? d is the grating constant, the spacing between the scratches on the grating.
    Θ is the angle a light ray of certain color is deflected.
    For the first order n=1.
    You solved the problem at last.
     
  10. Mar 25, 2017 #9
    Thanks for the help. I was making that waaaaaay to difficult for some reason.
     
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