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Diffraction Grating and Dispersion

  1. Feb 9, 2015 #1
    1. The problem statement
    A diffraction grating has a slit separation of 2094.0 nm. What is the dispersion of the 2nd order lines at an angle of 30 degrees (in degrees per micrometer)?

    2. Relevant Equations
    dsinΘ=mλ
    y=(mDλ)/d

    3. Attempt at a solution
    I began by finding λ using the first equation, so...
    2λ=.002094(sin(.30))
    λ=.00000548205
    Note: I changed d from nm to mm before I began any calculations

    We haven't gone over this in class yet, so I'm confused on where to go from here. I believe it has something to do with the second equation listed above, but I don't know what variable represents the dispersion or if these are really even the equations I need. I would like some guidance as for what steps to take next or what equations I should be using- I think just a push in the right direction could help immensely.
     
  2. jcsd
  3. Feb 10, 2015 #2

    BvU

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    Hello LL, welcome to PF :)

    Check your calculations ! sin(.30) is not the sine of 30 degrees ! I don't see how you can end up with 5.5 nm wavelength (assuming you report it in mm -- why do you do that if they want an answer in degrees per micrometer ?).

    And the angular dispersion is nothing else than ##d\theta\over d\lambda##.
    So you're better off with the first eqn than with the second (that is -- if I am not mistaken -- a distance y off-axis on a screen at distance D).

    Now, what is ##d\theta\over d\lambda## at ##\ \ \theta = \pi/6\ \ ## if ## \ d\sin\theta = 2\lambda\ ## ?

    It's more math than physics...
     
  4. Feb 10, 2015 #3
    Hi BvU!

    Alright, I was able to work on the problem in class- I was missing a vital equation that made the whole process really easy.

    The equation should be D=m/(d)cosΘ

    So if...
    D=dispersion
    m=2
    d=2.094 micrometers
    Θ=30 degrees

    Then the equation reads...
    D=(2)/(2.094cos(30))
    D=1.10 deg/micrometer

    I'm sorry about the confusion, I really appreciated your help! You were right, it was just figuring out what math is the right math!
     
  5. Feb 11, 2015 #4

    BvU

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    And here's me thinking you were supposed to derive this vital equation. You were nearly there!

    The "math" (physicists have their own ideas about how to deal with differentials :) ) is real easy: $$
    \ d\sin\theta = 2\lambda\quad\Rightarrow\ d \cos\theta \; d\theta = 2\; d\lambda\quad \Rightarrow\ {d\theta\over d\lambda} = {2\over d\cos\theta}$$
     
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