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Diffraction grating spectrometer

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that you have a reflection diffraction grating with n=140 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen. How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59 nanometers, which are a well known pair of lines for sodium, in the second order (m=2)?

    2. Relevant equations

    maxima for reflection diffraction grating: sinθ=(mλ)/d where m=2 in this instance

    Rayleigh's resolution criterion: sinθ=1.22*(λ/d)

    Spectrometer: λ=(d/m)*sinθ

    d=distance between slits

    3. The attempt at a solution

    m=2 and d is the unknown. I tried setting the above equations equal to each other, eliminating sinθ, but I couldn't get any further because I don't know which λ to use since we are trying to separate two different wavelengths.
     
  2. jcsd
  3. Nov 4, 2011 #2

    SammyS

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    Use one λ, and then the other and compare.

    Since they're so close together, the answers will be very nearly the same.
     
  4. Nov 4, 2011 #3
    Where am I supposed to use both λ's? Also, the question asks to create a situation in which the two λ's are far enough apart that they are indistinguishable, so they must be related somehow.
     
  5. Nov 5, 2011 #4

    SammyS

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    589.00 and 589.59 nanometers
     
  6. Nov 5, 2011 #5
    Sorry, I meant I don't understand which equations to use and where to plug the two λ in.
     
  7. Nov 5, 2011 #6

    SammyS

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    How is this different than the newer thread you started, except for the little extra you wrote:
    "m=2 and d is the unknown. I tried setting the above equations equal to each other, eliminating sinθ, but I couldn't get any further because I don't know which λ to use since we are trying to separate two different wavelengths. I'm also a bit confused on which equation(s) is relevant. Also, I think my setting-them-equal strategy is flawed because d will always cancel out when you do that. I don't really know where to start."​
    ?

    https://www.physicsforums.com/showthread.php?t=547558"
     
    Last edited by a moderator: Apr 26, 2017
  8. Nov 5, 2011 #7

    SammyS

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    You need a relationship involving the total number of lines in the grating and the resolving power.
     
    Last edited: Nov 5, 2011
  9. Nov 5, 2011 #8
    Sorry, I was just trying to clarify what I have tried already. That post has been deleted. Our book doesn't have anything about resolving power and we haven't covered that in class (not sure why it's on a homework problem then). Someone in my class mentioned the equation sinθ=1.22(λ/d) but I'm not really sure what it applies to and it doesn't involve the number of lines. We covered diffraction grating in class but we haven't covered resolution at all, so I'm really lost on this problem. I'll look around online and see what I can learn.
     
  10. Nov 5, 2011 #9

    Redbelly98

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    That applies to a circular aperture; irrelevant here.

    It's odd that your book and class lectures have not talked about this, yet you were assigned this problem. In any event, you might try a google search on: grating resolution
     
  11. Nov 5, 2011 #10
    Ok, I had to go outside the book to find the resolution equations R=λ/Δλ and for a spectrometer R=mN where N=d*n and setting R=R got d=3.565 mm, which was correct.

    It is pretty frustrating that our book contains a question about resolution and yet doesn't even mention those equations. I think I might write to the publisher about that. Thanks for the help everyone! At least now I understand resolution for the next test!
     
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