1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    3. Attempt at a solution
    I began by finding λ using the first equation, so...
    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


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    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=2.094 micrometers
    Θ=30 degrees

    Then the equation reads...
    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


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    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: $$
    \ d\sin\theta = 2\lambda\quad\Rightarrow\ d \cos\theta \; d\theta = 2\; d\lambda\quad \Rightarrow\ {d\theta\over d\lambda} = {2\over d\cos\theta}$$
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted