1. Not finding help here? Sign up for a free 30min 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!

Diffraction grating, distances between maxima

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data

    A grating has a line density of 1010 cm−1, and a screen perpendicular to the ray that makes the central peak of the diffraction pattern is 2.5 m from the grating. If light of two wavelengths, 590 nm and 680 nm, passes through the grating, what is the separation on the (flat) screen between the second-order maxima for the two wavelengths?

    2. Relevant equations

    [tex]dsin\theta = m \lambda[/tex]

    3. The attempt at a solution

    I thought I would find the angular distance to the second order maxima covered by each wavelength:

    [tex]\theta = arcsin(\frac{2 \lambda}{d})[/tex] where d is [tex]\frac{1}{1010}\times 10^{-2}[/tex]

    Then I would take the difference, and plug it into the equation [tex]sin\theta = \frac{y}{L}[/tex] and solve for y. This gives me the wrong answer but I don't see why this wouldn't work. Can anyone see my mistake?

    Thanks!
     
  2. jcsd
  3. Oct 17, 2015 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Take the difference of what???
    You have two wavelengths, two angles and two y positions. You need the separation between the positions, that is, the difference of the y-s.
     
  4. Oct 17, 2015 #3
    What I meant was that I would solve for the angular spread of the second order maxima for one wavelength, then the other, and the take the difference of those two values. What I get should be the angular distance between the two maxima. I could then use the equation I gave in my OP and solve. This gives the same answer that I got: [tex]2.5sin(7.896)-2.5sin(6.845)[/tex] The idea is that if I calculate the angle to one maxima, I should be able to tell what vertical distance it covers. Then I can do the same for the other maxima, and take the difference between those two distances as the distance from one maxima to another.
     
  5. Oct 17, 2015 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is a good method, what result did you get? In principle, the distance of the maximum from the centre is L tan(θ). If θ is really small, the sine and the tangent are nearly equal. But the angles you got are not small enough.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted