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Homework Help: 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?

  2. jcsd
  3. Oct 17, 2015 #2


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    Homework Helper

    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


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    Homework Helper

    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.
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