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Homework Help: Diffraction grating

  1. Mar 14, 2006 #1
    A beam of light is incident normally on a diffraction grating of width 2cm. It is found that at 30 degrees, the nth order diffraction maximum for [itex]\lambda_1 = 5 \times 10^{-5}[/itex]cm is superimposed on the (n + 1)th order of [itex]\lambda_2 = 4 \times 10^{-5}[/itex]cm.

    1]How many lines per cm does the grating have?
    2]Find out whether the first order spectrum from such a grating can be used to resolve the wavelengths [itex]\lambda_3 = 5800[/itex] Angstrom units & [itex]\lambda_4 = 5802[/itex] Angstrom units.

    My work:

    1]If N ruling occupy a total width W, then slit width d=W/N.

    [tex]d\sin \theta = n\lambda_1 = (n+1)\lambda_2[/tex]

    [tex]{2\over N}{1\over 2} = n\lambda_1 = (n+1)\lambda_2[/tex]

    [itex]5000n= 4000(n+1)[/tex] (in Angstrom units).

    So, I got: n = 4; which I substituted in the first equation and I got the total number of rulings N = 0.5 x 104

    So, number of rulings per cm is: N/Total width = N/2 = 0.25 x 104

    Is this part correct?

    2]For this part, I can find the difference between the 2 wavelengths:

    [itex]\Delta \lambda = 2[/itex] Angstrom units.
    How do I determine whether the grating has good resolving power or not?
  2. jcsd
  3. Mar 14, 2006 #2
    i think for your scond one you need to find the distance between n=1 nodal lines for each of the wavelengths and compare with the power of resolution of the human eye (look it up)
  4. Mar 16, 2006 #3
    You mean I have to find the angular separation between the two lines in order to determine whether the resolution is good or not?
  5. Mar 16, 2006 #4
    Hey someone please help me here....I am unable to get any breakthroughs. Different sources give different resolution power for human eye and I don't think it is needed in this problem. Is there some other way of determining whether the resolution is good or not?
  6. Mar 18, 2006 #5


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    In this problem, the beam of light is inclined at an angle of 30 degrees to the grating.


    [tex]d\sin \theta = n\lambda_1[/tex]

    won't be right as you will have to factor in the difference in path length because the beam of light is inclined at an angle of 30 degrees to the grating.
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