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Optics : Diffraction gratings.

  1. Apr 24, 2008 #1
    1. I am attempting a question from a textbook but the wording or perhaps the question itself is confusing me.
    *Light falls at perpendicular incidence on a transmission diffraction grating. The second order diffracted light leaving the grating is examined.
    The grating has 600 slits per mm, a total width of 10 cm, and is being used to examine spectral features near a wavelength of 450 nm. How close ( in nm) can the wavelength of two spectral lines be, for the two to still be seen as two, rather than blended into a single intensity peak?

    2. Relevant equations
    Ok so I have done question son diffraction gratings before, but all straightforward, and using the equation d*sin theta =m*lamda

    3. The attempt at a solution

    i have worked out theta to be 32.6 degrees, and (not sure if this is right) but used the equation for double-slit diffraction: y=m*lamda*D/d and worked out the spacing between the maximum and the first minimum, y, to be 0.05389m. Is this at all on the right track or am I totally lost?

    I fear the latter. Any help much appreciated!
  2. jcsd
  3. Apr 25, 2008 #2


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    Homework Helper

    I think you want to look up the formulas dealing with resolving power. There is one that gives the resolving power needed to differentiate two close wavelengths in this experiment; and there is another that gives the resolving power of a specific diffraction grating that depends on the order number.
  4. Apr 25, 2008 #3
    Thanks, i found the equation R=lamda/deltalamda
    and R=mN, where N is the number of gratings and m the order.
    So i basically worked out R from the second equation where I have both m and N, then I plugged it into the first and got 3.75x10^-12 m for delta lamda, is that on the right track?
  5. Apr 25, 2008 #4


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    That looks right to me.
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