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Diffraction Grating Grooves

  1. Apr 18, 2007 #1
    A reflection grating is required that can resolve wavelengths as close as 0.02A (angstroms) in second order for the spectral region around 350 nm. The grating is to be installed in an instrument where light from the entrance slit is incident normally on the grating. If the manufacturer provides rulings over a 10cm grating width, determine
    (a) the minimum number of grooves/cm required
    (b) the optimum blaze angle for work in this region
    (c) the angle of diffraction where irradiance is maximum
    (d) the dispersion in nanometers per degree



    Attempt
    a) the minimum number of grooves/cm required
    Pretty sure part a) is correct
    R = mN = λ/Δλ = (350e-9)/(0.002e-9) = 175000
    m = 2
    So N = 87500
    Grooves/cm => 87500/10 = 8750 grooves/cm

    b) the optimum blaze angle for work in this region
    unsure if i'm correct.
    Useful formulas
    θ_i = incidient angle, θ_b = blaze angle, θ direction of blaze peak
    θ_i - θ_b = θ + θ_b

    d*sin θ = mλ

    However i didn't know how to apply it using these as i didn't see the connection, but did find a formula on google (part 1.6.1 of http://www.jobinyvon.com/usadivisions/OOS/oos1.htm)
    2*sin θ_b = m*n*λ_b
    2*sin θ_b = 2*350*8750*10^-7
    sin θ_b = 0.30625
    θ_b = 17.83 degrees
    although i think this is wrong please help.

    c) the angle of diffraction where irradiance is maximum
    Need help...
    Do i somehow have to use this formula?
    I(θ) = I(0) * ( (sin β)/β )² * ( sin (Nα) / (N*sin α) )²


    d) the dispersion in nanometers per degree
    the formula is dθ/dλ = m/(d*cos θ)
    but how do you use this? i know m = 2, but what's d or θ


    Thank you for helping me.
     
  2. jcsd
  3. Apr 20, 2007 #2
    any help is appreciated.
     
  4. Apr 26, 2007 #3
    your method for part b) seems a bit off
     
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