What Are the Key Parameters for Designing a Reflection Diffraction Grating?

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SUMMARY

The discussion focuses on designing a reflection diffraction grating capable of resolving wavelengths as close as 0.02 Å in second order for the spectral region around 350 nm. The minimum number of grooves required is calculated to be 8750 grooves/cm based on the formula R = mN = λ/Δλ. The optimum blaze angle is determined to be approximately 17.83 degrees using the equation 2*sin θ_b = m*n*λ_b. The angle of diffraction for maximum irradiance and the dispersion in nanometers per degree remain unresolved, with participants seeking further clarification on the relevant formulas.

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  • Understanding of diffraction grating principles
  • Familiarity with the equations governing wavelength resolution
  • Knowledge of blaze angle calculations
  • Ability to apply trigonometric functions in optical contexts
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  • Study the principles of blaze angle optimization in diffraction gratings
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  • Investigate the dispersion formula dθ/dλ = m/(d*cos θ) and its application in grating design
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Qyzren
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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.
 
Physics news on Phys.org
any help is appreciated.
 
your method for part b) seems a bit off
 

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