# Homework Help: Diffraction Grating Grooves

1. Apr 18, 2007

### Qyzren

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

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. Apr 20, 2007

### Qyzren

any help is appreciated.

3. Apr 26, 2007

### Spastik_Relativity

your method for part b) seems a bit off