Answer to Question #255610 in Optics for Xola

Question

A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence. (i) What is the angular separation, in the third-order spectrum, between the 400 nm and 600 nm lines? [5] (ii) Water (of refractive index 1.33) now fills the whole space between the grating and the screen? What is the angular separation, in the first-order spectrum, between the 400 nm and 600 nm lines? [5]

Answer 1

(i)

$d= \frac{1}{N_p} = \frac{10^{-3}}{300} \\ N_p = 300 \times 10^3 \\ λ_1 = 400 \times 10^{-9} \;m \\ λ_2 = 600 \times 10^{-9} \;m \\ dsin θ_1 = nλ_1 \\ \frac{10^{-3}}{300} \times sin θ_1 = 3 \times 400 \times 10^{-9} \\ sin θ_1 = 12 \times 3 \times 10^{-2} \\ θ_1 = sin^{-1} (0.36) \\ θ_1=21.1 \;° \\ \frac{10^{-3}}{300} sin θ_2 = 3 \times λ_2 \\ \frac{10^{-3}}{300} sin θ_2 = 3 \times 600 \times 10^{-9} \\ θ_2 = sin^{-1} (0.54) \\ θ_2 = 32.68 \;° \\ Δ θ = θ_2- θ_1 \\ Δ θ = 32.68-21.1 = 11.58 \;°$

(ii)

$dsin θ^*_1 = \frac{mλ_1}{n} \\ n = 1.33 \\ m=3 \\ θ^*_1 = sin^{-1} (\frac{3 \times 400 \times 10^{-9} \times 300 \times 10^3}{1.33}) \\ θ^*_1 = sin^{-1} (0.2706) = 15.699 \;° \\ dsin θ^*_2 = \frac{mλ_2}{n} \\ λ_2 = 600 \times 10^{-9} \; m \\ θ^*_2 = sin^{-1} (\frac{3 \times 600 \times 10^{-9} \times 300 \times 10^3}{1.33}) \\ θ^*_2 = sin^{-1} (0.406) = 23.953 \;° \\ Δ θ^* = θ^*_2- θ^*_1 \\ Δ θ^* = 23.953-15.699 = 8.254 \;°$