Answer to Question #255610 in Optics for Xola

Question #255610
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]
1
Expert's answer
2021-10-24T18:27:30-0400

(i)

d=1Np=103300Np=300×103λ1=400×109  mλ2=600×109  mdsinθ1=nλ1103300×sinθ1=3×400×109sinθ1=12×3×102θ1=sin1(0.36)θ1=21.1  °103300sinθ2=3×λ2103300sinθ2=3×600×109θ2=sin1(0.54)θ2=32.68  °Δθ=θ2θ1Δθ=32.6821.1=11.58  °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)

dsinθ1=mλ1nn=1.33m=3θ1=sin1(3×400×109×300×1031.33)θ1=sin1(0.2706)=15.699  °dsinθ2=mλ2nλ2=600×109  mθ2=sin1(3×600×109×300×1031.33)θ2=sin1(0.406)=23.953  °Δθ=θ2θ1Δθ=23.95315.699=8.254  °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 \;°


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