Maxwell's Rainbow — Question 3

Section 33-1 · Physics pages 5376–5377

Question 3 A wavelength passband reverses its endpoints when converted to frequency An ideal filter transmits vacuum wavelengths from 500 nm through 600 nm. Use c = 3.00 × 108 m/s and define fractional frequency bandwidth as ∆f/fc, with arithmetic frequency center fc. λ (nm) T 500 600 Schematic; labels, not drawing lengths, specify the data. (a) Find the transmitted frequency interval and its width. (b) Find fc and the fractional bandwidth; compare fc with the frequency corresponding to 550 nm.
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Question 3 - Solution A wavelength passband reverses its endpoints when converted to frequency λ (nm) T 500 600 (a) Method and calculation fmin = c/(600 nm) = 5.0 × 1014 Hz, fmax = 6.0 × 1014 Hz. ∆f = 1.0 × 1014 Hz . (b) Interpretation and further calculation fc = 5.5 × 1014 Hz, ∆f/fc = 2/11 = 0.1818 . c/(550 nm) = 5.45455 × 1014 Hz̸ = fc . Physical check The longest transmitted wavelength corresponds to the lowest transmitted frequency.

Question 3

A wavelength passband reverses its endpoints when converted to frequency

An ideal filter transmits vacuum wavelengths from 500 nm through 600 nm. Use c=3.00×108c=3.00\times10^8 m/s and define fractional frequency bandwidth as Δf/fc\Delta f/f_c, with arithmetic frequency center fc.

Diagram shown in the original SVG worksheet.

  1. Find the transmitted frequency interval and its width.

  2. Find fc and the fractional bandwidth; compare fc with the frequency corresponding to 550 nm.

Question 3 - Solution

A wavelength passband reverses its endpoints when converted to frequency

Diagram shown in the original SVG worksheet.

(a) Method and calculation

fmin=c/(600nm)=5.0×1014Hz,fmax=6.0×1014Hz.f_{\min}=c/(600\,\mathrm{nm})=5.0\times10^{14}\,\mathrm{Hz},\quad f_{\max}=6.0\times10^{14}\,\mathrm{Hz}.Δf=1.0×1014Hz.\boxed{\Delta f=1.0\times10^{14}\,\mathrm{Hz}}.

(b) Interpretation and further calculation

fc=5.5×1014Hz,Δf/fc=2/11=0.1818.\boxed{f_c=5.5\times10^{14}\,\mathrm{Hz},\quad\Delta f/f_c=2/11=0.1818}.c/(550nm)=5.45455×1014Hz≠fc.\boxed{c/(550\,\mathrm{nm})=5.45455\times10^{14}\,\mathrm{Hz}\ne f_c}.

Physical check

The longest transmitted wavelength corresponds to the lowest transmitted frequency.