Maxwell's Rainbow — Question 4

Section 33-1 · Physics pages 5378–5379

Question 4 Equal cycle counts do not imply equal pulse durations across the spectrum Two approximately monochromatic vacuum pulses each contain N=120 carrier cycles. Pulse A has wavelength 2.5 mm; pulse B has wavelength 500 nm. Treat the envelope length as Nλ and use c = 3.00 × 108 m/s. x E c ℓ= Nλ Schematic; labels, not drawing lengths, specify the data. (a) Find each pulse duration and spatial length. (b) Find the duration ratio and the number of B cycles in the duration of pulse A.
Show solutionHide solution
Question 4 - Solution Equal cycle counts do not imply equal pulse durations across the spectrum x E c ℓ= Nλ (a) Method and calculation τ = N/f = Nλ/c, ℓ= cτ = Nλ. τA = 1.00 ns, ℓA = 0.300 m; τB = 0.200 ps, ℓB = 60.0 µm . (b) Interpretation and further calculation τA/τB = λA/λB = 5000 . NB in τA = fBτA = 6.00 × 105 . Physical check A finite pulse has a bandwidth; the carrier-cycle description is an approximation suitable for a many-cycle pulse.

Question 4

Equal cycle counts do not imply equal pulse durations across the spectrum

Two approximately monochromatic vacuum pulses each contain N=120 carrier cycles. Pulse A has wavelength 2.5 mm; pulse B has wavelength 500 nm. Treat the envelope length as NλN\lambda and use c=3.00×108c=3.00\times10^8 m/s.

Diagram shown in the original SVG worksheet.

  1. Find each pulse duration and spatial length.

  2. Find the duration ratio and the number of B cycles in the duration of pulse A.

Question 4 - Solution

Equal cycle counts do not imply equal pulse durations across the spectrum

Diagram shown in the original SVG worksheet.

(a) Method and calculation

τ=N/f=Nλ/c,ℓ=cτ=Nλ.\tau=N/f=N\lambda/c,\quad\ell=c\tau=N\lambda.τA=1.00ns,ℓA=0.300m;τB=0.200ps,ℓB=60.0μm.\boxed{\tau_A=1.00\,\mathrm{ns},\ \ell_A=0.300\,\mathrm m;\quad\tau_B=0.200\,\mathrm{ps},\ \ell_B=60.0\,\mu\mathrm m}.

(b) Interpretation and further calculation

τA/τB=λA/λB=5000.\boxed{\tau_A/\tau_B=\lambda_A/\lambda_B=5000}.NB in τA=fBτA=6.00×105.\boxed{N_{B\text{ in }\tau_A}=f_B\tau_A=6.00\times10^5}.

Physical check

A finite pulse has a bandwidth; the carrier-cycle description is an approximation suitable for a many-cycle pulse.