The Traveling Electromagnetic Wave, Quantitatively — Question 2

Section 33-3 · Physics pages 5414–5415

Question 2 A phase measurement needs a wavelength bound to resolve cycle ambiguity At one instant, the phase increase between two probes separated by d=0.300 m along +x is pi/2 modulo 2pi. The wave travels +x and its wavelength is known to lie between 0.80 m and 1.50 m. A fixed probe measures period T=4.00 ns. phase comparison d Schematic; labels, not drawing lengths, specify the data. (a) List the possible positive wave numbers and use the wavelength bound to select one. (b) Find the wavelength and phase speed, and test consistency with vacuum propagation.
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Question 2 - Solution A phase measurement needs a wavelength bound to resolve cycle ambiguity phase comparison d (a) Method and calculation kd = π/2 + 2πm ⇒km = π/2 + 2πm d , m = 0, 1, 2, . . . . λm = 2π km = 4d 1 + 4m ⇒m = 0 . (b) Interpretation and further calculation λ = 1.20 m, k = 5π/3 rad/m . vϕ = λ/T = 3.00 × 108 m/s consistent with vacuum. Physical check Without the wavelength bound, the phase difference alone cannot distinguish an added whole number of cycles.

Question 2

A phase measurement needs a wavelength bound to resolve cycle ambiguity

At one instant, the phase increase between two probes separated by d=0.300 m along +x is pi/2 modulo 2pi. The wave travels +x and its wavelength is known to lie between 0.80 m and 1.50 m. A fixed probe measures period T=4.00 ns.

Diagram shown in the original SVG worksheet.

  1. List the possible positive wave numbers and use the wavelength bound to select one.

  2. Find the wavelength and phase speed, and test consistency with vacuum propagation.

Question 2 - Solution

A phase measurement needs a wavelength bound to resolve cycle ambiguity

Diagram shown in the original SVG worksheet.

(a) Method and calculation

kd=π/2+2πm⇒km=π/2+2πmd,m=0,1,2,….kd=\pi/2+2\pi m\Rightarrow k_m=\frac{\pi/2+2\pi m}{d},\quad m=0,1,2,\ldots.λm=2πkm=4d1+4m⇒m=0.\lambda_m=\frac{2\pi}{k_m}=\frac{4d}{1+4m}\Rightarrow\boxed{m=0}.

(b) Interpretation and further calculation

λ=1.20m,k=5π/3rad/m.\boxed{\lambda=1.20\,\mathrm m,\quad k=5\pi/3\,\mathrm{rad/m}}.vϕ=λ/T=3.00×108m/sconsistent with vacuum.\boxed{v_\phi=\lambda/T=3.00\times10^8\,\mathrm{m/s}}\quad\text{consistent with vacuum}.

Physical check

Without the wavelength bound, the phase difference alone cannot distinguish an added whole number of cycles.