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The dispersion relation of a waveguide, which relates the wavenumber \(k\) to the angular frequency \(\omega\), is \(k(\omega)=\frac{1}{c}\sqrt{\omega^2-\omega_0^2}\), where the speed of light is \(c=3\times10^8\,\text{m/s}\) and \(\omega_0\) is a constant. If the group velocity is \(2\times10^8\,\text{m/s}\), then the phase velocity is

1. 1.5 × 10^8 m/s
2. 2 × 10^8 m/s
3. 3 × 10^8 m/s
4. 4.5 × 10^8 m/s

Correct answer: 4.5 × 10^8 m/s

Solution

For \(k=\frac{1}{c}\sqrt{\omega^2-\omega_0^2}\), the phase velocity is \(v_p=\omega/k\) and the group velocity is \(v_g=d\omega/dk\). These satisfy \(v_pv_g=c^2\). With \(v_g=2\times10^8\,\text{m/s}\), \(v_p=\frac{(3\times10^8)^2}{2\times10^8}=4.5\times10^8\,\text{m/s}\).

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