The dispersion equation of a waveguide, which relates the wavenumber k to the frequency ω, is \(k\left( \omega \right) = \left( {1/\;c} \right)\;\sqrt {{\omega ^2} - \omega _o^2} \) where the speed of light c = 3 × 10^{8} m/s, and ω_{o} is a constant. If the group velocity is 2 × 10^{8} m/s, then the phase velocity is

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GATE EC 2019 Official Paper

Option 4 : 4.5 × 10^{8} m/s

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept__:

*Phase velocity:*

The rate at which the phase of the wave propagates in space

\({V_p} = \frac{\omega }{\beta }\)

*Group Velocity:*

The velocity with which overall envelope of the wave travels

\({V_g} = \frac{{d\omega }}{{d\beta }} = \frac{{d\omega }}{{dk}}\)

\(K\left( \omega \right) = \frac{1}{c}\sqrt {\omega - \omega _0^2} \ldots \left( 1 \right)\)

V_{g} = 2 × 10^{8} m/s (green)

\({V_g} = \frac{{d\omega }}{{dk}}\)

\(\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}}\)

differentiating (1)

\(\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}} = \frac{{1 \times 2\omega }}{{2C\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{{V_g}}}\)

\(\Rightarrow \frac{\omega }{{3 \times {{10}^8}\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{2 \times {{10}^8}}}\)

\(\sqrt {{\omega ^2} - \omega _0^2} = \frac{{2\omega }}{3}\)

\({V_p} = \frac{\omega }{K} = \frac{{\omega c}}{{\sqrt {{\omega ^2} - \omega _0^2} }}\)

\(= \frac{{\omega c}}{{\left( {\frac{{2\omega }}{3}} \right)}}\)

\({V_p} = \frac{{3c}}{2} = \frac{{3 \times {{10}^8} \times 3}}{2} = 4.5 \times {10^8}m/s\)

__Shortcut__:

Group velocity × Phase velocity = c^{2}

2 × 10^{8} × phase velocity = (3 × 10^{8})^{2}

\({V_p} = \frac{{9 \times {{10}^{16}}}}{{2 \times {{10}^8}}}\)

= 4.5 × 10