GATE 2023 EC Q.20: solution and answer

GATE 2023 · ECQ.20 · 1 mark · MCQ

Consider a narrow band signal, propagating in a lossless dielectric medium ( $ε_{r} = 4, μ_{r} = 1$ ), with phase velocity $v_{p}$ and group velocity $v_{g}$ . Which of the following statement is true? ( $c$ is the velocity of light in vacuum.)

(A)
$v_{p} > c, v_{g} > c$
(B)
$v_{p} < c, v_{g} > c$
(C)
$v_{p} > c, v_{g} < c$
(D)
$v_{p} < c, v_{g} < c$

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Answer and solution

ANSWER

(D) $v_{p} < c, v_{g} < c$ . In a lossless dielectric with $ε_{r} = 4$ and $μ_{r} = 1$ , the medium is non-dispersive, so $v_{p} = v_{g} = \frac{c}{μ _{r} ε _{r}} = \frac{c}{2}$ .

Official answer: DSource: Official answer key, GATE 2023 EC (IIT Kanpur)

For a uniform plane wave in an unbounded, lossless dielectric, the phase constant is $β = ω μ ε = ω μ_{0} μ_{r} ε_{0} ε_{r} = \frac{ω}{c} μ_{r} ε_{r},$ where $c = \frac{1}{μ _{0} ε _{0}} \approx 3 \times 1 0^{8} m/s$ , with $μ_{0} = 4 π \times 1 0^{- 7} H/m$ and $ε_{0} \approx 8.854 \times 1 0^{- 12} F/m$ .
Phase velocity: $v_{p} = \frac{ω}{β} = \frac{c}{μ _{r} ε _{r}} = \frac{c}{1 \times 4} = \frac{c}{2} \approx 1.5 \times 1 0^{8} m/s .$
Group velocity: $v_{g} = (\frac{d β}{d ω})^{- 1}$ . With $ε_{r}$ and $μ_{r}$ constant, $β$ is directly proportional to $ω$ , so $\frac{d β}{d ω} = \frac{μ _{r} ε _{r}}{c} \Rightarrow v_{g} = \frac{c}{μ _{r} ε _{r}} = \frac{c}{2} .$
The medium is non-dispersive (the ratio $ω / β$ does not depend on frequency), so every frequency component of the narrow-band signal moves at the same speed and $v_{p} = v_{g} = \frac{c}{2}$ . Their product is $v_{p} v_{g} = \frac{c ^{2}}{4}$ , not $c^{2}$ . The result $v_{p} v_{g} = c^{2}$ belongs to a dispersive guided wave such as an air-filled hollow waveguide, where $v_{p} > c$ and $v_{g} < c$ . It does not apply to an unbounded plane wave.
Both velocities are $\frac{c}{2}$ , so $v_{p} < c$ and $v_{g} < c$ . Option (D) is correct.

KEY IDEA

In an unbounded lossless dielectric with constant $ε_{r}$ and $μ_{r}$ , $β = \frac{ω}{c} μ_{r} ε_{r}$ is linear in $ω$ . The medium is therefore non-dispersive, and the phase and group velocities are equal: $v_{p} = v_{g} = \frac{c}{μ _{r} ε _{r}}$ . With $μ_{r} ε_{r} = 4 > 1$ , both velocities equal $c /2$ , which is less than $c$ .

WATCH OUT FOR

Using the waveguide result $v_{p} v_{g} = c^{2}$ and so picking (C), $v_{p} > c$ and $v_{g} < c$ . That relation comes from the dispersive cut-off behaviour of an air-filled hollow waveguide. A plane wave in an unbounded lossless dielectric has no cut-off and no dispersion, so $v_{p} = v_{g} = c /2$ . The words 'narrow band' only make group velocity meaningful. They do not make the medium dispersive.

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GATE 2023 EC Q.20: solution and answer

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