Question:
Consider a narrow band signal, propagating in a lossless dielectric medium $(\varepsilon_r = 4, \mu_r = 1)$, 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.)
Consider a narrow band signal, propagating in a lossless dielectric medium $(\varepsilon_r = 4, \mu_r = 1)$, 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.)
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In a lossless, non-dispersive medium, phase velocity equals group velocity. Both are reduced by a factor of $\sqrt{\varepsilon_r \mu_r}$ compared to $c$.
Updated On: Feb 12, 2026
- $v_p>c, \; v_g>c$
- $v_p<c, \; v_g>c$
- $v_p>c, \; v_g<c$
- $v_p<c, \; v_g<c$
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The Correct Option is D
Solution and Explanation
Step 1: Phase velocity in a medium.
For a non-magnetic dielectric with $\mu_r = 1$, \[ v_p = \frac{c}{\sqrt{\varepsilon_r \mu_r}} = \frac{c}{\sqrt{4 . 1}} = \frac{c}{2}. \] Thus, $v_p = \tfrac{c}{2}<c$.
Step 2: Group velocity in a non-dispersive medium.
In a lossless, non-dispersive dielectric medium, \[ v_g = v_p. \] Hence, $v_g = \tfrac{c}{2}<c$.
Step 3: Conclusion.
Both $v_p$ and $v_g$ are less than $c$.
Therefore, the correct statement is: \[ v_p<c, v_g<c. \] \[ \boxed{(D)} \]
For a non-magnetic dielectric with $\mu_r = 1$, \[ v_p = \frac{c}{\sqrt{\varepsilon_r \mu_r}} = \frac{c}{\sqrt{4 . 1}} = \frac{c}{2}. \] Thus, $v_p = \tfrac{c}{2}<c$.
Step 2: Group velocity in a non-dispersive medium.
In a lossless, non-dispersive dielectric medium, \[ v_g = v_p. \] Hence, $v_g = \tfrac{c}{2}<c$.
Step 3: Conclusion.
Both $v_p$ and $v_g$ are less than $c$.
Therefore, the correct statement is: \[ v_p<c, v_g<c. \] \[ \boxed{(D)} \]
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