GATE PH 2025 • Q16
MCQ • 1 Mark
Section 3: Electromagnetic Theory
Electromagnetic Waves in Media
Subtopic: Wave Propagation in Dielectrics & Conductors
Question Statement:
Which one of the following is correct for the phase velocity vp and group velocity
vg? (c is the speed of light in vacuum)
Options:
-
(A) For matter waves in the relativistic case, vpvg > c2
-
(B) For electromagnetic waves in a medium, vp represents the speed with which energy
propagates
-
(C) For electromagnetic waves in a medium, both vp and vg can be more than c
-
(D) For matter waves in free space, vp ≠ vg
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.16)**
**1. Wave Mechanics Formulation**:
- Phase velocity: $v_p = \frac{\omega}{k}$
- Group velocity: $v_g = \frac{d\omega}{dk}$
**2. Analytical Derivation for Non-Relativistic Matter Waves**:
- For a free non-relativistic particle of mass $m$ and momentum $p = \hbar k$:
$E = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}$
- Using de Broglie relation $E = \hbar\omega$:
$\hbar\omega = \frac{\hbar^2 k^2}{2m} \implies \omega(k) = \frac{\hbar k^2}{2m}$
- Step 1: Compute phase velocity:
$v_p = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m} = \frac{v}{2}$
- Step 2: Compute group velocity:
$v_g = \frac{d\omega}{dk} = \frac{d}{dk}\left(\frac{\hbar k^2}{2m}\right) = \frac{\hbar k}{m} = \frac{p}{m} = v$
- Step 3: Compare velocities:
$v_p = \frac{1}{2}v_g \implies v_p \ne v_g$
- For matter waves in free space, phase velocity and group velocity are strictly unequal ($v_p = v_g/2$).
**3. Official Verification**:
The verified official IIT answer is **Option (D)** (`For matter waves in free space, vp ≠ vg`).
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