Related Concepts: 5.02 Plane Waves in Lossless vs. Lossy Media & Skin Depth Calculations | 5.03 Poynting Vector, Power Flow & Normal Incidence Reflection (Gamma, Tau) | 5.04 Wave Polarization & Ionospheric Sky-Wave Radio Propagation | 5.07 Solved PYQ Numerical Bank - Waves & Propagation

5.05 Dispersion, Phase & Group Velocity, Doppler Effect & Brewster’s Angle

Core Idea

Three loosely related “extra” topics that between them account for roughly 20 marks per paper:

• Dispersion — why a signal made of many frequencies smears out as it travels, and the $u_p$ vs $u_g$ distinction that explains it.
• Doppler effect — why a moving source shifts the observed frequency, and why receding stars look red.
• Brewster’s angle — the oblique-incidence angle at which parallel-polarized light is not reflected at all.

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1. Phase Velocity, Group Velocity & Dispersion

[PYQ: 2024 — 10 Marks] — “Define phase velocity and group velocity of a plane wave, and explain the concept of dispersion in wave propagation.”

1.1 Definitions

Phase velocity ( $u_p$)

The speed at which a surface of constant phase of a single-frequency wave advances through the medium: $u_p=\frac{\omega }{\beta }[\text{m/s}]$ It describes how fast the individual wave crests move. It carries no information — a pure sinusoid of infinite extent conveys nothing.

Group velocity ( $u_g$)

The speed at which the envelope of a wave packet — a group of closely spaced frequencies — travels: $u_g=\frac{d\omega }{d\beta }=\frac{1}{d\beta /d\omega }[\text{m/s}]$ Since all real signals are modulated (finite bandwidth), the group velocity is the speed at which information and energy actually travel.

Dispersion

Dispersion is the distortion suffered by a signal when its different frequency components propagate at different phase velocities, so the components arrive out of their original relative phase and the pulse shape spreads. A medium is dispersive if $u_p$ depends on $\omega$ — equivalently, if $\beta$ is not a linear function of $\omega$.

1.2 The general relation

Differentiate $\beta =\omega /u_p$ with respect to $\omega$:

$\frac{d\beta }{d\omega }=\frac{1}{u_p}-\frac{\omega }{u_p^2}\frac{d{u}_p}{d\omega }$

Invert to get:

$\boxed{u_g=\frac{u_p}{1-\frac{\omega }{u_p}\frac{d{u}_p}{d\omega }}}$

1.3 The three regimes

Regime	Condition	Relation	Behaviour
No dispersion	$\frac{d{u}_p}{d\omega }=0$	$u_g=u_p$	All components travel together; pulse shape is preserved perfectly.
Normal dispersion	$\frac{d{u}_p}{d\omega }<0$	$u_g<u_p$	$u_p$ falls with frequency; higher frequencies lag.
Anomalous dispersion	$\frac{d{u}_p}{d\omega }>0$	$u_g>u_p$	$u_p$ rises with frequency; higher frequencies lead.

[GRAPH: Two curves of β versus ω on the same axes — a straight line through the origin (non-dispersive, u_p = u_g = constant slope) and a curved line (dispersive, local slope ≠ chord slope). Governing relations: u_p = ω/β is the chord slope from the origin; u_g = dω/dβ is the local tangent slope. — source: L 14.pdf]

Why $u_g=u_p$ means no dispersion [PYQ: 2018]

If $u_p$ is independent of $\omega$, then $\beta =\omega /u_p$ is directly proportional to $\omega$ — a straight line through the origin. Every frequency component then travels at the identical speed, so their relative phases are preserved and the envelope arrives undistorted. The tangent slope equals the chord slope, i.e. $u_g=u_p$. Dispersion is precisely the failure of that proportionality.

1.4 High-yield proof: $u_p{u}_g=c^2$

[PYQ: 2018 — 03+08 Marks] — “Investigate why there will be no dispersion when group velocity and phase velocity are equal. Prove that ${\mu }_p{\mu }_g=c^2$.”

In an ionised medium (plasma) or a waveguide, the phase constant is:

$\beta =\frac{\omega }{c}\sqrt{1-{\left(\frac{{\omega }_p}{\omega }\right)}^2}=\frac{1}{c}\sqrt{{\omega }^2-{\omega }_p^2}$

Square both sides: $c^2{\beta }^2={\omega }^2-{\omega }_p^2$. Differentiate with respect to $\omega$ (noting ${\omega }_p$ is a constant of the medium):

$2c^2\beta \frac{d\beta }{d\omega }=2\omega ⟹\beta \frac{d\beta }{d\omega }=\frac{\omega }{c^2}$

Substitute $\beta =\omega /u_p$ and $d\beta /d\omega =1/u_g$:

$\frac{\omega }{u_p}\cdot \frac{1}{u_g}=\frac{\omega }{c^2}$

Key Result

$\boxed{u_p{u}_g=c^2}$

The consequence people find surprising

Since $u_g<c$ always (information cannot outrun light), this forces $u_p>c$ in a plasma or waveguide. The phase velocity legitimately exceeds the speed of light. No relativity violation occurs, because phase velocity carries no information — only $u_g$ does, and it stays below $c$.

Anomalous vs normal dispersion — graphical distinction (syllabus item, source L 14.pdf)

• Normal: refractive index $n$ increases with frequency; $d{u}_p/d\omega <0$; $u_g<u_p$. Typical of transparent dielectrics away from absorption bands (glass, ordinary optical fibre).
• Anomalous: $n$ decreases with frequency; $d{u}_p/d\omega >0$; $u_g>u_p$. Occurs near a strong absorption resonance, and in plasmas and waveguides.

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2. The Doppler Effect

[PYQ: 2015, 2016, 2017, 2019, 2021, 2024] — ⭐⭐⭐⭐⭐

2.1 Definition and mechanism

Doppler effect

The Doppler effect is the apparent shift in the observed frequency of a wave caused by relative motion between the source and the observer along the line joining them.

Mechanism. Each successive wavefront is emitted from a slightly different position. If the source approaches, each new crest has a shorter distance to travel than the last, so the crests arrive more often — wavelength compresses, frequency rises. If the source recedes, each crest has further to travel, crests arrive less often — wavelength stretches, frequency falls.

[FIGURE: A moving transmitter emitting concentric circular wavefronts, bunched together ahead of the motion and spread apart behind it, with a stationary observer on each side. — source: Sadiku 7th Ed., Fig. 10.19 / L 14.pdf]

2.2 Mathematics

For a source and receiver in relative motion with radial velocity $v$ along the line of sight, with $v\ll c$:

$\boxed{f^{\prime }=f\left(1\pm \frac{v}{c}\right)}$

Motion	Sign	Observed frequency	Wavelength
Source approaching receiver	$+$	$f^{\prime }=f\left(1+\frac{v}{c}\right)>f$	Compressed (blue shift)
Source receding from receiver	$-$	$f^{\prime }=f\left(1-\frac{v}{c}\right)<f$	Stretched (red shift)

The Doppler shift itself is:

$\Delta f=f^{\prime }-f=\pm \frac{v}{c}f$

Derivation sketch for "with proper mathematical illustration"

Let the source move away at speed $v$. In one period $T=1/f$ the wave advances $cT$ while the source retreats $vT$, so the emitted wavelength is stretched to ${\lambda }^{\prime }=(c+v)T$. The observed frequency is $f^{\prime }=\frac{c}{{\lambda }^{\prime }}=\frac{c}{(c+v)/f}=\frac{f}{1+v/c}\approx f\left(1-\frac{v}{c}\right)\text{for }v\ll c$ The approaching case follows with $v\to -v$.

2.3 Red shift of a receding star

[PYQ: 2020, 2024]

The answer

Light from a star receding from Earth is Doppler-shifted downward in frequency, i.e. toward longer wavelengths. Since red lies at the long-wavelength / low-frequency end of the visible spectrum, the star’s characteristic spectral lines appear displaced toward red — the red shift. The larger the recession velocity, the greater the shift. Hubble’s observation that essentially every distant galaxy is red-shifted, with shift proportional to distance, is the primary evidence that the universe is expanding.

The converse, a blue shift, indicates an approaching object.

2.4 Practical examples

[PYQ: 2020 — “State two more practical examples of this particular effect”]

Application	How the Doppler effect is used
Doppler radar / police speed guns	Vehicle speed is computed from the frequency shift of the radar signal reflected off the moving car.
Satellite tracking	A satellite’s orbit and velocity are determined from the shift in its radio carrier as it passes over a ground station.
Doppler weather radar	Detects rotation in storm cells (and therefore tornado formation) from the shift of returns off moving raindrops.
Medical Doppler ultrasound	Measures blood flow velocity from the shift of ultrasound reflected off moving red blood cells.
Astronomical red shift	Measures recession velocities of galaxies.

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3. Brewster’s Angle (Oblique Incidence)

3.1 Definition

[PYQ: 2020 — 04 Marks]

Brewster angle ( ${\theta }_B$)

The Brewster angle is the particular angle of incidence at which a parallel-polarized (p-polarized) electromagnetic wave striking a dielectric boundary experiences zero reflection — all of the incident power is transmitted into the second medium. It is also called the polarizing angle, because an unpolarized wave incident at ${\theta }_B$ produces a purely perpendicular-polarized reflected wave. $\tan {\theta }_B=\sqrt{\frac{{ϵ}_{r2}}{{ϵ}_{r1}}}=\frac{n_2}{n_1}$

S- and P-polarization (Instructor 2 notes concept)

• P-polarization (parallel, TM): $\vec{E}$ lies in the plane of incidence.
• S-polarization (perpendicular, TE): $\vec{E}$ is perpendicular to the plane of incidence (German senkrecht).

The plane of incidence is the plane containing the incident ray and the surface normal. The two polarizations reflect differently, which is exactly why Brewster’s angle exists for one and not the other.

3.2 Master proof: Brewster’s angle exists only for parallel polarization

[PYQ: 2022 — 13 Marks] — “Mathematically prove that in case of non-magnetic media, Brewster’s angle exists only for parallel polarizations rather than for perpendicular polarizations.”

For non-magnetic media, ${\mu }_1={\mu }_2={\mu }_0$, so $\eta ={\eta }_0/\sqrt{{ϵ}_r}$.

Case 1 — Perpendicular (S) polarization: no solution

${\Gamma }_{\perp }=\frac{{\eta }_2\cos {\theta }_i-{\eta }_1\cos {\theta }_t}{{\eta }_2\cos {\theta }_i+{\eta }_1\cos {\theta }_t}$

Set ${\Gamma }_{\perp }=0\Rightarrow {\eta }_2\cos {\theta }_{iB}={\eta }_1\cos {\theta }_t$. With $\eta \propto 1/\sqrt{{ϵ}_r}$:

$\frac{\cos {\theta }_{iB}}{\sqrt{{ϵ}_{r2}}}=\frac{\cos {\theta }_t}{\sqrt{{ϵ}_{r1}}}⟹{ϵ}_{r1}{\cos }^2{\theta }_{iB}={ϵ}_{r2}{\cos }^2{\theta }_t$

Apply Snell’s law, $\sin {\theta }_t=\sqrt{{ϵ}_{r1}/{ϵ}_{r2}}\sin {\theta }_{iB}$, so ${\cos }^2{\theta }_t=1-\frac{{ϵ}_{r1}}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}$:

${ϵ}_{r1}(1-{\sin }^2{\theta }_{iB})={ϵ}_{r2}\left(1-\frac{{ϵ}_{r1}}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}\right)$

${ϵ}_{r1}-{ϵ}_{r1}{\sin }^2{\theta }_{iB}={ϵ}_{r2}-{ϵ}_{r1}{\sin }^2{\theta }_{iB}$

$⟹{ϵ}_{r1}={ϵ}_{r2}$

Conclusion for S-polarization

The ${\sin }^2{\theta }_{iB}$ terms cancel identically, leaving a condition that contains no angle at all. It demands ${ϵ}_{r1}={ϵ}_{r2}$ — i.e. the two media are identical and there is no boundary. No Brewster angle exists for perpendicular polarization in non-magnetic media.

Case 2 — Parallel (P) polarization: a valid solution exists

${\Gamma }_{\parallel }=\frac{{\eta }_2\cos {\theta }_t-{\eta }_1\cos {\theta }_i}{{\eta }_2\cos {\theta }_t+{\eta }_1\cos {\theta }_i}$

Set ${\Gamma }_{\parallel }=0\Rightarrow {\eta }_2\cos {\theta }_t={\eta }_1\cos {\theta }_{iB}$, giving:

$\sqrt{{ϵ}_{r1}}\cos {\theta }_t=\sqrt{{ϵ}_{r2}}\cos {\theta }_{iB}⟹{ϵ}_{r1}(1-{\sin }^2{\theta }_t)={ϵ}_{r2}(1-{\sin }^2{\theta }_{iB})$

Substitute Snell’s law again:

${ϵ}_{r1}\left(1-\frac{{ϵ}_{r1}}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}\right)={ϵ}_{r2}(1-{\sin }^2{\theta }_{iB})$

${ϵ}_{r1}-\frac{{ϵ}_{r1}^2}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}={ϵ}_{r2}-{ϵ}_{r2}{\sin }^2{\theta }_{iB}$

${\sin }^2{\theta }_{iB}\left({ϵ}_{r2}-\frac{{ϵ}_{r1}^2}{{ϵ}_{r2}}\right)={ϵ}_{r2}-{ϵ}_{r1}$

Multiply through by ${ϵ}_{r2}$:

${\sin }^2{\theta }_{iB}\left({ϵ}_{r2}^2-{ϵ}_{r1}^2\right)={ϵ}_{r2}({ϵ}_{r2}-{ϵ}_{r1})$

${\sin }^2{\theta }_{iB}=\frac{{ϵ}_{r2}({ϵ}_{r2}-{ϵ}_{r1})}{({ϵ}_{r2}-{ϵ}_{r1})({ϵ}_{r2}+{ϵ}_{r1})}=\frac{{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}$

Convert to a tangent using ${\tan }^2\theta =\frac{{\sin }^2\theta }{1-{\sin }^2\theta }$:

${\tan }^2{\theta }_{iB}=\frac{{ϵ}_{r2}/({ϵ}_{r1}+{ϵ}_{r2})}{1-{ϵ}_{r2}/({ϵ}_{r1}+{ϵ}_{r2})}=\frac{{ϵ}_{r2}}{{ϵ}_{r1}}$

Conclusion for P-polarization

$\boxed{\tan {\theta }_{iB}=\sqrt{\frac{{ϵ}_{r2}}{{ϵ}_{r1}}}}$ Unlike the S-polarization case, the ${\sin }^2{\theta }_{iB}$ terms do not cancel, and the result is a genuine, real, always-solvable angle for any pair of positive permittivities. Brewster’s angle therefore exists strictly for parallel polarization. Proved.

The one-line reason, if you are short on time

In the perpendicular case the angle-dependent terms cancel algebraically and the zero-reflection condition degenerates to ${ϵ}_{r1}={ϵ}_{r2}$ (no boundary). In the parallel case they survive and yield $\tan {\theta }_B=\sqrt{{ϵ}_{r2}/{ϵ}_{r1}}$.

3.3 Reflection characteristics: S vs P (Instructor 2 notes concept)

[GRAPH: |Γ| versus angle of incidence θ_i from 0° to 90° for both polarizations at a dielectric boundary. The S-curve (perpendicular) rises monotonically from |Γ₀| at normal incidence to 1 at grazing. The P-curve (parallel) falls to exactly zero at θ_B, then rises to 1 at grazing. Governing equations: Γ⊥ = (η₂cosθᵢ − η₁cosθₜ)/(η₂cosθᵢ + η₁cosθₜ); Γ∥ = (η₂cosθₜ − η₁cosθᵢ)/(η₂cosθₜ + η₁cosθᵢ). — source: L 15_updated.pdf]

Feature	S-polarized (perpendicular)	P-polarized (parallel)
$\vec{E}$ orientation	Perpendicular to plane of incidence	In the plane of incidence
$\Vert \Gamma \Vert$ vs ${\theta }_i$	Increases monotonically to 1	Dips to zero at ${\theta }_B$, then rises to 1
Brewster angle	Does not exist (non-magnetic media)	$\tan {\theta }_B=\sqrt{{ϵ}_{r2}/{ϵ}_{r1}}$
At grazing incidence	Total reflection	Total reflection

Everyday application

Glare off water, roads and glass is predominantly S-polarized, because near Brewster’s angle the P-component is barely reflected. Polarized sunglasses are cut to block the horizontal (S) component and so remove most of the glare.

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4. Common Mistakes That Cost Marks

Avoid these

1. Saying phase velocity carries energy. It does not — only group velocity does. This distinction is the whole point of the $u_p{u}_g=c^2$ question.
2. Panicking at $u_p>c$. It is legitimate and expected in a plasma or waveguide. Say why it is not a relativity violation.
3. Confusing normal and anomalous dispersion. Normal: $u_g<u_p$. Anomalous: $u_g>u_p$.
4. Getting the Doppler sign backwards. Approaching → frequency up. Receding → frequency down → red shift.
5. In the Brewster proof, not showing that the S-case degenerates. The whole question is why one polarization has a solution and the other does not — showing the cancellation is the answer.
6. Quoting $\tan {\theta }_B=\sqrt{{ϵ}_{r1}/{ϵ}_{r2}}$ upside down. It is medium-2 over medium-1.

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5. PYQ Bank — Verbatim Questions & Answer Plans

Q1 — Phase, group velocity and dispersion [PYQ: 2024 — 10 Marks]

“Define phase velocity and group velocity of a plane wave, and explain the concept of dispersion in wave propagation.”

Answer plan: §1.1 three definitions with formulas → §1.2 general relation → §1.3 regime table → sketch of $\beta$ vs $\omega$ for dispersive and non-dispersive media.

Q2 — No dispersion when $u_g=u_p$; prove $u_p{u}_g=c^2$ [PYQ: 2018 — 03+08 Marks]

“Investigate why there will be no dispersion when group velocity and phase velocity are equal. Prove that ${\mu }_p{\mu }_g=c^2$.”

Answer plan: Part 1 (3 marks) → the §1.3 success callout: $u_p$ independent of $\omega \Rightarrow \beta \propto \omega \Rightarrow$ all components travel together. Part 2 (8 marks) → the full §1.4 differentiation.

Q3 — Doppler with mathematical illustration [PYQ: 2015, 2017, 2019 — 07/10 Marks]

“Explain Doppler effect with proper mathematical illustration.”

Answer plan: §2.1 definition and mechanism → the wavefront figure → §2.2 derivation sketch and the $f^{\prime }=f(1\pm v/c)$ formula → the approach/recede table.

Q4 — Doppler + red shift [PYQ: 2024 — 13 Marks]

“Explain Doppler effect in electromagnetics with proper illustration. How does the Doppler effect relate to the red-shift of a receding star?”

Answer plan: Q3 content (≈8 marks) then §2.3 in full (≈5 marks), including the cosmological expansion point.

Q5 — Red shift + two examples [PYQ: 2020 — 06 Marks]

“Which effect causes the ‘red shift’ of the light spectrum emitted by a receding distant star. State two more practical examples of this particular effect.”

Answer plan: Name the Doppler effect, give the §2.3 explanation briefly, then pick two from the §2.4 table with a sentence of mechanism each.

Q6 — Define Brewster angle [PYQ: 2020 — 04 Marks]

“Define Brewster angle.”

Answer plan: §3.1 definition callout plus the formula. Mention the polarizing-angle property — it is the natural 4-mark extension.

Q7 — Brewster exists only for parallel polarization [PYQ: 2022 — 13 Marks]

“Mathematically prove that in case of non-magnetic media, Brewster’s angle exists only for parallel polarizations rather than for perpendicular polarizations.”

Answer plan: §3.2 — both cases, in full. Case 1 must end at ${ϵ}_{r1}={ϵ}_{r2}$ with the explicit statement that this means “no boundary, therefore no solution.” Case 2 must end at the boxed tangent formula. State ${\mu }_1={\mu }_2={\mu }_0$ at the start and use Snell’s law by name.

Q8 — Doppler as a short-note term [PYQ: 2016, 2021 — part of 09/10 Marks]

“Explain the following terms: i) Doppler effect in electromagnetics, ii) Skin depth of a conductor, iii) Plasma frequency and plasma oscillation.”

Answer plan: Doppler part → §2.1 + the formula, 4 lines. Other parts route to 5.02 Plane Waves in Lossless vs. Lossy Media & Skin Depth Calculations and 5.04 Wave Polarization & Ionospheric Sky-Wave Radio Propagation.

Q9 — Dispersion inside the lossy-medium numerical [PYQ: 2017, 2019, 2021]

The narrow-band signal problem ends with “Determine $u_p$ and $u_g$. Is the medium dispersive?” — you compute both velocities and compare. Full working in 5.07 Solved PYQ Numerical Bank - Waves & Propagation.

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6. Self-Check Before Moving On

• Can you define $u_p$ and $u_g$ and say which one carries information?
• Can you state the condition for a medium to be dispersive in terms of $\beta (\omega )$?
• Can you distinguish normal from anomalous dispersion by the sign of $d{u}_p/d\omega$?
• Can you derive $u_p{u}_g=c^2$ from the plasma dispersion relation?
• Can you explain why $u_p>c$ is not a relativity violation?
• Can you derive $f^{\prime }=f(1\pm v/c)$ from the wavefront-stretching argument?
• Can you explain red shift and name two other Doppler applications?
• Can you run both halves of the Brewster proof and say exactly where the S-case fails?

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Source: 05 electromagnetic_waves_master_notes.md §3 and §6 (master dump), ECE 2105 Syllabus Weeks 10 & 12, PYQ bank 2015–2025, Sadiku Ch. 10, L 14.pdf, L 15_updated.pdf