Electromagnetic Waves | Physics

2.1 Properties of EM Waves

From Maxwell’s equations, the following properties hold for plane EM waves:

Transversality: $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to $\mathbf{k}$ and to each other.
Orthogonality: $\mathbf{E} \perp \mathbf{B}$ And $|\mathbf{E}| = c|\mathbf{B}|$ .
In-phase: $\mathbf{E}$ and $\mathbf{B}$ oscillate in phase.
Dispersion relation: $\omega = ck$ in vacuum.

Proof of transversality. For a plane wave $\mathbf{E} = \mathbf{E}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}$ Gauss’s law gives $i\mathbf{k}\cdot\mathbf{E}_0 = 0$ So $\mathbf{k} \perp \mathbf{E}_0$ . Similarly From $\nabla \cdot \mathbf{B} = 0$ : $\mathbf{k} \perp \mathbf{B}_0$ . $\blacksquare$

Proof of $|\mathbf{E}| = c|\mathbf{B}|$ . From Faraday’s law for a plane wave: $\mathbf{k} \times \mathbf{E}_0 = \omega\mathbf{B}_0$ . Taking magnitudes: $kE_0 = \omega B_0$ So $E_0/B_0 = \omega/k = c$ . $\blacksquare$

Worked Example: Plane wave fields and intensity

Problem. A plane wave in vacuum has $\mathbf{E} = (20\hat{\mathbf{x}} + 30\hat{\mathbf{y}})\cos(kz - \omega t)$ V/m with $\lambda = 500$ nm. Find $\mathbf{B}$ The intensity, and describe the polarisation state.

Solution. $|\mathbf{E}_0| = \sqrt{20^2 + 30^2} = \sqrt{1300} \approx 36.1$ V/m. $B_0 = E_0/c = 36.1/(3 \times 10^8) = 1.20 \times 10^{-7}$ T.

Since $\mathbf{k} = k\hat{\mathbf{z}}$ and $\mathbf{B}_0 = \hat{\mathbf{k}} \times \mathbf{E}_0/c$ : $\mathbf{B} = (-20\hat{\mathbf{y}} + 30\hat{\mathbf{x}})B_0/E_0 \cdot \cos(kz - \omega t)/c$ $= (30\hat{\mathbf{x}} - 20\hat{\mathbf{y}})(1/c)\cos(kz - \omega t)$ T.

Intensity: $I = \frac{1}{2}c\varepsilon_0 E_0^2 = \frac{1}{2}(3 \times 10^8)(8.854 \times 10^{-12})(1300) = 1.73$ W/m $^2$ .

Polarisation: $\mathbf{E}_0$ has components along $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ with a Constant phase relationship ( $\delta = 0$ ), so the wave is linearly polarised at angle $\theta = \arctan(30/20) = 56.3°$ from the $x$ -axis.

2.2 Energy and Intensity

The Poynting vector:

$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$

Represents the energy flux (W/m $^2$ ). The time-averaged intensity for a plane wave:

$I = \langle S \rangle = \frac{1}{2}c\varepsilon_0 E_0^2$

The energy density of an EM field is:

$u = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2 = \varepsilon_0 E^2$

(the electric and magnetic contributions are equal for a plane wave). The intensity is related to The energy density by $I = uc$ .

Radiation pressure. For a perfectly absorbing surface: $P_{\mathrm{rad} = I/c}$ . For a perfectly Reflecting surface: $P_{\mathrm{rad} = 2I/c}$ .

Worked Example: Radiation pressure from a laser

Problem. A 5 mW laser beam ( $\lambda = 632.8$ nm) is normally incident on a perfectly reflecting Mirror. The beam has a diameter of 1 mm. Find the radiation pressure and the force on the mirror.

Solution. Beam area: $A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}$ m $^2$ . Intensity: $I = P/A = 5 \times 10^{-3}/(7.85 \times 10^{-7}) = 6.37 \times 10^3$ W/m $^2$ .

Radiation pressure (reflecting): $P_{\mathrm{rad} = 2I/c = 2(6.37 \times 10^3)/(3 \times 10^8) = 4.25 \times 10^{-5}}$ Pa.

Force: $F = P_{\mathrm{rad} \cdot A = (4.25 \times 10^{-5})(7.85 \times 10^{-7}) = 3.34 \times 10^{-11}}$ N.

2.3 EM Waves in Matter

In a linear, isotropic, non-magnetic medium with refractive index $n$ :

$v = \frac{c}{n}, \quad \mathbf{k} = n\frac{\omega}{c}\hat{\mathbf{k}}$

The index of refraction is related to the relative permittivity and permeability:

$n = \sqrt{\varepsilon_r \mu_r}$

For non-magnetic materials ( $\mu_r \approx 1$ ): $n \approx \sqrt{\varepsilon_r}$ .

The wavelength inside a medium of refractive index $n$ is $\lambda_n = \lambda_0/n$ Where $\lambda_0$ Is the vacuum wavelength. The frequency remains unchanged across the boundary.

Worked Example: EM wave propagation in glass

Problem. A plane wave of wavelength $\lambda_0 = 600$ nm in vacuum enters a glass slab ( $n = 1.50$ ) at normal incidence. Find (a) the wavelength and wave speed inside the glass, (b) the Frequency, and (c) the ratio of intensities inside and outside the glass, accounting for reflection At the front surface.

Solution.

(a) $\lambda_n = \lambda_0/n = 600/1.50 = 400$ nm. $v = c/n = 2.0 \times 10^8$ m/s.

(b) $f = c/\lambda_0 = (3 \times 10^8)/(600 \times 10^{-9}) = 5.0 \times 10^{14}$ Hz (unchanged).

(c) At normal incidence: $R = [(n_1 - n_2)/(n_1 + n_2)]^2 = [(1 - 1.5)/(1 + 1.5)]^2 = (0.5/2.5)^2 = 0.04$ . Transmittance: $T = 1 - R = 0.96$ . The intensity inside the glass is $I_{\mathrm{inside} = 0.96\,I_0}$ But the power per unit area Referenced to the vacuum intensity is $I_{\mathrm{inside} = (n_2/n_1)\,T\,I_0 = 1.5 \times 0.96 \times I_0 = 1.44\,I_0}$ If we compare the electric field amplitudes squared times the respective impedances.

2.4 Boundary Conditions and Snell’s Law

At a planar interface between two linear, isotropic media, the tangential components of $\mathbf{E}$ and $\mathbf{H}$ and the normal components of $\mathbf{D}$ and $\mathbf{B}$ are Continuous across the boundary.

Consider a plane wave incident from medium 1 ( $n_1$ ) onto medium 2 ( $n_2$ ), with the interface at $z = 0$ and the plane of incidence the $xz$ -plane.

The phase matching condition requires the phases of all three waves (incident, reflected, Transmitted) to match at $z = 0$ for all $x$ and $t$ . This gives:

$k_1\sin\theta_i = k_1\sin\theta_r = k_2\sin\theta_t$

From the first equality: $\theta_i = \theta_r$ (law of reflection). From the second equality: $n_1\sin\theta_i = n_2\sin\theta_t$ (Snell’s law).

Proof. The incident, reflected, and transmitted fields are:

$E_i \propto e^{i(k_{1x}x + k_{1z}z - \omega t)}, \quad E_r \propto e^{i(k_{1x}'x + k_{1z}'z - \omega t)}, \quad E_t \propto e^{i(k_{2x}x + k_{2z}z - \omega t)}$

At $z = 0$ The tangential field must be continuous for all $x$ and $t$ : $k_{1x} = k_{1x}' = k_{2x}$ I.e., $k_1\sin\theta_i = k_1\sin\theta_r = k_2\sin\theta_t$ . Since $k = n\omega/c$ This yields Snell’s law. $\blacksquare$

2.5 Fresnel Equations

Applying the boundary conditions for the tangential fields yields the Fresnel equations for the Amplitude reflection and transmission coefficients.

s-polarisation ( $\mathbf{E}$ perpendicular to the plane of incidence, along $\hat{\mathbf{y}}$ ): The tangential components of $\mathbf{E}$ and $\mathbf{H}$ give:

$r_s = \frac{E_{0r}}{E_{0i}} = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}$

$t_s = \frac{E_{0t}}{E_{0i}} = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t}$

p-polarisation ( $\mathbf{E}$ parallel to the plane of incidence): The tangential components of $\mathbf{E}$ and $\mathbf{H}$ give:

$r_p = \frac{E_{0r}}{E_{0i}} = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}$

$t_p = \frac{E_{0t}}{E_{0i}} = \frac{2n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t}$

Reflectance and transmittance (energy fractions):

$R_s = |r_s|^2, \quad T_s = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t_s|^2, \quad R_s + T_s = 1$

$R_p = |r_p|^2, \quad T_p = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t_p|^2, \quad R_p + T_p = 1$

At normal incidence ( $\theta_i = 0$ ): $r_s = r_p = (n_1 - n_2)/(n_1 + n_2)$ and $R = [(n_1 - n_2)/(n_1 + n_2)]^2$ .

Worked Example: Fresnel coefficients at a glass-air interface

Problem. Light is incident from air ( $n_1 = 1.00$ ) onto glass ( $n_2 = 1.50$ ) at $\theta_i = 30°$ . Calculate $r_s$ , $r_p$ , $R_s$ And $R_p$ .

Solution. From Snell’s law: $\sin\theta_t = \sin 30°/1.50 = 0.333$ So $\theta_t = 19.47°$ . $\cos\theta_i = \cos 30° = 0.866$ , $\cos\theta_t = \cos 19.47° = 0.943$ .

$r_s = \frac{1.00 \times 0.866 - 1.50 \times 0.943}{1.00 \times 0.866 + 1.50 \times 0.943} = \frac{0.866 - 1.414}{0.866 + 1.414} = \frac{-0.549}{2.280} = -0.241$

$R_s = r_s^2 = 0.0580$

$r_p = \frac{1.50 \times 0.866 - 1.00 \times 0.943}{1.50 \times 0.866 + 1.00 \times 0.943} = \frac{1.299 - 0.943}{1.299 + 0.943} = \frac{0.356}{2.242} = 0.159$

$R_p = r_p^2 = 0.0252$

At this angle, p-polarised light is reflected less efficiently than s-polarised light. The Negative sign of $r_s$ indicates a phase shift of $\pi$ upon reflection.

error is to swap the $n_1\cos\theta_i$ and $n_2\cos\theta_t$ terms. Remember: for $r_s$ The numerator starts with $n_1\cos\theta_i$ ; for $r_p$ The numerator starts with $n_2\cos\theta_i$ . Also, $r$ and $t$ are Amplitude coefficients, while $R$ and $T$ are energy coefficients — they are related but not Interchangeable.

2.6 Total Internal Reflection and Evanescent Waves

When light travels from a denser to a rarer medium ( $n_1 \gt n_2$ ) and the angle of incidence Exceeds the critical angle:

$\theta_c = \arcsin\!\left(\frac{n_2}{n_1}\right)$

Snell’s law gives $\sin\theta_t = (n_1/n_2)\sin\theta_i \gt 1$ So $\theta_t$ becomes complex. Writing $\cos\theta_t = i\sqrt{\sin^2\theta_t - 1}$ The Fresnel coefficients become complex with $|r_s|^2 = |r_p|^2 = 1$ : all energy is reflected.

The transmitted field becomes an evanescent wave:

$E_t \propto e^{-\kappa z}\, e^{i(k_x x - \omega t)}$

Where:

$\kappa = k_0\sqrt{n_1^2\sin^2\theta_i - n_2^2}, \quad k_x = k_0 n_1\sin\theta_i$

The field decays exponentially with penetration depth $\delta = 1/\kappa$ into the second medium, But propagates without loss along the interface. No net energy is transported across the boundary ( $T = 0$ ).

Frustrated total internal reflection (FTIR). If a third medium (with $n_3 \geq n_2$ ) is brought Within a distance comparable to $\delta$ of the interface, the evanescent wave can couple into it, Allowing energy transmission across the gap. This is the optical analogue of quantum mechanical Tunnelling.

Worked Example: Critical angle and evanescent wave penetration

Problem. Light travels from glass ( $n_1 = 1.50$ ) to air ( $n_2 = 1.00$ ) at $\theta_i = 50°$ . Find (a) the critical angle, (b) the penetration depth for $\lambda = 500$ nm, and (c) the Propagation constant along the interface.

Solution.

(a) $\theta_c = \arcsin(n_2/n_1) = \arcsin(1/1.50) = 41.8°$ . Since $50° \gt 41.8°$ TIR occurs.

(b) $\kappa = k_0\sqrt{n_1^2\sin^2\theta_i - n_2^2}$ $= \frac{2\pi}{\lambda}\sqrt{(1.50)^2\sin^2 50° - 1.00^2}$ $= \frac{2\pi}{500 \times 10^{-9}}\sqrt{2.25 \times 0.587 - 1.00}$ $= (1.257 \times 10^7)\sqrt{0.320}$ $= (1.257 \times 10^7)(0.566) = 7.11 \times 10^6$ m $^{-1}$

Penetration depth: $\delta = 1/\kappa = 1.41 \times 10^{-7}$ m $= 141$ nm.

Cross-References

Diffraction: Applies the wave nature of electromagnetic radiation to explain bending through apertures, building on the plane wave solutions derived here.
Polarization: Explores the vector nature of the electric field, including linear, circular, and elliptical polarisation states.
Geometric Optics: Takes the high-frequency limit of electromagnetic wave propagation, reducing wave equations to ray tracing.
Calculus
Linear Algebra
Vector Calculus
Quantum Computing

flowchart TD
    A[2_Electromagnetic Waves] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Total internal reflection occurs when light tries to pass from a denser medium to a less dense medium at too steep an angle. Instead of refracting, the light bounces back completely, with an evanescent wave that decays exponentially into the less dense medium. This evanescent field is real but carries no net energy away from the interface. The critical angle depends only on the ratio of refractive indices. This principle underpins fiber optic communication, where light is trapped inside a glass core by repeated total internal reflections, traveling long distances with minimal loss. The penetration depth of the evanescent wave can be tuned by changing the angle of incidence.

Common Mistakes

Mistake 1: Forgetting that frequency is conserved when light crosses a boundary When light passes from one medium to another, the frequency remains constant while the wavelength and speed change. Students often mistakenly assume the wavelength is conserved, leading to incorrect calculations of refraction. The correct approach is $v = c/n$ , $\lambda_n = \lambda_0/n$ , with $f = c/\lambda_0$ unchanged across the boundary.

Mistake 2: Confusing amplitude coefficients with energy coefficients in Fresnel equations The Fresnel coefficients $r$ and $t$ are amplitude ratios, while $R = |r|^2$ and $T = (n_2\cos\theta_t/n_1\cos\theta_i)|t|^2$ are energy fractions. Students frequently use $R = r^2$ instead of $R = |r|^2$ (which matters when $r$ is negative, indicating a phase shift) and forget the cos factor in the transmittance formula.

Mistake 3: Assuming $|\mathbf{E}| = c|\mathbf{B}|$ implies $\mathbf{E}$ and $\mathbf{B}$ have the same units The relation $E = cB$ is in SI units, where $E$ is in V/m and $B$ is in tesla. The numerical values differ by a factor of $c \approx 3 \times 10^8$ . In Gaussian units, $E = B$ numerically. Students sometimes forget this distinction when comparing results from different unit systems.

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