Oblique incidence at a lucite–air interface: 22-Elec-A7 Electromagnetics, December 2019, Q2

Question 2: Oblique incidence at a lucite–air interface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A plane wave travels through lucite and strikes the plane boundary with free space at $30^{\circ}$ from the normal, with its electric field along $\hat{\mathbf{a}}_y$ — out of the plane of incidence, so this is perpendicular (TE) polarisation.

Given data
Quantity	Symbol	Value
Medium 1 (region $z \lt 0$)	$\varepsilon_{r1}$	$2.8$ (lucite, lossless, $\mu_r = 1$)
Medium 2 (region $z \gt 0$)	$\varepsilon_{r2}$	$1.0$ (free space)
Frequency	$f$	$2\ \text{GHz}$
Angle of incidence	$\theta_i$	$30^{\circ}$
Incident field strength	$\|E^{i}\|$	$10\ \text{V/m}$
Polarisation	—	$\hat{\mathbf{a}}_y$ (perpendicular / TE)

Find. The refraction angle, a fully numerical time-domain expression for the incident field, the perpendicular-polarisation Fresnel coefficients, and the critical angle with its physical meaning.

[Figure not reproduced: Figure 2 — the interface geometry redrawn from the printed figure. E is out of the page for all three waves (perpendicular / TE polarisation); the plane of incidence is the x–z plane. See the official exam paper.]

Approach. Snell's law fixes the geometry, the wavenumber in lucite fixes the phase term, the perpendicular-polarisation Fresnel formulas give the coefficients, and the dense-to-rare ordering of the media makes a critical angle exist.

Set up the refractive indices. With $\mu_r = 1$ in both media the index is just the square root of the relative permittivity: $$n_1 = \sqrt{\varepsilon_{r1}} = \sqrt{2.8} = 1.6733, \qquad n_2 = 1.$$ Because $n_1 \gt n_2$ the wave is going from the optically denser medium into the rarer one, which is exactly the situation in which a critical angle exists (part d).
Apply Snell's law of refraction. Phase matching along the boundary requires $n_1\sin\theta_i = n_2\sin\theta_t$, so $$\sin\theta_t = \frac{n_1}{n_2}\sin\theta_i = 1.6733\,(0.5000) = 0.8367 \;\Longrightarrow\; \theta_t = \boxed{56.79^{\circ}}.$$ The refracted ray bends away from the normal, as it must when entering a rarer medium.
Compute the wavenumber in lucite. The phase velocity in medium 1 is $v_{p1} = c/n_1 = 1.7928\times 10^{8}\ \text{m/s}$, and with $\omega = 2\pi f = 1.2566\times 10^{10}\ \text{rad/s}$, $$k_1 = \frac{\omega}{v_{p1}} = 70.10\ \text{rad/m} \qquad (\lambda_1 = 2\pi/k_1 = 8.96\ \text{cm}).$$
Resolve the propagation vector into components. With $\theta_i$ measured from the normal ($z$ axis), the incident unit propagation vector is $\hat{\mathbf{k}}^{i} = \sin\theta_i\,\hat{\mathbf{a}}_x + \cos\theta_i\,\hat{\mathbf{a}}_z$, so $$k_x = k_1\sin 30^{\circ} = 35.05\ \text{rad/m}, \qquad k_z = k_1\cos 30^{\circ} = 60.70\ \text{rad/m}.$$
Write the time-domain incident field. Combining the amplitude, the polarisation direction and the phase term gives $$\boxed{\mathbf{E}^{i}(x,z,t) = \hat{\mathbf{a}}_y\,10 \cos\!\left(1.2566\times 10^{10}\,t - 35.05\,x - 60.70\,z\right)\ \text{V/m}}$$ with $x$ and $z$ in metres and $t$ in seconds. The accompanying magnetic field is $\mathbf{H}^{i} = (\hat{\mathbf{k}}^{i}\times \mathbf{E}^{i})/\eta_1$, which lies in the $x$–$z$ plane and points into the second quadrant, matching the arrows on the exam figure.
Evaluate the intrinsic impedances. For a non-magnetic dielectric $\eta = \eta_0/\sqrt{\varepsilon_r}$, so $$\eta_1 = \frac{376.7}{1.6733} = 225.1\ \Omega, \qquad \eta_2 = 376.7\ \Omega.$$
Apply the perpendicular-polarisation Fresnel formulas. For E perpendicular to the plane of incidence, $$\Gamma_{\perp} = \frac{\eta_2\cos\theta_i - \eta_1\cos\theta_t}{\eta_2\cos\theta_i + \eta_1\cos\theta_t} = \frac{326.2 - 123.3}{326.2 + 123.3} = \boxed{0.4514}$$ and, since the tangential E must be continuous, $$\tau_{\perp} = 1 + \Gamma_{\perp} = \boxed{1.4514}.$$
Check the power balance. A transmission coefficient greater than unity always deserves a check, and it is not a violation: the transmitted power fraction carries the impedance and obliquity factors. $$R = |\Gamma_{\perp}|^{2} = 0.2038, \qquad T = |\tau_{\perp}|^{2}\,\frac{\eta_1\cos\theta_t}{\eta_2\cos\theta_i} = 0.7962,$$ and $R + T = 1.0000$ exactly. About 20 % of the incident power is reflected back into the lucite.
Find the critical angle and say what happens there. A critical angle exists because $n_1 \gt n_2$; it is the incidence angle at which the refracted ray grazes the boundary ($\theta_t = 90^{\circ}$): $$\theta_c = \arcsin\!\left(\frac{n_2}{n_1}\right) = \arcsin\!\left(\frac{1}{1.6733}\right) = \boxed{36.70^{\circ}}.$$ For $\theta_i \ge \theta_c$ Snell's law would demand $\sin\theta_t \gt 1$: no propagating transmitted wave exists, $|\Gamma_{\perp}| = 1$ (total internal reflection, with a phase shift rather than a loss), and the field in free space becomes evanescent — it decays exponentially away from the boundary while still travelling along it, carrying no net power into medium 2. Note that the $30^{\circ}$ used in parts (a)–(c) is comfortably sub-critical, which is why those answers came out real.

Question 2 — final results
Part	Quantity	Result
(a)	Angle of refraction	$\theta_t = 56.79^{\circ}$
(b)	Angular frequency	$\omega = 1.2566\times 10^{10}\ \text{rad/s}$
(b)	Wavenumber in lucite	$k_1 = 70.10\ \text{rad/m}$ ($k_x = 35.05$, $k_z = 60.70\ \text{rad/m}$)
(b)	Incident field	$\mathbf{E}^{i} = \hat{\mathbf{a}}_y 10\cos(1.2566\times 10^{10}t - 35.05x - 60.70z)\ \text{V/m}$
(c)	Reflection coefficient	$\Gamma_{\perp} = 0.4514$
(c)	Transmission coefficient	$\tau_{\perp} = 1.4514$
(c)	Power split	$R = 0.204$, $T = 0.796$ ($R+T = 1$)
(d)	Critical angle	$\theta_c = 36.70^{\circ}$ (total internal reflection beyond)

Topic. Oblique incidence of a perpendicular-polarised plane wave on a lossless dielectric boundary.

Key relations. Snell: $n_1\sin\theta_i = n_2\sin\theta_t$ with $n = \sqrt{\varepsilon_r\mu_r}$; $\Gamma_{\perp} = (\eta_2\cos\theta_i - \eta_1\cos\theta_t)/(\eta_2\cos\theta_i + \eta_1\cos\theta_t)$; $\tau_{\perp} = 1 + \Gamma_{\perp}$; $\theta_c = \arcsin(n_2/n_1)$ (exists only for $n_1 \gt n_2$); $R + T = 1$ with $T = |\tau|^2\eta_1\cos\theta_t/(\eta_2\cos\theta_i)$.

Why this works. The boundary conditions must hold at every point of the interface and at every instant, which forces the tangential components of all three wave vectors to be equal — that single statement gives both the law of reflection and Snell's law. Matching the tangential E and H amplitudes then gives the Fresnel ratios; for perpendicular polarisation E is entirely tangential, so continuity of E is simply $1 + \Gamma = \tau$. When the phase-matching condition cannot be satisfied by a real angle, the transverse wavenumber in medium 2 turns imaginary and the wave becomes evanescent.

Common pitfall. Quoting $\tau \gt 1$ as an error, or “fixing” it by renormalising. The field ratio genuinely exceeds one when the wave enters a higher-impedance medium; it is the Poynting flux, not the field, that is conserved. A second trap is using the parallel (TM) Fresnel formula because the figure “looks” in-plane — check where E points, not where k points.

Source. Ulaby & Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed., §8-2 and §8-3; Hayt & Buck, Engineering Electromagnetics, 9th ed., §12.5–12.6; Sadiku, Elements of Electromagnetics, 7th ed., §10.8.

Question 2 of 7: Oblique incidence at a lucite–air interface

Question 2: Oblique incidence at a lucite–air interface (20 marks)