Question 1 of 10: Oblique incidence at a free-space / silica interface
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A7 Electromagnetics (the archive copy carries no date on its cover; every page header reads “16-ELEC-A7 Electromagnetics / May 2019”). Three hours, closed book, one double-sided 8.5” × 11” aid sheet and an approved Casio or Sharp calculator permitted. Ten questions, each of equal value, of which five constitute a complete exam paper — the first five completed answers are the ones marked. A full marking scheme is printed on page 8. All ten questions are worked below, because the set is a study resource rather than a timed attempt.
Reference texts. The following are the standard works for this subject and are cited by chapter throughout: F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics; M. N. O. Sadiku, Elements of Electromagnetics; W. H. Hayt and J. A. Buck, Engineering Electromagnetics; D. M. Pozar, Microwave Engineering; C. A. Balanis, Antenna Theory: Analysis and Design.
Where the paper itself omits data or contradicts its own marking scheme, the gap is flagged in a check callout rather than filled silently.
All arithmetic below;s own numbers imply: $c = 3\times10^{8}$ m/s (Q3 gives 3 m in 20 ns, exactly $c/2$, and Q2 gives $2\times10^{8}$ m/s), $\eta_0 = 120\pi = 376.99\ Ω$, $\mu_0 = 4\pi\times10^{-7}$ H/m and $\epsilon_0 = 8.854\times10^{-12}$ F/m.
Question 1: Oblique incidence at a free-space / silica interface
Given. A 1 GHz plane wave in free space strikes a lossless silica half-space at 30° from the normal, with its electric field polarised along $y$.
Given data
Quantity
Symbol
Value
Frequency
$f$
1 GHz
Incident field amplitude
$E^i$
2 V/m
Angle of incidence (from the normal)
$\theta_i$
30°
Medium 1 (region $z \lt 0$)
$\epsilon_{r1}$
1 (free space)
Medium 2 (region $z \gt 0$)
$\epsilon_{r2}$
3.8 (silica)
Relative permeability, both media
$\mu_r$
1
Loss
$\sigma$
0 in both media
Find. The magnetic-field strength on each side of the interface, the refraction angle $\theta_t$, and the reflection and transmission coefficients of the electric field.
[Figure not reproduced: Figure 1 (redrawn). Perpendicular (TE) incidence at the free-space / silica boundary: E lies along +y (out of the page) for all three waves, so the plane of incidence is the x–z plane. See the official exam paper.]
Approach. With $\mathbf{E}$ along $y$ and the plane of incidence the $x$–$z$ plane, the wave is perpendicularly (TE) polarised, so the Fresnel coefficients for perpendicular polarisation apply; parts (b) and (c) are settled first because part (a)(ii) needs the transmission coefficient.
Classify the polarisation. The interface normal is $\hat{a}_z$ and the incident ray lies in the $x$–$z$ plane, so that plane is the plane of incidence. The electric field points along $\hat{a}_y$, which is perpendicular to it, so this is perpendicular (TE, or horizontal) polarisation — not the parallel case, and the two have different Fresnel formulas.
Evaluate the intrinsic impedances. For a lossless, non-magnetic medium $\eta = \sqrt{\mu_0/(\epsilon_0\epsilon_r)} = \eta_0/\sqrt{\epsilon_r}$, with $\eta_0 = 120\pi = 376.99\ \Omega$: $$\eta_1 = 376.99\ \Omega, \qquad \eta_2 = \frac{376.99}{\sqrt{3.8}} = \frac{376.99}{1.9494} = 193.39\ \Omega$$ The phase constants follow from $\beta = \omega\sqrt{\mu\epsilon}$: $\beta_1 = 2\pi f/c = 20.944$ rad/m and $\beta_2 = \beta_1\sqrt{3.8} = 40.827$ rad/m.
(b) Apply Snell's law of refraction. Phase matching along the interface requires $\beta_1\sin\theta_i = \beta_2\sin\theta_t$, so with $\mu_r = 1$ in both media $$\sin\theta_t = \sqrt{\frac{\epsilon_{r1}}{\epsilon_{r2}}}\, \sin\theta_i = \frac{\sin 30^\circ}{\sqrt{3.8}} = \frac{0.5}{1.9494} = 0.25649$$ and therefore $$\boxed{\theta_t = \arcsin(0.25649) = 14.86^\circ}$$ The ray bends toward the normal on entering the denser medium, as it must.
(c) Evaluate the perpendicular-polarisation Fresnel coefficients. Matching the tangential $E$ and $H$ at $z = 0$ for TE polarisation gives $$\Gamma_\perp = \frac{\eta_2\cos\theta_i - \eta_1\cos\theta_t}{\eta_2\cos\theta_i + \eta_1\cos\theta_t}, \qquad \tau_\perp = \frac{2\eta_2\cos\theta_i}{\eta_2\cos\theta_i + \eta_1\cos\theta_t}$$ Substituting $\cos 30^\circ = 0.86603$ and $\cos 14.86^\circ = 0.96655$: $\eta_2\cos\theta_i = 167.48\ \Omega$ and $\eta_1\cos\theta_t = 364.38\ \Omega$, so $$\boxed{\Gamma_\perp = \frac{E^r}{E^i} = \frac{167.48 - 364.38}{167.48 + 364.38} = -0.3702} \qquad \boxed{\tau_\perp = \frac{E^t}{E^i} = \frac{2(167.48)}{531.86} = 0.6298}$$ The check $\tau_\perp = 1 + \Gamma_\perp$ holds exactly, which it must because the tangential electric field is continuous. In absolute terms $|E^r| = 0.740$ V/m (reversed in sign, i.e. a $180^\circ$ phase reversal) and $|E^t| = 1.260$ V/m.
(a)(i) Magnetic field in the region $z \lt 0$. A uniform plane wave carries $|H| = |E|/\eta$, so the incident wave has $$\boxed{H^i = \frac{E^i}{\eta_1} = \frac{2}{376.99} = 5.305\ \text{mA/m}}$$ Its direction is $\hat{a}_H = \hat{a}_k \times \hat{a}_E$; with $\hat{a}_k = \sin\theta_i\,\hat{a}_x + \cos\theta_i\,\hat{a}_z$ and $\hat{a}_E = \hat{a}_y$ this gives $\mathbf{H}^i = 5.305(-0.8660\,\hat{a}_x + 0.5000\,\hat{a}_z)$ mA/m, i.e. $(-4.594\,\hat{a}_x + 2.653\,\hat{a}_z)$ mA/m. Region 1 also carries the reflected wave, of strength $H^r = |\Gamma_\perp| E^i/\eta_1 = 1.964$ mA/m, so the two superpose into a standing-wave pattern along the normal whose peak tangential value is $(H^i + H^r)\cos\theta_i = 5.427$ mA/m.
(a)(ii) Magnetic field in the region $z \gt 0$. Only the transmitted wave exists there, and it sees the silica impedance: $$\boxed{H^t = \frac{E^t}{\eta_2} = \frac{\tau_\perp E^i}{\eta_2} = \frac{1.2596}{193.39} = 6.513\ \text{mA/m}}$$ Note that $H^t$ is larger than $H^i$ even though $E^t$ is smaller: the silica impedance is roughly half the free-space value, so the same power flows with a larger magnetic and smaller electric field. In vector form $\mathbf{H}^t = 6.513(-0.9666\,\hat{a}_x + 0.2565\,\hat{a}_z)$ mA/m.
Verify by conserving power across the boundary. The component of the Poynting vector normal to the interface must balance, $S^i\cos\theta_i = S^r\cos\theta_i + S^t\cos\theta_t$, with $S = |E|^2/2\eta$. Numerically $5.305\times10^{-3}$ W/m$^2$ arrives, $7.27\times10^{-4}$ returns and $4.578\times10^{-3}$ continues; the two sides agree to machine precision, confirming both coefficients.
