Question 2 of 7: Oblique Incidence at a Lucite–Free-Space Interface

Question 2: Oblique Incidence at a Lucite–Free-Space Interface (20 marks)

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Given. Interface at $z=0$: lucite ($\varepsilon_{r1}=2.8$, lossless, $\mu_{r1}=1$) for $z<0$, free space ($\varepsilon_{r2}=1$) for $z>0$. Plane wave at $f=2$ GHz, $\theta_i=30^\circ$, $\mathbf{E}^i$ polarized along $\hat y$ (perpendicular to the plane of incidence, i.e. TE/perpendicular polarization), $|E^i|=10$ V/m.

Find. $\theta_t$; the time-domain $\mathbf{E}^i(x,z,t)$; the field reflection/transmission coefficients $\Gamma_\perp$, $\tau_\perp$; the critical angle and its significance.

Approach. Snell's law for $\theta_t$; build $\mathbf E^i$ from $k_1=\omega\sqrt{\mu_0\varepsilon_1}$ resolved along $x$ and $z$; use the perpendicular-polarization Fresnel coefficients in terms of intrinsic impedances; find $\theta_c$ from the condition $\theta_t=90^\circ$.

1. Part (a) — Angle of refraction (Snell's law). $n_1=\sqrt{\varepsilon_{r1}}=\sqrt{2.8}=1.673$, $n_2=1$: $$n_1\sin\theta_i = n_2\sin\theta_t \Rightarrow \sin\theta_t = 1.673\sin30^\circ = 0.8367 \Rightarrow \boxed{\theta_t = 56.8^\circ}.$$
2. Part (b) — Incident field, time-domain expression. $\omega=2\pi f = 2\pi(2\times10^9)=1.257\times10^{10}$ rad/s; the wavenumber in lucite is $$k_1=\omega\sqrt{\mu_0\varepsilon_0\varepsilon_{r1}}=\frac{\omega n_1}{c}=\frac{(1.257\times10^{10})(1.673)}{3\times10^8}=70.1\ \text{rad/m}.$$ With propagation direction $\hat{\mathbf k}_i=\sin\theta_i\,\hat x+\cos\theta_i\,\hat z$, so $k_1\sin\theta_i=35.05$ rad/m and $k_1\cos\theta_i=60.71$ rad/m: $$\boxed{\mathbf E^i(x,z,t)=\hat y\,(10)\cos\!\big(1.257\times10^{10}\,t-35.05\,x-60.71\,z\big)\ \text{V/m}}.$$
3. Part (c) — Intrinsic impedances. $$\eta_1=\frac{\eta_0}{\sqrt{\varepsilon_{r1}}}=\frac{376.7}{1.673}=225.1\ \Omega,\qquad \eta_2=\eta_0=376.7\ \Omega.$$
4. Perpendicular (TE) Fresnel coefficients. For perpendicular polarization, wave going from medium 1 into medium 2: $$\Gamma_\perp=\frac{\eta_2\cos\theta_i-\eta_1\cos\theta_t}{\eta_2\cos\theta_i+\eta_1\cos\theta_t}=\frac{376.7\cos30^\circ-225.1\cos56.8^\circ}{376.7\cos30^\circ+225.1\cos56.8^\circ}=\boxed{0.451}$$ $$\tau_\perp=\frac{2\eta_2\cos\theta_i}{\eta_2\cos\theta_i+\eta_1\cos\theta_t}=\boxed{1.451}.$$ Check: $1+\Gamma_\perp=1.451=\tau_\perp$ — the perpendicular-polarization boundary condition ($E^i+E^r=E^t$ at $z=0$) is satisfied exactly.
5. Part (d) — Critical angle. Total internal reflection is possible here because the wave travels from the OPTICALLY DENSER medium (lucite, $n_1=1.673$) toward the less dense one (free space, $n_2=1$). Setting $\theta_t=90^\circ$ in Snell's law: $$\sin\theta_c=\frac{n_2}{n_1}=\frac{1}{1.673}=0.598 \Rightarrow \boxed{\theta_c=36.7^\circ}.$$ Since $\theta_i=30^\circ<\theta_c$, the wave in this problem is NOT totally internally reflected; at $\theta_c$ the refracted ray would graze the interface ($\theta_t=90^\circ$) and for $\theta_i>\theta_c$ all incident power would reflect ($|\Gamma_\perp|=1$) with the transmitted field becoming an evanescent surface wave.