Evanescent Field above a Totally Reflecting Water Surface: 22-Elec-A7 Electromagnetics, December 2013, Q8

Question 8: Evanescent Field above a Totally Reflecting Water Surface (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 wave inside lossless freshwater strikes the water–air boundary from below, at an angle well past the critical angle, so no power escapes into the air and only an evanescent field exists above the surface.

Given data
Quantity	Symbol	Value
Frequency	$f$	20 kHz
Relative permittivity of water	$\epsilon_{r1}$	81
Conductivity of water	$\sigma$	0 (lossless)
Refractive index of water	$n_1$	9
Angle of incidence (from the normal)	$\theta_i$	$30^{\circ}$
Medium above	$n_2$	1 (air)
Required amplitude	—	10 % of the incident value

Find. The height in air at which the field amplitude has fallen to one tenth of the incident amplitude.

Figure 8.1 — Total internal reflection at the water–air surface. Above the surface the field does not propagate but decays exponentially with height.

Approach. Confirm that the incidence exceeds the critical angle, then obtain the evanescent decay constant from the phase-matching condition and solve the exponential decay for the required height.

Critical angle. With $n_1 = \sqrt{81} = 9$ and $n_2 = 1$,$$\theta_c = \arcsin\!\left(\frac{n_2}{n_1}\right) = \arcsin\!\left(\tfrac{1}{9}\right) = 6.38^{\circ}.$$Since $\theta_i = 30^{\circ} \gt \theta_c$, the wave is totally internally reflected: no power crosses into the air, and the field above the surface is evanescent.
Phase matching along the surface. The tangential wavenumber must be continuous across the boundary, so in air$$k_x = n_1 k_0 \sin\theta_i = 9(0.5)k_0 = 4.5\,k_0 \gt k_0 .$$The air-side wave must therefore satisfy$k_z^{2} = k_0^{2} - k_x^{2} \lt 0$, i.e. $k_z$ is purely imaginary and the field varies as $e^{-\alpha z}$ rather than as a travelling wave.
Evanescent decay constant. Taking the magnitude,$$\alpha = k_0\sqrt{\epsilon_{r1}\sin^{2}\theta_i - 1},\qquad k_0 = \frac{2\pi f}{c} = \frac{2\pi(2\times10^{4})}{3\times10^{8}} = 4.189\times10^{-4}\ \text{rad/m}.$$Substituting,$$\sqrt{81(0.5)^{2}-1} = \sqrt{19.25} = 4.387,$$$$\alpha = (4.189\times10^{-4})(4.387) = 1.838\times10^{-3}\ \text{Np/m}.$$The decay is extremely gentle because the free-space wavelength at 20 kHz is 15 km.
Height for a tenth of the amplitude. Setting $e^{-\alpha z} = 0.10$,$$z = \frac{\ln 10}{\alpha} = \frac{2.3026}{1.838\times10^{-3}},$$$$\boxed{z = 1.25\times10^{3}\ \text{m} \approx 1.25\ \text{km}}$$
Sense check. The evanescent field is not a radiating wave — it carries no time-average power away from the surface, and it stretches over kilometres here only because the wavelength itself is 15 km. At optical frequencies the same geometry would confine the field to well under a micrometre. A receiver at 1.25 km altitude would detect a field, but it is a stored reactive field bound to the interface, not a signal propagating upward.

Check: the boxed height measures the decay from the field value at the surface, taking that value as the incident amplitude, which is the standard reading of the question. Including the interface transmission coefficient shifts the answer by a few hundred metres and depends on the polarization, which the question does not state: for perpendicular (TE) polarization $|t| = 1.74$ gives $z = 1.56$ km, while for parallel (TM) polarization $|t| = 0.395$ gives $z = 0.75$ km. All three follow from the same decay constant $\alpha = 1.838\times10^{-3}$ Np/m.

Final Results
Quantity	Symbol	Value
Refractive index of freshwater	$n_1$	9.00
Critical angle	$\theta_c$	$6.38^{\circ}$
Free-space wavenumber	$k_0$	$4.189\times10^{-4}$ rad/m
Evanescent decay constant	$\alpha$	$1.838\times10^{-3}$ Np/m
Height for 10 % amplitude	$z$	1.25 km
Same height, TE polarization (refinement)	$z_{\perp}$	1.56 km
Same height, TM polarization (refinement)	$z_{\parallel}$	0.75 km

Topic. Total internal reflection and the evanescent field.

Key relations. $\theta_c = \arcsin(n_2/n_1)$; phase matching $k_x = n_1k_0\sin\theta_i$; $\alpha = k_0\sqrt{n_1^{2}\sin^{2}\theta_i - n_2^{2}}$; amplitude $\propto e^{-\alpha z}$; penetration depth $1/\alpha$.

Why this works. Beyond the critical angle the boundary cannot supply a transmitted wave whose tangential wavenumber matches the incident one, because that would require the normal wavenumber to be imaginary. Maxwell's equations resolve this with a field that clings to the surface, oscillating along it at the incident trace velocity while decaying exponentially away from it. No net power flows across the boundary, yet the field is genuinely present and can be tapped by a nearby object — the principle behind frustrated total reflection and behind every optical waveguide.

Common pitfall. The decay constant uses $n_1^{2}\sin^{2}\theta_i$, not $\sin^{2}\theta_i$ alone; dropping the refractive index makes the argument of the square root negative and the problem appear unsolvable. It is also wrong to look for a refraction angle from Snell's law here, since $\sin\theta_t = 9\sin 30^{\circ} = 4.5$ has no real solution.

Source. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed., §8.2–8.3 (oblique incidence and total internal reflection); Hayt & Buck, 9th ed., §12.5.

Question 8 of 8: Evanescent Field above a Totally Reflecting Water Surface

Question 8: Evanescent Field above a Totally Reflecting Water Surface (20 marks)