Question 5 of 8: Detecting a Thin Middle Layer — Three Methods, and the Equivalence Principle

Question 5: Detecting a Thin Middle Layer — Three Methods, and the Equivalence Principle (10 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 3-layer horizontal earth: layer 1 $v_1=800$ m/s, $\rho_1=100\ \Omega\cdot\text{m}$; layer 2 (the "thin" target) $v_2=1500$ m/s, $\rho_2=2\ \Omega\cdot\text{m}$; layer 3 (basement/half-space) $v_3=3000$ m/s, $\rho_3=50\ \Omega\cdot\text{m}$. Note layer 2 is both a low-velocity AND a low-resistivity layer sandwiched between higher-velocity, higher-resistivity layers — a "H-type" resistivity sequence ($\rho_1>\rho_2<\rho_3$) and a low-velocity seismic layer, which is the hardest case for every method below.

(a)(i) Vertical electrical sounding (VES). A conductive middle layer sandwiched between two more resistive layers produces an H-type sounding curve: apparent resistivity $\rho_a$ falls from $\rho_1$ toward $\rho_2$ as AB/2 increases past roughly the depth to layer 2, then rises back toward $\rho_3$ once AB/2 is large enough that current flow is dominated by the basement. As layer 2 becomes very THIN, the current has less and less "room" to be channelled through the low-resistivity layer before AB/2 also senses layer 3, so the dip toward $\rho_2$ becomes shallower and narrower and, for a sufficiently thin layer, may never reach anywhere near the true $\rho_2=2\ \Omega\cdot\text{m}$ at all — the curve can look almost like a simple 2-layer $\rho_1\to\rho_3$ curve, making the presence of layer 2 easy to miss entirely.

(a)(ii) Seismic refraction. Refraction relies on layer 2 being a DISTINCT, DETECTABLE segment on the first-arrival time-distance graph. Layer 2's velocity (1500 m/s) is higher than $v_1$'s, so it CAN refract in principle, but as an intermediate layer its refracted branch only exists over the narrow offset range between its own crossover distance (where it first becomes the fastest path) and the larger offset at which the DEEPER, faster layer-3 refraction overtakes it. As layer 2 thins, that offset window shrinks and eventually the layer-2 branch is never the first arrival at any offset — it is completely "hidden" beneath the direct-to-layer-3 refraction, called a BLIND ZONE (or hidden layer problem), and the time-distance graph simply jumps from the $v_1$ direct-wave segment almost straight to the $v_3$ refracted segment, exactly as if layer 2 did not exist.

(a)(iii) Seismic reflection. A synthetic seismogram at any one trace location shows a reflection wavelet from the top of layer 2 (positive reflection coefficient here, since $Z_2\lt Z_1$: low velocity into the layer) and a second wavelet from its base (comparable-magnitude, opposite-sign reflection coefficient, since $Z_3\gt Z_2$), separated in two-way time by $\Delta t=2h_2/v_2$. As layer 2 thins, $\Delta t$ shrinks; once it becomes smaller than roughly one quarter of the dominant seismic wavelet's period (the classic seismic "tuning thickness"), the two wavelets can no longer be resolved as separate events — they interfere (constructively or destructively depending on the exact phase), and the pair collapses into one composite, distorted wavelet whose amplitude no longer represents either interface's true reflection coefficient. Below tuning thickness, reflection seismic can still detect that SOMETHING changed at that time but can no longer resolve layer 2's actual thickness from the waveform alone.

(b) Why VES resistivity of the middle layer is ambiguous even when it is thick enough to detect — the principle of equivalence. For a layer bounded by higher-resistivity layers above and below (an H-type sequence, exactly the case here), a VES sounding is sensitive mainly to the layer's transverse resistance $T=\rho_2 h_2$ (current flows predominantly ALONG the conductive layer, since it is the path of least resistance), not to $\rho_2$ and $h_2$ independently. Any combination of $\rho_2$ and $h_2$ that keeps the PRODUCT $\rho_2 h_2$ (and, more precisely, the two dimensionless ratios that control the sounding curve's shape) unchanged produces a curve that is indistinguishable, within normal field noise, from the true one — so the inversion is fundamentally non-unique: a thinner, more conductive layer and a thicker, less conductive layer with the same transverse resistance both fit the data equally well. This can only be resolved with independent information (a borehole log, or a seismic/other geophysical constraint on $h_2$), which is precisely the trade-off already illustrated numerically for the gravity cylinder in Question 7(c).