A vertical contact separating two resistive domains, \(\rho_1\) and \(\rho_2\) (with \(\rho_2 > \rho_1\)), is shown in the figure below, with the strike direction along X, the profile direction (perpendicular to strike) along Y - where the electric field component \(E_y\) is discontinuous across the contact - and depth along Z. The magnitude of the discontinuity in the apparent resistivity, when the electric field, \(E\), is perpendicular to the strike, is
Show Hint
E perpendicular to strike is the H-polarization (TM) mode - use continuity of tangential H and of normal current density across the contact.
Check the same result with a consistency/limiting-case argument instead of writing out the full boundary-value algebra again.
Step 1: What must be true when there is no contrast. If \(\rho_1=\rho_2\) there is obviously no discontinuity, so the answer must reduce to 1 when \(\rho_1=\rho_2\) - this rules out nothing by itself, but it confirms the answer must have the pure-ratio form \((\rho_1/\rho_2)^n\) or \((\rho_2/\rho_1)^n\), not a mixed expression.
Step 2: Fix the power from the physics. In H-polarization, the current density perpendicular to strike is forced to stay continuous (it cannot pile up at the contact), so the electric field itself scales linearly with the local resistivity for a fixed current density, \(E \propto \rho\). Apparent resistivity, derived from \(E^2\) with the tangential \(H\) unchanged across the boundary, must then scale quadratically with \(\rho\), not as a square root - this rules out the two square-root options.
Step 3: Fix the direction of the ratio. Evaluating apparent resistivity on the \(\rho_1\) side relative to the \(\rho_2\) side (the field on the low-resistivity side is suppressed relative to the high-resistivity side, by continuity of \(J\)) gives \(\rho_{a1}/\rho_{a2}=(\rho_1/\rho_2)^2\), confirming option (A) without repeating the field algebra.
