Reflection, Refraction, Bounded Waves, Ray Theory and Seismic Tomography

This chapter answers the second half of the heading “Seismology and Interior of the Earth” of Part B2 of the Geology and Geophysics (GG) paper: “Reflection and refraction of elastic waves; Inhomogeneous and evanescent waves and bounded waves; Eikonal Equation and Ray theory; Wave propagation in elastic media, Seismic Tomography.” The wave types and the wave equation were set out in Chapter 1. This chapter teaches Snell’s law with the ray parameter and critical angle, amplitudes at an interface with the Zoeppritz equations in outline, travel-time curves, evanescent, inhomogeneous and bounded waves with Rayleigh and Love waves, the eikonal equation and ray theory, and seismic tomography. The numericals are a refraction angle, the energy reflected at normal incidence, layer thickness from a head wave, turning depth in a velocity gradient and a four-cell tomography problem.

1. Snell’s law, the ray parameter and the critical angle

At a plane interface between media of speeds v₁ and v₂ the horizontal slowness of the wave is continuous, which gives Snell’s law sin i₁/v₁ = sin i₂/v₂ = p. The constant p = sin i / v is the ray parameter (s/km), and it equals the reciprocal of the apparent speed along the interface, v_app = v/sin i = 1/p. In a stack of flat layers p is the same in every layer. A P wave meeting a solid–solid boundary also creates converted S waves, and the law covers all of them: sin i/α₁ = sin j/β₁ = sin i′/α₂ = sin j′/β₂ = p. In any one medium the S ray lies closer to the vertical than the P ray, since β < α.

When v₂ > v₁ there is a critical angle θ_c = arcsin(v₁/v₂), at which the refracted ray runs along the interface (p = 1/v₂). Beyond it no homogeneous wave enters the lower medium and the reflection is total. A head wave (refracted wave) travels along the top of the faster layer and radiates back up at θ_c. Worked example. A P wave in a medium of 3.0 km/s meets one of 5.0 km/s at i = 30°: sin j = (5.0/3.0) × 0.5 = 0.833, so j = 56.4°, while θ_c = arcsin(0.6) = 36.9°. An array that sees a P wave with apparent speed 10 km/s, in crust of P speed 6 km/s, has sin i = 6/10, so the ray is 36.9° from the vertical.

2. Amplitudes at an interface: the Zoeppritz equations in outline

Amplitudes follow from boundary conditions. At a welded solid–solid interface the displacement and the traction (σ_zz, σ_zx, σ_zy) are continuous. For an incident P or SV wave, four waves are involved (reflected P and SV, transmitted P and SV), so four unknown amplitudes are fixed by four equations: the Zoeppritz equations. SH motion decouples, with only reflected and transmitted SH and two equations. At a liquid–solid boundary shear traction vanishes and no S wave exists in the liquid, and at a free surface all traction vanishes.

The results depend on the contrast and on the angle of incidence, which is the basis of AVO (amplitude versus offset) analysis. At normal incidence nothing is converted, and with the acoustic impedance Z = ρv the amplitude reflection coefficient is R = (Z₂ − Z₁)/(Z₂ + Z₁) (the exploration convention, positive for an increase of impedance). The reflected energy fraction is R² and the transmitted fraction is 1 − R². A negative R means a polarity reversal, as at the top of a gas sand. Worked example. Z₁ = 2000 × 2000 = 4.0×10⁶ and Z₂ = 2500 × 3000 = 7.5×10⁶ give R = 3.5/11.5 = 0.304, so R² = 0.0926: 9.26% of the energy is reflected.

⚠️ Amplitude and energy are different, and conventions differ
A reflection coefficient depends on whether it is defined for pressure, particle velocity or displacement, and on its sign convention, so the stem must state it. The energy fraction R² does not depend on the convention. Beyond the critical angle |R| = 1 but the reflected wave has a phase shift.

3. Travel-time curves: direct, reflected and head waves, and the Earth’s curve

  • Direct wave: t = x/v₁, a straight line through the origin.
  • Reflected wave from a flat interface at depth h: t = √(x² + 4h²)/v₁, a hyperbola, t² = t₀² + x²/v₁² with t₀ = 2h/v₁.
  • Head wave: t = x/v₂ + t_i, with intercept time t_i = 2h cosθ_c / v₁ = 2h √(v₂² − v₁²)/(v₁ v₂). It first arrives beyond the crossover distance x_c = 2h √((v₂ + v₁)/(v₂ − v₁)). The slope of this branch is 1/v₂ and the intercept then gives h.

Worked example. v₁ = 2000 m/s, v₂ = 4000 m/s, h = 500 m. θ_c = 30°, t_i = 2 × 500 × cos 30° / 2000 = 0.433 s and x_c = 1000 × √(6000/2000) = 1732 m, where the direct and head-wave times are both 0.866 s. A layer with a lower speed than the one above it gives no head wave and stays hidden in refraction work.

For the whole Earth the travel-time curve T(Δ) has slope dT/dΔ = p. In a sphere the ray parameter p = r sin i / v is constant along a ray, and at the deepest point (i = 90°) p = r_t / v(r_t); a larger distance Δ means a smaller p and a deeper turning point. The Herglotz–Wiechert inversion converts T(Δ) into v(r) if speed increases steadily with depth, whereas a low-velocity zone gives no first arrivals and stays hidden. The core’s low velocity produces the P shadow zone from about 103° to 142°.

4. Inhomogeneous, evanescent and bounded waves

A homogeneous plane wave has the same amplitude over each surface of constant phase. An inhomogeneous plane wave has an amplitude that varies across that surface, for example u ∝ exp(−ωηz) exp[iω(px − t)] with η = √(p² − 1/v²). When p > 1/v, η is real and the wave is evanescent: it runs along the interface with apparent speed 1/p < v and decays exponentially with distance from it, over about a wavelength or less except very near the critical angle. Beyond the critical angle the transmitted wave in the faster medium is evanescent. It carries no net energy away from the interface, so the reflection is total.

  • Rayleigh waves exist at the free surface of a half-space as coupled evanescent P and SV waves. They occur at all frequencies, are non-dispersive in a homogeneous half-space, move in a retrograde ellipse at the surface and travel at about 0.92 β (0.919 β for a Poisson solid).
  • Love waves are SH waves trapped in a surface layer of speed β₁ over a faster half-space β₂ by repeated post-critical reflection. They cannot exist in a homogeneous half-space, are dispersive, and have a phase speed c between β₁ and β₂ (near β₁ at short periods, near β₂ at long periods).
  • Stoneley waves are guided along a solid–solid or solid–liquid interface, and channel waves in a low-speed seam. All of these are bounded (guided) waves: inhomogeneous waves whose energy is held near a boundary.

Dispersion. The phase speed is c = ω/k and the group speed, at which energy travels, is U = dω/dk = c + k dc/dk. Longer periods penetrate deeper into faster rock, so for periods of tens of seconds the longer ones arrive first, and measured dispersion curves of Rayleigh and Love waves give the shear-speed structure of crust and upper mantle. Surface waves fall off as 1/√r, more slowly than body waves.

5. The eikonal equation and ray theory

Ray theory is the high-frequency solution of the wave equation, valid when the wavelength is much smaller than the scale over which velocity changes. Try u = A(x) exp[iω(T(x) − t)]. To leading order in ω the wave equation gives the eikonal equation |∇T|² = 1/v² = s², where T(x) is the travel time of the wavefront and s the slowness. The next order gives the transport equation, 2∇A·∇T + A∇²T = 0 for constant density, which keeps the energy flux constant along a ray tube. Rays are the curves perpendicular to the wavefronts T = constant and obey the ray equation d/ds (s dx/ds) = ∇s, which expresses Fermat’s principle: the travel time is stationary along the true path.

Ray theory fails at caustics, in shadow zones, at sharp boundaries and where heterogeneity is as small as the wavelength (diffraction). In a medium with a constant velocity gradient v = v₀ + g z, rays are circular arcs of radius R = 1/(p g) that turn where v = 1/p, at depth z_t = (1/p − v₀)/g. Worked example. v₀ = 2.0 km/s, g = 0.5 s⁻¹ and take-off angle i₀ = 30° from the vertical give p = sin 30°/2.0 = 0.25 s/km, R = 1/(0.25 × 0.5) = 8 km and turning depth z_t = (4.0 − 2.0)/0.5 = 4 km. The ray returns to the surface at range X = 2R cos i₀ = 13.9 km.

6. Seismic tomography

Tomography images the interior from many crossing rays. After linearising about a reference model, the travel-time residual of ray i is δt_i = Σ_j L_ij δs_j, where L_ij is the length of ray i in cell j and δs_j is the slowness perturbation (δs = −δv/v²). In matrix form this is d = G m, solved by least squares, m = (GᵀG)⁻¹Gᵀd, in practice damped, m = (GᵀG + ε²I)⁻¹Gᵀd, or by iterative sparse methods (ART, SIRT, LSQR), with the rays re-traced and the process repeated because paths depend on the model. The data can be body-wave delays, surface-wave dispersion or ambient-noise correlations, and full-waveform inversion uses the whole seismogram.

Resolution depends on ray coverage. Where no rays pass, or all pass in one direction, parts of the model lie in the null space and smear along the rays; a checkerboard test shows what can be recovered. Worked example. Take a 2 × 2 grid of cells 10 km wide, numbered 1 2 over 3 4, with four rays along row 1, row 2, column 1 and column 2. G has rank 3: the pattern (+, −, −, +) changes no ray sum, so these rays cannot recover it. Let the residuals be 0.10, 0.30, 0.20 and 0.20 s, and let cell 1 be known to be unperturbed. Column 1 gives 10 δs₃ = 0.20, so δs₃ = 0.020 s/km, row 1 gives δs₂ = 0.010 s/km, and row 2 gives 10(δs₃ + δs₄) = 0.30, so δs₄ = 0.010 s/km; column 2 then checks. With a reference speed of 5.0 km/s (0.200 s/km) cell 4 has 0.210 s/km, or 4.76 km/s.

⚠️ Slow is not always hot, and fast is not always cold
Speed depends on composition, melt, water and anisotropy as well as on temperature. Fast anomalies often mark cold subducted slabs and cratonic keels, and slow ones hot or partially molten mantle, but the image alone does not prove the cause, and smearing along rays makes anomalies look larger than they are.

Key takeaways

  • Snell’s law: sin i/v = p is constant along a ray and equals 1/(apparent speed). The critical angle is arcsin(v₁/v₂) for v₂ > v₁, beyond which the reflection is total and a head wave runs along the interface.
  • The Zoeppritz equations come from continuity of displacement and traction: four equations for P–SV, two for SH. At normal incidence R = (Z₂ − Z₁)/(Z₂ + Z₁) with Z = ρv, and the reflected energy is R²; state the convention.
  • Travel-time curves: direct t = x/v₁, reflected hyperbola t² = t₀² + x²/v₁², head wave t = x/v₂ + t_i with t_i = 2h cosθ_c/v₁. In the Earth dT/dΔ = p, and a low-velocity zone or the core produces hidden layers and shadow zones.
  • Beyond the critical angle the transmitted wave is evanescent (exponential decay, no net energy flux). Bounded waves are trapped near boundaries: Rayleigh waves at about 0.92β on a free surface, and dispersive Love waves in a slow surface layer with β₁ < c < β₂.
  • Ray theory is the high-frequency limit: the eikonal equation |∇T|² = 1/v² gives wavefronts, the transport equation gives amplitude, and Fermat’s principle gives the paths. In v = v₀ + g z rays are circles that turn at z_t = (1/p − v₀)/g.
  • Tomography solves d = G m for slowness perturbations by damped least squares or iteration. Poor coverage leaves a null space (a row-and-column grid cannot see a checkerboard), and slow or fast anomalies need not mean hot or cold.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A seismic array records a P wave that crosses it with an apparent horizontal speed of 10 km/s. The P speed in the crust beneath the array is 6 km/s. What angle does the ray make with the vertical?

    1. 31°
    2. 37°
    3. 53°
    4. 59°
    Show answer

    Answer: B — 37°

    The apparent speed is v/sin i, so sin i = 6/10 = 0.6 and i = 37° from the vertical. The tempting 53° is arccos 0.6, the angle from the horizontal, and the remaining values do not follow from the ratio 0.6.

  2. A P wave in a slow medium meets a faster medium beyond the critical angle. What is the transmitted wave in the faster medium?

    1. A homogeneous plane wave refracted at 90° that carries away a fixed fraction of the incident energy along the boundary
    2. No wave at all, because the boundary conditions cannot then be satisfied by any wave field
    3. A converted S wave only, which takes up all the energy that is not reflected as P
    4. An evanescent wave that decays exponentially away from the interface and carries no net energy into the medium
    Show answer

    Answer: D — An evanescent wave that decays exponentially away from the interface and carries no net energy into the medium

    The boundary conditions are still satisfied, by an inhomogeneous (evanescent) wave with exp(−ωηz) decay that travels along the interface. Its time-averaged normal energy flux is zero, which is why the reflection is total. A refracted wave at exactly 90° applies only at the critical angle, and a converted S wave cannot account for the missing transmitted P energy.

  3. A P wave is obliquely incident on a welded solid–solid interface. How many wave amplitudes are unknown, and what conditions fix them?

    1. Four: reflected P, reflected SV, transmitted P and transmitted SV, fixed by continuity of displacement and traction
    2. Two: reflected P and transmitted P only, fixed by continuity of pressure and of the normal particle velocity across it
    3. Four: reflected and transmitted P and SH waves, fixed by continuity of displacement and of the pressure across it
    4. Six: three reflected and three transmitted waves, fixed by continuity of the stress tensor across the boundary alone
    Show answer

    Answer: A — Four: reflected P, reflected SV, transmitted P and transmitted SV, fixed by continuity of displacement and traction

    In the P–SV system the incident P wave generates reflected P and SV and transmitted P and SV; continuity of the two displacement components and the two traction components in the plane gives four equations, the Zoeppritz equations. SH is decoupled from P and SV, and the two-wave option with pressure continuity describes a fluid–fluid interface.

  4. A P pulse travels in rock (ρ = 2500 kg/m³, v = 3500 m/s) and meets, at normal incidence, a gas-bearing sand (ρ = 2000 kg/m³, v = 2500 m/s). Take R = (Z₂ − Z₁)/(Z₂ + Z₁) with Z = ρv. Which statement is correct?

    1. R ≈ +0.27: the reflected pulse has the same polarity and carries about 7.4% of the incident energy
    2. R ≈ −0.27: the reflected pulse has reversed polarity and carries about 27% of the incident energy
    3. R ≈ −0.27: the reflected pulse has reversed polarity and carries about 7.4% of the incident energy
    4. R ≈ −0.73: the reflected pulse has reversed polarity and carries about 53% of the incident energy
    Show answer

    Answer: C — R ≈ −0.27: the reflected pulse has reversed polarity and carries about 7.4% of the incident energy

    Z₁ = 2500 × 3500 = 8.75×10⁶ and Z₂ = 2000 × 2500 = 5.0×10⁶, so R = (5.0 − 8.75)/(5.0 + 8.75) = −0.273. The sign is negative because impedance decreases, a polarity reversal. The energy fraction is R² = 0.0744, about 7.4%, not 27% (which confuses amplitude with energy).

  5. Which description of Love and Rayleigh waves is correct?

    1. Love waves are horizontally polarised shear waves that need a slow layer; Rayleigh waves are retrograde ellipses in the vertical plane
    2. Rayleigh waves are horizontally polarised shear waves that need a slow layer; Love waves are retrograde ellipses in the vertical plane
    3. Both are compressional waves, with Love waves travelling at about 0.92 times the P speed and Rayleigh waves faster than S waves
    4. Both exist only in a homogeneous half-space and both are non-dispersive, with speeds always above the S speed of that medium
    Show answer

    Answer: A — Love waves are horizontally polarised shear waves that need a slow layer; Rayleigh waves are retrograde ellipses in the vertical plane

    Love waves are SH-type and exist only when a slow layer overlies a faster medium; Rayleigh waves are coupled P–SV motion at a free surface, travelling at about 0.92 β. The option that makes Rayleigh waves horizontally polarised swaps the two sets of properties; surface waves are slower than S waves, and Love waves need layering and are dispersive.

  6. On a long continental path, surface waves of period 100 s arrive before those of period 20 s. What is the reason?

    1. The 100 s waves are generated earlier at the source, because large earthquakes release long periods first
    2. The 100 s waves are body waves, which always outrun surface waves in layered media
    3. The 100 s waves have the higher frequency, and high-frequency waves always travel faster
    4. The long periods sample deeper, faster layers, so their speed is higher, which is normal dispersion
    Show answer

    Answer: D — The long periods sample deeper, faster layers, so their speed is higher, which is normal dispersion

    Penetration depth grows with wavelength, and speed generally increases with depth, so longer periods have higher speed and arrive first (normal dispersion). Frequency is the inverse of period, so the 100 s waves have the lower frequency, and their arrival order is set by propagation, not by the time of generation.

  7. The eikonal equation |∇T|² = 1/v² follows from the elastic wave equation in the limit in which

    1. the wave frequency tends to zero, so that the wavelength is much larger than any structure in the medium it crosses
    2. the wave frequency is high, so that the wavelength is much smaller than the scale over which the velocity varies
    3. the medium is homogeneous, so that the velocity is the same everywhere and rays are straight lines
    4. the receiver is far from the source, so that the wavefront is effectively planar over the whole array
    Show answer

    Answer: B — the wave frequency is high, so that the wavelength is much smaller than the scale over which the velocity varies

    Ray theory is the high-frequency limit: substituting u = A exp[iω(T − t)] and keeping the leading order in ω gives the eikonal equation, valid when the wavelength is small compared with the scale of velocity variation. The equation holds in heterogeneous media, so homogeneity is not required, and the low-frequency limit is the opposite regime.

  8. A 2 × 2 grid of equal cells is crossed only by rays along its two rows and its two columns. Which perturbation pattern changes none of the four travel times and so cannot be resolved?

    1. The same slowness change in all four cells, since a uniform change cancels along each ray
    2. The checkerboard pattern (+, −, −, +), because every row sum and every column sum is zero
    3. A change in one corner cell only, since that cell is crossed by only one ray
    4. A change in the two cells of one row only, since the row ray averages them
    Show answer

    Answer: B — The checkerboard pattern (+, −, −, +), because every row sum and every column sum is zero

    With rays along rows and columns, G has rank 3, and the vector (+, −, −, +) lies in its null space because each row and column ray gets +1 and −1 in equal measure. A uniform change adds to every travel time, a single cell changes two rays, and a change in one row alters that row’s ray and both column rays.

  9. On a refraction travel-time plot, the branch of the head wave is a straight line that does not pass through the origin. What do its slope and intercept give?

    1. The slope is 1/v₁ of the upper layer, and the intercept is the one-way travel time to the interface
    2. The slope is 1/v₂ of the lower layer, and with v₁ and v₂ the intercept time gives the layer thickness h
    3. The slope is 1/v₂ of the lower layer, and the intercept is the critical distance where the head wave starts
    4. The slope is 2h/v₁, and the intercept is the speed of the lower layer
    Show answer

    Answer: B — The slope is 1/v₂ of the lower layer, and with v₁ and v₂ the intercept time gives the layer thickness h

    The head-wave time is t = x/v₂ + t_i, so the slope is 1/v₂, and t_i = 2h cosθ_c/v₁ with sin θ_c = v₁/v₂ lets h be solved once v₁ (from the direct wave) and v₂ are known. The direct wave, not the head wave, has slope 1/v₁, and the intercept is a time, not a distance or a speed.

  10. Which of the following statements about bounded and evanescent waves are correct?

    1. A Rayleigh wave is non-dispersive at the surface of a homogeneous half-space
    2. Surface waves decay with distance faster than body waves, as 1/r² against 1/r
    3. Love waves cannot exist at the surface of a homogeneous half-space
    4. The amplitude of an evanescent wave decays exponentially with distance from the interface
    Show answer

    Answer: A — A Rayleigh wave is non-dispersive at the surface of a homogeneous half-space; C — Love waves cannot exist at the surface of a homogeneous half-space; D — The amplitude of an evanescent wave decays exponentially with distance from the interface

    The three correct statements are standard. Surface waves spread over a surface and fall off as 1/√r, which is slower than the 1/r of body waves, so the claim of 1/r² is wrong.

  11. Which of the following statements about ray theory are correct?

    1. Ray theory remains accurate at caustics, where neighbouring rays converge to a focus
    2. The eikonal equation gives the travel-time field T(x) and hence the wavefronts
    3. The transport equation governs how the amplitude changes along a ray, keeping the energy flux in a ray tube constant
    4. By Fermat’s principle the travel time along the true ray path is stationary with respect to small changes of path
    Show answer

    Answer: B — The eikonal equation gives the travel-time field T(x) and hence the wavefronts; C — The transport equation governs how the amplitude changes along a ray, keeping the energy flux in a ray tube constant; D — By Fermat’s principle the travel time along the true ray path is stationary with respect to small changes of path

    The eikonal and transport equations and Fermat’s principle are the core of ray theory. At a caustic the ray-theory amplitude becomes infinite, so the method breaks down there along with shadow zones and diffraction.

  12. Which of the following statements about seismic tomography are correct?

    1. After linearisation the travel-time residuals obey d = G m, with G holding the path lengths of the rays in the cells
    2. Poor ray coverage leaves part of the model in the null space, so damping or other regularisation is needed
    3. A positive slowness perturbation corresponds to a lower velocity than the reference model
    4. A fast anomaly always means cold material, because temperature is the only thing that changes speed
    Show answer

    Answer: A — After linearisation the travel-time residuals obey d = G m, with G holding the path lengths of the rays in the cells; B — Poor ray coverage leaves part of the model in the null space, so damping or other regularisation is needed; C — A positive slowness perturbation corresponds to a lower velocity than the reference model

    The linear system, the null space and the sign of slowness are all correct, since s = 1/v and a larger s means a smaller v. Composition, melt, water and anisotropy also alter speeds, so a fast anomaly is not always cold.

  13. A P wave travelling at 3.0 km/s meets a boundary with a medium of 5.0 km/s at an angle of incidence of 30° from the normal. Find the angle of the refracted P ray from the normal, in degrees, to the nearest degree.

    Numerical answer — type the value.

    Show answer

    Answer: 56

    Snell’s law: sin j = (v₂/v₁) sin i = (5.0/3.0) × 0.5 = 0.8333, so j = 56.4°, which rounds to 56°. The critical angle is arcsin(0.6) = 36.9°, and 30° is below it, so a transmitted wave exists.

  14. At normal incidence a P wave goes from a layer of ρ = 2000 kg/m³ and v = 2000 m/s into a layer of ρ = 2500 kg/m³ and v = 3000 m/s. Find the percentage of the incident energy that is reflected, to two decimal places. Keep at least four significant figures in R.

    Numerical answer — type the value.

    Show answer

    Answer: 9.26

    Z₁ = 2000 × 2000 = 4.0×10⁶ and Z₂ = 2500 × 3000 = 7.5×10⁶. R = (7.5 − 4.0)/(7.5 + 4.0) = 0.3043, and the energy fraction is R² = 0.0926, which is 9.26%. Quoting R × 100 = 30.4% would confuse amplitude with energy.

  15. In a refraction survey the upper layer has v₁ = 2000 m/s and the lower layer v₂ = 4000 m/s. The head-wave branch has an intercept time of 0.433 s. Use t_i = 2h cosθ_c / v₁ with sin θ_c = v₁/v₂. Find the thickness h of the upper layer in metres, to the nearest metre.

    Numerical answer — type the value.

    Show answer

    Answer: 500

    sin θ_c = 2000/4000 = 0.5, so θ_c = 30° and cos θ_c = 0.866. Then h = t_i v₁ / (2 cosθ_c) = 0.433 × 2000 / (2 × 0.866) = 500 m. Using sin θ_c in place of cos θ_c would give 866 m.

  16. The velocity increases linearly with depth as v = 2.0 + 0.5 z (v in km/s, z in km). A ray leaves the surface at 30° from the vertical. Find the depth in km at which the ray reaches its deepest point.

    Numerical answer — type the value.

    Show answer

    Answer: 4

    The ray parameter is p = sin 30° / 2.0 = 0.25 s/km. The ray turns where sin i = 1, that is where v = 1/p = 4.0 km/s. Then 2.0 + 0.5 z = 4.0 gives z = 4 km. Using the take-off speed 2.0 km/s at the turning point, instead of 1/p, is the usual slip.

  17. A 2 × 2 grid has cells of side 10 km, numbered 1 2 over 3 4. Four rays run along row 1 (cells 1, 2), row 2 (cells 3, 4), column 1 (cells 1, 3) and column 2 (cells 2, 4). The travel-time residuals are 0.10, 0.30, 0.20 and 0.20 s, respectively. Cell 1 is known to be unperturbed. Find the slowness perturbation of cell 4 in units of 10⁻³ s/km.

    Numerical answer — type the value.

    Show answer

    Answer: 10

    Each ray gives 10 km × (sum of the δs of its cells) = residual. Column 1: 10 (δs₁ + δs₃) = 0.20 with δs₁ = 0 gives δs₃ = 0.020 s/km. Row 2: 10 (δs₃ + δs₄) = 0.30 gives δs₄ = 0.030 − 0.020 = 0.010 s/km = 10 × 10⁻³ s/km. Column 2 checks: 10 (0.010 + 0.010) = 0.20 s.