Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 40C / b / i

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- Solution i

Solution

The reflected P-wave is propagating only while its normal wavenumber is real, equivalently while

∣ sin ϕ ∣ \leq 1

. Since

sin ϕ = \frac{c _{P}}{c _{S}} sin θ,

it is evanescent when

sin θ > \frac{c _{S}}{c _{P}} .

Put

s = \frac{c _{P}}{c _{S}} sin θ > 1, q = s^{2} - 1, cos ϕ = i q .

Then its spatial factor is

e^{i k_{P} s x + k_{P} q z},

which decays as

z \to - \infty

. The reflected SV amplitude becomes

R = - \frac{i q c o s θ - s s i n θ}{i q c o s θ + s s i n θ} .

The numerator and denominator have equal complex modulus, so

∣ R ∣ = 1 .

Thus evanescent mode conversion changes the reflected SV phase but not its amplitude, as described by Unit-modulus SV reflection with evanescent P conversion.

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