Question

A plane P-wave is incident at an angle of \(60^\circ\) with respect to the normal to a horizontal reflector. If the incident medium is a homogeneous Poisson solid (\(\nu=0.25\)), the angle of the reflected, mode-converted S-wave with respect to the normal (in degrees) is _______ (rounded off to one decimal place).

Hint:

For mode-converted reflections, keep the horizontal slowness \(p=\sin\theta/V\) constant across wave types. In a Poisson solid with \(\nu=0.25\), \(V_P/V_S=\sqrt{3}\) is a handy number.

Answer 1

Step 1: Use the velocity ratio for a Poisson solid.
For an isotropic elastic solid, \[ \frac{V_P}{V_S}=\sqrt{\frac{2(1-\nu)}{1-2\nu}}. \] With \(\nu=0.25\): \[ \frac{V_P}{V_S}=\sqrt{\frac{2(1-0.25)}{1-2(0.25)}}=\sqrt{\frac{2(0.75)}{0.5}}=\sqrt{3}. \] Step 2: Apply Snell's law (ray parameter conservation).
For reflection with mode conversion at a horizontal interface in the same medium, \[ \frac{\sin\theta_P}{V_P}=\frac{\sin\theta_S}{V_S}. \] Given \(\theta_P=60^\circ\) (from the normal), and \(V_S/V_P=1/\sqrt{3}\), \[ \sin\theta_S=\frac{V_S}{V_P}\sin\theta_P=\frac{1}{\sqrt{3}}\times \sin60^\circ =\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=\frac{1}{2}. \] Step 3: Compute the S-wave angle.
\[ \theta_S=\sin^{-1}\!\left(\frac{1}{2}\right)=30^\circ \;\Rightarrow\; \boxed{30.0^\circ}. \]