A contact angle beam transducer that produces a 45 degree shear wave in steel would produce what angle in aluminum?

• A. Greater than 45 degrees
• B. Less than 45 degrees
• C. Unknown, more information is required
• D. 45 degrees

Answer: (B)

When a shear wave crosses from one solid to another, its path bends according to Snell's law for shear waves: sin θ_in divided by the velocity in the first material equals sin θ_out divided by the velocity in the second. The transducer produces a 45-degree shear wave in steel, so θ_in is 45° and Vs_steel is the shear wave speed there. Moving into aluminum, the shear wave velocity is slightly lower than in steel, so the ratio Vs_aluminum / Vs_steel is less than one. That makes sin θ_out = (Vs_aluminum / Vs_steel) × sin 45° smaller than sin 45°, which means θ_out is less than 45°. With typical velocities, the angle in aluminum ends up a bit under 45 degrees, roughly in the low 40s. So the angle in aluminum would be less than 45 degrees.