Under what condition can total internal reflection occur in ultrasound, and what would be observed at the interface?

• A. When moving from a slower to a faster medium and the incidence angle exceeds the critical angle
• B. When moving from a slower to a faster medium and incidence angle is small
• C. When speeds are equal in both media
• D. When moving from faster to slower medium and the incidence angle exceeds the critical angle; most energy reflects back into the first medium

Answer: (D)

Total internal reflection happens when a wave travels from a faster medium into a slower medium and hits the boundary at an angle above a certain threshold, called the critical angle. At and beyond that angle, the transmitted wave cannot propagate into the second medium; instead, all or almost all of the energy stays in the first medium and is reflected back. In ultrasound terms, if the first medium has a higher speed of sound than the second and the incidence angle exceeds the critical angle, the refracted ray would have to travel along or outside the boundary, which isn’t possible for a traveling wave in the second medium. So the interface sends back a strong echo with little to no transmission into the second medium.

The critical angle is determined by the speeds: sin(theta_c) equals the speed in the second medium divided by the speed in the first medium. Practically, you’d observe a sharp reflection at the interface when the angle is large enough, with very little energy transmitted beyond it. Moving from slower to faster, or having equal speeds, or choosing a small incidence angle would not produce total internal reflection.