A pulse travels normally through uniform tissue at 1540 m/s and returns from a boundary after 64 μs. The acoustic impedances on the two sides are Z₁ = 1.50 × 10⁶ and Z₂ = 2600000 kg m⁻² s⁻¹. Ignore attenuation and multiple reflections. Use intensity reflection coefficient R = [(Z₂−Z₁)/(Z₂+Z₁)]². (a) Find boundary depth and reflected intensity fraction. (b) Explain why coupling gel is used and why a shorter pulse improves depth resolution.

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Depth 0.04928 m; reflected intensity fraction 0.07198 (7.198%); gel reduces impedance mismatch and short pulses separate close echoes. The echo delay covers travel to the boundary and back. Convert 64 μs to 0.000064 s, then depth = ct/2 = 1540 × 0.000064/2 = 0.04928 m. The impedance ratio is dimensionless. Substituting the values gives R = [(2600000 − 1.50 × 10⁶)/(2600000 + 1.50 × 10⁶)]² = 0.07198, corresponding to 7.198% of incident intensity. An air gap at the skin has a large acoustic-impedance mismatch and would reflect much of the incident ultrasound. Coupling gel replaces that gap with a medium more closely matched to tissue, improving transmission into the body. Two boundaries at nearby depths produce echoes separated by a short time. A shorter pulse reduces overlap between the returned signals, so the boundaries are easier to resolve separately. This timing resolution is distinct from the echo strength governed by impedance contrast and attenuation.

An ultrasound echo from a boundary 7

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