The elapsed time between the transmitted pulse and the detected echo is 26 microseconds. How far is the interface from the transducer?

The elapsed time between the transmitted pulse and the detected echo is 26 microseconds. How far is the interface from the transducer?

• A. 2 cm
• B. 1 cm
• C. 3 cm
• D. 4 cm

Answer: (A)

The time given is the round-trip time for the ultrasound pulse. The pulse travels to the interface and back, so the depth is half the distance the sound covered in that time. In soft tissue, the speed of sound is about 1540 m/s (roughly 1.54 mm/μs). Use d = (c × t) / 2. Convert the time: t = 26 μs = 26 × 10^-6 s. Then c × t = 1540 × 26e-6 ≈ 0.04004 m. Halving gives d ≈ 0.02002 m ≈ 2.0 cm. So the interface is about 2 cm from the transducer.

The time given is the round-trip time for the ultrasound pulse. The pulse travels to the interface and back, so the depth is half the distance the sound covered in that time. In soft tissue, the speed of sound is about 1540 m/s (roughly 1.54 mm/μs). Use d = (c × t) / 2.

Convert the time: t = 26 μs = 26 × 10^-6 s. Then c × t = 1540 × 26e-6 ≈ 0.04004 m. Halving gives d ≈ 0.02002 m ≈ 2.0 cm.

So the interface is about 2 cm from the transducer.