If the time of flight is 52 μs, the reflector depth is ____ cm and the total distance traveled is ____ cm.

If the time of flight is 52 μs, the reflector depth is ____ cm and the total distance traveled is ____ cm.

• A. 2 cm; 4 cm
• B. 4 cm; 8 cm
• C. 3 cm; 6 cm
• D. 5 cm; 10 cm

Answer: (B)

In ultrasound, time of flight is the round-trip time the pulse takes to reach a reflector and return. The total distance the pulse travels during that time is speed × time. The reflector depth is half of that distance because the pulse covers the path to the reflector and back. Using the standard soft-tissue speed of about 1540 m/s and a time of flight of 52 microseconds: - total distance = 1540 m/s × 52 × 10^-6 s ≈ 0.08008 m ≈ 8.0 cm - reflector depth = 0.08008 m / 2 ≈ 0.04004 m ≈ 4.0 cm So the reflector is about 4 cm deep, and the total distance traveled is about 8 cm.

In ultrasound, time of flight is the round-trip time the pulse takes to reach a reflector and return. The total distance the pulse travels during that time is speed × time. The reflector depth is half of that distance because the pulse covers the path to the reflector and back.

Using the standard soft-tissue speed of about 1540 m/s and a time of flight of 52 microseconds:

• total distance = 1540 m/s × 52 × 10^-6 s ≈ 0.08008 m ≈ 8.0 cm

• reflector depth = 0.08008 m / 2 ≈ 0.04004 m ≈ 4.0 cm

So the reflector is about 4 cm deep, and the total distance traveled is about 8 cm.