If the time of flight is 26 μs, the reflector depth is _____ and the total distance traveled is _____.

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Multiple Choice

If the time of flight is 26 μs, the reflector depth is _____ and the total distance traveled is _____.

Explanation:
Time-of-flight in ultrasound is the round-trip distance the pulse travels to the reflector and back. The depth is the one-way distance, and it relates to TOF by d = (c × TOF) / 2, where c is the speed of sound in tissue (about 1540 m/s). Convert TOF to seconds: 26 μs = 26 × 10^-6 s. Multiply by c: 1540 × 26 × 10^-6 = 0.04004 m. Half of that gives the depth: 0.04004 / 2 ≈ 0.02002 m ≈ 2 cm. The total distance traveled (round trip) is twice the depth: ≈ 4 cm.

Time-of-flight in ultrasound is the round-trip distance the pulse travels to the reflector and back. The depth is the one-way distance, and it relates to TOF by d = (c × TOF) / 2, where c is the speed of sound in tissue (about 1540 m/s).

Convert TOF to seconds: 26 μs = 26 × 10^-6 s. Multiply by c: 1540 × 26 × 10^-6 = 0.04004 m. Half of that gives the depth: 0.04004 / 2 ≈ 0.02002 m ≈ 2 cm. The total distance traveled (round trip) is twice the depth: ≈ 4 cm.

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