Production and use of ultrasound — practice questions

1. Identify the given values: Acoustic impedance of muscle ($Z_1$) = 1.70 × 10$^6$ kg m$^{-2}$ s$^{-1}$ Acoustic impedance of bone ($Z_2$) = 6.00 × 10$^6$ kg m$^{-2}$ s$^{-1}$
2. Recall the intensity reflection coefficient formula: $I_r / I_i = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$
3. Substitute the values into the formula: $I_r / I_i = \left(\frac{6.00 \times 10^6 - 1.70 \times 10^6}{6.00 \times 10^6 + 1.70 \times 10^6}\right)^2$
4. Calculate the numerator and denominator: Numerator: $(6.00 - 1.70) \times 10^6 = 4.30 \times 10^6$ Denominator: $(6.00 + 1.70) \times 10^6 = 7.70 \times 10^6$
5. Perform the division and square the result: $I_r / I_i = \left(\frac{4.30 \times 10^6}{7.70 \times 10^6}\right)^2 = \left(\frac{4.30}{7.70}\right)^2$ $I_r / I_i = (0.55844...)^2 \approx 0.31186$
6. State the final answer: The intensity reflection coefficient is approximately 0.312 (to 3 significant figures).

This problem has two parts: calculating the final intensity using the attenuation formula, and finding the half-value thickness.

Part 1: Calculate the final intensity

1. Identify the given values: Initial intensity ($I_0$) = 5.0 W cm⁻² Linear attenuation coefficient ($\mu$) = 0.23 cm⁻¹ Distance travelled ($x$) = 4.0 cm
2. Recall the attenuation formula: $I = I_0 e^{-\mu x}$
3. Substitute the values into the formula: $I = 5.0 \times e^{-(0.23 \times 4.0)}$
4. Calculate the exponent: Exponent = $-(0.23 \times 4.0) = -0.92$
5. Calculate the final intensity: $I = 5.0 \times e^{-0.92}$ $I = 5.0 \times 0.3985...$ $I \approx 1.9926$ W cm⁻²
6. State the final answer for Part 1: The intensity of the beam after 4.0 cm is approximately 2.0 W cm⁻² (to 2 significant figures).

Part 2: Determine the half-value thickness

1. Recall the formula for half-value thickness: $x_{1/2} = \frac{\ln(2)}{\mu}$
2. Substitute the value of μ: $x_{1/2} = \frac{\ln(2)}{0.23 \text{ cm}^{-1}}$
3. Calculate the result: $x_{1/2} = \frac{0.6931...}{0.23} \approx 3.013$ cm
4. State the final answer for Part 2: The half-value thickness for this tissue is approximately 3.0 cm (to 2 significant figures).