Question 32.WE.2: A beam of ultrasound is normally incident on the boundary be...
A beam of ultrasound is normally incident on the boundary between muscle and bone. Use Table 32.3 to determine the fraction of its intensity which is reflected.
| Material | Density / kg m^{−3} | Speed of sound / m s^{−1} | Acoustic impedance / 10^{6} kg m^{-2} s^{-1} |
| air | 1.3 | 330 | 0.0004 |
| water | 1000 | 1500 | 1.50 |
| Biological | |||
| blood | 1060 | 1570 | 1.66 |
| fat | 925 | 1450 | 1.34 |
| soft tissue (average) | 1060 | 1540 | 1.63 |
| muscle | 1075 | 1590 | 1.71 |
| bone (average; adult) | 1600 | 4000 | 6.40 |
| Transducers | |||
| barium titanate | 5600 | 5500 | 30.8 |
| lead zirconate titanate | 7650 | 3790 | 29.0 |
| quartz | 2650 | 5700 | 15.1 |
| polyvinylidene difluoride | 1780 | 2360 | 4.20 |
Table 32.3 The density (ρ), speed of sound in air (c) and acoustic impedance (Z) of some materials important in medical scanning
Step 1 Write down the values of Z_{1} (for muscle) and Z_{2} (for bone).
Z_{1} = 1.71 × 10^{6} kg m^{−2} s^{−1}
Z_{2} = 6.40 × 10^{6} kg m^{−2} s^{−1}
Step 2 Substitute these values in the equation for \frac{I_{r}}{I_{0}} .
\frac{I_{r}}{I_{0}} = \frac{(Z_{2} − Z_{1})^{2}}{(Z_{2} + Z_{1})^{2}}
Hint: We can use this equation because we know that the angle of incidence = 0°.
\frac{I_{r}}{I_{0}} = \frac{(6.40 − 1.71)^{2}}{(6.40 + 1.71)^{2}}
= 0.33
Hint: We can ignore the factor of 10^{6} in the Z values because this is a factor common to all the values, so they cancel out.
So 33% of the intensity of ultrasound will be reflected at the muscle–bone boundary.