Question

An ultrasound wave is incident normal to the interface of muscle and bone. The acoustic impedances of muscle and bone are \(1.7 \times 10^{-4}\) kg m^-2 sec^-1 and \(6.1 \times 10^{-4}\) kg m^-2 sec^-1, respectively. The fraction of the incident energy that is reflected is

• A. 0.98
• B. 0.32
• C. 0.50
• D. 0.65

Hint:

Use R = ((Z2-Z1)/(Z2+Z1))^2 for the fraction of energy reflected at normal incidence.

Answer 1

The Correct Option is B

Step 1: Recall the reflection formula at a normal-incidence interface.
When an ultrasound wave hits a boundary between two tissues straight on (normal incidence), part of the energy reflects back and part transmits through. The fraction of incident intensity that reflects, the intensity reflection coefficient, is
\[ R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 \]
where \(Z_1\) and \(Z_2\) are the acoustic impedances of the two tissues on either side of the boundary.

Step 2: Note down the given values.
Muscle impedance, \(Z_1 = 1.7 \times 10^{-4}\) kg m\(^{-2}\) sec\(^{-1}\). Bone impedance, \(Z_2 = 6.1 \times 10^{-4}\) kg m\(^{-2}\) sec\(^{-1}\). Both share the same power of ten, so it cancels neatly in the ratio.

Step 3: Find the difference and the sum.
\[ Z_2 - Z_1 = 6.1 - 1.7 = 4.4 \]
\[ Z_2 + Z_1 = 6.1 + 1.7 = 7.8 \]
(both in units of \(10^{-4}\), which cancels in the ratio).

Step 4: Compute the ratio and square it.
\[ \frac{Z_2 - Z_1}{Z_2 + Z_1} = \frac{4.4}{7.8} = 0.564 \]
\[ R = (0.564)^2 = 0.318 \approx 0.32 \]

Step 5: Why the other options are wrong.
Option (A) \(0.98\) would need the two tissues to have a very large impedance mismatch (almost total reflection), which is not the case here since the impedances differ by less than a factor of \(4\).
Options (C) \(0.50\) and (D) \(0.65\) do not follow from squaring the actual impedance ratio; they would need a bigger mismatch than given.

Final Answer:
The fraction of incident energy reflected at the muscle-bone interface is \(0.32\).
\[ \boxed{0.32} \]