A parallel beam of ultrasound is incident normally on the surface of a layer of fat of thickness 1.1 cm, as shown in Fig. 6.1.

Question 12

A parallel beam of ultrasound is incident normally on the surface of a layer of fat of thickness 1.1 cm, as shown in Fig. 6.1.

Fig. 6.1

For the ultrasound,

I1 is the intensity just after entering the surface of the fat layer,

I2 is the intensity incident on the fat-muscle boundary,

I3 is the intensity reflected from the fat-muscle boundary,

I4 is the intensity received back at the surface of the fat layer.

Some data for the fat are given in Fig. 6.2.

specific acoustic impedance Z                       1.4 × 10⁶ kg m^-2 s^-1

density ρ                                 940 kg m^-3

absorption (attenuation) coefficient μ             48 m^-1

Fig. 6.2

(a) Calculate the time interval between a short pulse of ultrasound initially entering the layer of fat and then returning back to the surface of the fat layer. [3]

(b) Calculate the ratio I2 / I1. [2]

(c) Intensity I4 is 0.33% of intensity I1.

Determine the ratio I3 / I2. [2]

(d) The specific acoustic impedance of the muscle is greater than that of the fat.

State the effect, if any, on the value of the ratio I3 / I2 of an increase in the difference between the specific acoustic impedance of the muscle and that of the fat. [1]

 [Total: 8]

Reference: Past Exam Paper – March 2016 Paper 42 Q6

Solution:

(a)

{Specific acoustic impedance Z = ρc}

speed = Z / ρ

speed = 1.4 × 10⁶ / 940 (=1490)

{Speed = distance / time

Time = distance / speed

The pulse moves a distance 1.1×10^-2 m in the fat to reach the boundary, and then another 1.1×10^-2 m  from the boundary back to the surface of the fat layer.}

time = (1.1 × 10^–2 × 2) / 1490            

time = 1.5 × 10^–5 s

(b)

{As the pulse moves through the fat layer, some of its intensity is absorbed by the fat while the remaining intensity is transmitted.}

I = I₀ exp (–μx)            or I₂ = I₁ exp(–μx)

{Ratio = I₂ / I₁ = exp(-μx) }

ratio = exp (– 48 × 1.1×10^–2)

ratio = 0.59

(c)

{As stated above, as the pulse moves from the fat surface to the boundary, some of it gets attenuated in the fat while the remaining is transmitted.

Fraction transmitted = I₂ / I₁ = 0.59                (as calculated in part (a))

At the boundary, some of the pulse is transmitted while the remaining (intensity of the pulse arriving at the boundary) is reflected.

Fraction reflected = I₃ / I₂

This is the fraction of (I₂ / I₁) that is reflected (that is, 0.59×(I₂ / I₁)).

This reflected pulse again moves through the fat (where it gets attenuated partly) back to the surface of the fat.

Again, the fraction transmitted = 0.59 (that is, 0.59×0.59×(I₂ / I₁))

The question tells us that this amount is 0.33%.}

0.33 / 100 = 0.59 × (I3 / I2) × 0.59

ratio = 9.5 × 10^–3

(d)

{ I₃ / I₂ = (Z2 − Z1)² / (Z2 + Z1)²

The greater the difference in values of Z, the greater is the ratio.}

ratio I₃ / I₂ increases