The thickness x of the layer of fat on an animal, as illustrated in Fig. 10.2, is to be investigated using ultrasound.

Question 13

(a) State what is meant by the acoustic impedance Z of a medium. [1]

(b) Two media have acoustic impedances Z1 and Z2.

The intensity reflection coefficient α for the boundary between the two media is given

by

α = (Z2 – Z1)² / (Z2 + Z1)²

.

Describe the effect on the transmission of ultrasound through a boundary where there is

a large difference between the acoustic impedances of the two media. [3]

(c) Data for the acoustic impedance Z and the absorption coefficient μ for fat and for muscle are shown in Fig. 10.1.

Z / kg m^-2 s^-1                     μ / m^-1

Fat                   1.3 × 10⁶                     48

Muscle             1.7 × 10⁶                     23

Fig. 10.1

The thickness x of the layer of fat on an animal, as illustrated in Fig. 10.2, is to be

investigated using ultrasound.

Fig. 10.2

The intensity of the parallel ultrasound beam entering the surface S of the layer of fat is I.

The beam is reflected from the boundary between fat and muscle.

The intensity of the reflected ultrasound detected at the surface S of the fat is 0.012 I.

Calculate

(i) the intensity reflection coefficient at the boundary between the fat and the muscle, [2]

(ii) the thickness x of the layer of fat. [3]

Reference: Past Exam Paper – June 2011 Paper 41 Q10

Solution:

(a) Acoustic impedance Z of a medium is the product of the density (of the medium) and the speed of sound (in the medium).

(b)

{This would make the numerator almost equal to the denominator in the formula.}

The intensity reflection coefficient α would be nearly equal to 1

EITHER The reflected intensity would be nearly equal to the incident intensity

OR the coefficient for the transmitted intensity = (1 – α).

{BUT the question asks for the transmission, not the reflection. If most of the ultrasound has been reflected, a very little amount would be transmitted.}

So, the transmitted intensity would be small.

(c)

(i)

{ α = (Z₂ – Z₁)² / (Z₂ + Z₁)² }

α = (1.7 – 1.3)² / (1.7 + 1.3)²  

α = 0.018                                

{ α = I_R / I = 0.018

So, 0.018 of the intensity has been reflected at the interface. Now, this reflected intensity will move through the fat towards surface S.

Considering this intensity from the boundary towards the surface S, we can take this reflected intensity as the new incident intensity. The ultrasound will move from the boundary towards surface S.

After being transmitted in the fat, the intensity at surface S is given to be 0.012 I.

So, the intensity was 0.018 I at the boundary, and as the ultrasound moves through the fat, it got attenuated and the new intensity at the surface S is now 0.012 I.}

(ii)

{The intensity of the ultrasound pulse is affected in 3 steps:

1 From the surface to the (fat-muscle) boundary: the pulse gets attenuated as it travels the thickness x of the fat.

I = I₀ exp (-μx)            where I₀ is the incident intensity and I is the intensity after attenuation

Fraction transmitted: I / I₀ = exp (-μx)

2 Reflection at the boundary: At the boundary, some of the (already attenuated) ultrasound is transmitted and some is reflected. It is this reflected intensity that will move back to the surface again.

Fraction reflected = 0.018

Intensity of reflected pulse = 0.018 I × exp (-μx)

3 Attenuation from the boundary back to the surface: After being reflected, the pulse travels back to the surface while again being attenuated through the thickness x.

In this case (from boundary back to surface):

Incident intensity in fat (I₀) = 0.018 I × exp (-μx)

Transmitted intensity (I) = 0.012 I

Formula: I = I₀ exp (-μx)

0.012 I = [0.018 I × exp (-μx)] × exp(-μx)

This can be simplified as follows:

0.012 = 0.018 exp (-μ×2x)

The term exp (-μ×2x) represents the total attenuation of the pulse in the fat accounting for both ‘outward’ and ‘return’ journeys.}

Attenuation in the fat = exp (-μx) = exp (-48 × 2x)               C1

[I = I_o exp(-μx)]

0.012 = 0.018 exp (-48 × 2x)                                                  C1

{0.012 / 0.018 = exp (-48 × 2x)

ln (0.012/0.018) = -48 × 2x

-0.405 = -48 × 2x

2x = -0.405 / (-48) = 0.008447 m

{Since the unit of μ is m^-1, the value of x obtained is in m.}

x = 0.008447 / 2 = 0.00422 m = 0.42 cm}

x = 0.42 cm                                                                             A1