Mini Progress Check – Self, Peer or Teacher Assess
Name:
Score/Grade:
CRO Ultrasound - Review Question
I think/reply.....
A*-E (12 mins)
1 (a) Acoustic impedance Z is the product of the density  of a medium and the speed of
ultrasound v.
The fraction f of ultrasound reflected at a boundary between two media of acoustic
impedances Z1 and Z2 is given by the equation
f=
(Z 2  Z1 ) 2
(Z 2  Z1 ) 2
medium
density / kg m–3
ultrasound velocity v / m
s–1
air
1.299
330
skin
1075
1590
coupling
medium
1090
1540
bone
1750
4080
Fig. 1
(i)
Use the data in Fig. 1 to find the fraction f of ultrasound reflected at an
air-skin boundary.
f = ...........................
[2]
Mr Powell 2013
1
Mini Progress Check – Self, Peer or Teacher Assess
(ii)
Hence explain the need for a coupling medium in ultrasound imaging.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
[2]
(b)
Fig. 2 is a CRO display showing the reflected ultrasound signal from the front edge
F and the rear edge R of a bone. The time-base setting is 1.0 × 10–5 s cm–1.
F
R
1 cm
1 cm
1.0 × 10–5 s
Fig. 2
Using appropriate data from Fig. 1 and Fig. 2, calculate the thickness of the bone.
thickness = ........................................... cm
[4]
[Total 8 marks]
Mr Powell 2013
2
Mini Progress Check – Self, Peer or Teacher Assess
Answers
13. (a) (i) Z for air is 429 (kg m–2 s–1) and
Z for skin is 1.71 × 106 (kg m–2s–1) (1)
Substitution into equation leading to F = 0.999 (1)
2
(ii) with gel, more ultrasound enters body / without gel,
most ultrasound is reflected (1)
most ultrasound is reflected (without gel) when the
difference in Z
is large
or
most ultrasound enters body when the different in Z is
small (1)
2
(b) 1.5 cm × 1 × 10–5 = 1.5 × 10–5 s (1)
s = vt or 4080 × 1.5 × 10–5 (1)
s = 6.12 cm (1) ecf if speed is wrong
/2 = 3.06 cm (1)
4
[8]
Mr Powell 2013
3