| Date | May 2015 | Marks available | 6 | Reference code | 15M.3.HL.TZ2.21 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Calculate, Determine, and Suggest | Question number | 21 | Adapted from | N/A |
Question
This question is about ultrasound.
Define acoustic impedance of a medium.
The acoustic impedances for various media are shown in the table.
Ultrasound is incident normally on a layer of soft tissue. Gel is placed between the skin and the transducer.
The fraction of the intensity of ultrasound that is reflected (reflection coefficient) at the boundary of two media of impedances Z1 and Z2 is given by the following equation.
\[{\left( {\frac{{{Z_2} - {Z_1}}}{{{Z_2} + {Z_1}}}} \right)^2}\]
(i) Suggest why the gel allows the ultrasound to enter the soft tissue without any reflection.
(ii) Calculate the reflection coefficient at the soft tissue–bone boundary.
(iii) The soft tissue between the skin and the bone absorbs 60% of the intensity of ultrasound travelling through it. The intensity of ultrasound leaving the transducer is I0. Determine, in terms of I0, the intensity of the ultrasound that is reflected back into the transducer from the bone.
Markscheme
the product of the speed of sound and the density of the medium/substance;
Accept as an equation with symbols defined.
(i) The data is on the previous page and most candidates do not realize it is to be used here. So two alternative MS are given which try to be fair to all candidates.
the impedance of gel and soft tissue are the same;
so the equation gives a reflection coefficient (OWTTE) of zero;
or (knowledge based answer)
gel replaces air which would cause unwanted reflection / air has a lower impedance than soft tissue;
the impedance of gel and soft tissue are the same/similar so reflection is reduced;
Do not reward bald “reflection will be less/reduced” for the second marking points as this is given in the question.
Do not accept “density” instead of impedance.
(ii) \(\left( {{{\left[ {\frac{{6.1 \times {{10}^6} - 1.6 \times {{10}^6}}}{{6.1 \times {{10}^6} + 1.6 \times {{10}^6}}}} \right]}^2}} \right) = 0.34\);
(iii) intensity reaching bone is 0.40I0 ;
intensity reflected from bone is 0.34×0.40I0 ; { (allow ECF from first marking point)
intensity reaching transducer is (0.40×0.34×0.40I0 )=0.054I0;
Award [2 max] for an answer of 0.16 I0 or 0.12 I0.
Award [3] for a bald correct answer.
Examiners report
Part (a) was an easy mark, although the speed of light was mentioned too often.
In (b)(i) many candidates ignored the data and answered using existing knowledge. The reflection coefficient was usually correctly calculated in (b)(ii). Part (b)(iii) was poorly answered. Most candidates did not take the time to analyse what was happening. Two attenuations and one reflection, so I = 0.4 × 0.34 × 0.4Io for 3 marks — but very few correct answers were seen.
