Date | May 2015 | Marks available | 5 | Reference code | 15M.3.HL.TZ1.19 |
Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
Command term | Deduce | Question number | 19 | Adapted from | N/A |
Question
This question is about ultrasound scanning.
Outline how ultrasound is generated for medical diagnostic purposes.
When ultrasound of intensity I0 travels in a medium of acoustic impedance Z1 and is incident on a medium of acoustic impedance Z2, the intensity IR that is reflected at the interface is given by the following equation.
\[{I_R} = {\left( {\frac{{{Z_1} - {Z_2}}}{{{Z_1} + {Z_2}}}} \right)^2}{I_0}\]
The following data are available.
Use the data to deduce why a layer of gel must be used between a transducer and the patient’s skin in medical ultrasound imaging.
In medical scanning, practitioners have the option of using A-scans or B-scans. Distinguish, with reference to the techniques used to produce the scans, between an A-scan and a B-scan.
Markscheme
crystal changes shape/vibrates when an alternating electric field is applied to it;
crystal dimensions are cut to achieve resonance at required frequency;
(acoustic impedance of air = ) 330 × 1.3 = ( =430 (kgm-2 s-1));
(acoustic impedance of skin =) 1500 × 1000 = ( =1.5×106(kgm-2 s-1));
\(\frac{{{I_R}}}{{{I_0}}} \approx 1/0.999\);
little/no transmission (because the reflected intensity is approximately equal to the incident intensity);
gel (has similar impedance to skin and so) is required as it excludes air from the interface (so transmission occurs);
Accept first two marking points if implied in the third marking point.
A-scan is a “graph” of returned intensity against time but B-scan is a 2-D image of section through patient;
A-scan: ultrasound probe is fixed in position;
B-scan: operator moves ultrasound generator and computer records echo returns;