Option: Medical physics – A-level Physics exam questions with mark scheme

1Standard6 marks

A student can see objects clearly only when they are between 0.12 m and 2.0 m from her eye.

(a) Name the defect of vision the student has and state the type of lens needed to correct it. [1]

(b) Calculate the power of the lens that allows her to see very distant objects clearly. Assume the lens is very close to the eye. [2]

(d) The student's optician also finds that she has astigmatism in one eye. State the shape of lens used to correct astigmatism. [1]

(a) myopia / short sight and diverging / concave lens (both needed)(1)
(b) uses $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ with $u = \infty$ and $v = -2.0$ m (allow $f = -2.0$ m / far point = focal length of diverging lens)(1)
power $= -0.50$ D (allow −0.5 D; minus sign required for this mark; allow m⁻¹ for D)(1)
(c) $\frac{1}{u} = -0.50 - \frac{1}{-0.12}$ / $\frac{1}{u} = 7.8(3)$ (ecf power from (b))(1)
$u = 0.13$ m (allow 0.128 m; allow 12.8–13 cm)(1)
(d) cylindrical lens (allow toric; do not allow 'converging' or 'diverging' alone)(1)
Answer: (a) Myopia; diverging lens. (b) −0.50 D. (c) 0.13 m. (d) Cylindrical lens.
Examiner tip: In (b) the image of a distant object must be virtual and at the far point, so $v$ is negative: dropping the minus sign loses the mark.

2Hard9 marks

A sound-level meter placed 1.5 m from a workshop machine gives an intensity level of 92 dB. Treat the machine as a point source that radiates sound equally in all directions, and ignore absorption and reflections.

Threshold of hearing intensity $I_0 = 1.0 \times 10^{-12}$ W m$^{-2}$

(a) Calculate the sound intensity at the meter. [2]

(b) Calculate the minimum distance from the machine at which the intensity level is 80 dB or less. [3]

(c) A second, identical machine is switched on at the same distance from the meter. Calculate the new intensity level. [2]

(d) A worker who has spent many years near such machines without ear protection has an equal-loudness curve measured. Describe how the threshold-of-hearing curve for this worker would differ from that of a person with normal hearing. [2]

(a) $92 = 10\log\left(\frac{I}{1.0\times10^{-12}}\right)$ / $I = 10^{9.2} \times 10^{-12}$(1)
$I = 1.6 \times 10^{-3}$ W m$^{-2}$ (allow 1.58 × 10⁻³)(1)
(b) intensity at 80 dB $= 1.0 \times 10^{-4}$ W m$^{-2}$ / intensity must fall by factor $10^{1.2}$ (= 15.8)(1)
uses $I \propto \frac{1}{r^2}$: $r = 1.5\sqrt{15.8}$ (ecf from (a))(1)
$r = 6.0$ m (allow 5.9–6.0 m)(1)
(c) total intensity doubled $= 3.2 \times 10^{-3}$ W m$^{-2}$ (ecf)(1)
intensity level $= 95$ dB (do not allow 184 dB)(1)
(d) threshold of hearing raised / higher intensity level needed to hear (at some frequencies)(1)
greatest increase (a peak / notch) at around 4 kHz (allow 3–6 kHz; ignore 'at high frequencies' alone for this mark)(1)
Answer: (a) $1.6 \times 10^{-3}$ W m$^{-2}$. (b) 6.0 m. (c) 95 dB. (d) Threshold raised, with a peak in hearing loss around 4 kHz.
Examiner tip: Two equal sources double the intensity, not the decibel value: add 3 dB, never 92 + 92.

3Standard6 marks

A flexible endoscope contains two bundles of optical fibres. Each fibre has a glass core of refractive index 1.52 surrounded by cladding of refractive index 1.48.

(a) Calculate the critical angle at the core–cladding boundary. [2]

(b) Explain why each fibre needs cladding when the fibres are packed together in a bundle. [2]

(c) One bundle carries light into the body and the other carries the image out. Explain why the fibres in the image bundle must be arranged coherently but those in the light-carrying bundle need not be. [2]

(a) $\sin\theta_c = \frac{1.48}{1.52}$(1)
$\theta_c = 77^\circ$ (allow 76.8°)(1)
(b) without cladding, fibres touching each other would allow light to pass (cross over) from one fibre to the next(1)
so image information is lost / image is blurred / degraded (allow cladding protects core surface from scratches that would let light escape)(1)
(c) coherent: the relative positions of the fibres are the same at both ends (allow fibres in the same order at each end)(1)
so each fibre carries one part (pixel) of the image to the correct position; the light (non-coherent) bundle only has to deliver light, so arrangement does not matter(1)
Answer: (a) 77°. (b) Stops light crossing between touching fibres, which would destroy the image. (c) Image needs fibre positions matched at both ends; illumination does not.
Examiner tip: 'Coherent' here means the fibre positions match at both ends, not that the light is in phase: using the wave meaning scores nothing.

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4Core5 marks

The diagram shows part of a normal ECG trace recorded from a patient at rest.

(a) Identify the feature, A, B or C, produced by:

(i) the depolarisation of the atria [1]

(ii) the repolarisation of the ventricles. [1]

(b) Determine the heart rate of the patient in beats per minute. [2]

(a)(i) A(1)
(a)(ii) C(1)
(b) time for one beat (between consecutive R peaks) $= 0.80$ s (allow 2 beats in 1.60 s)(1)
heart rate $= \frac{60}{0.80} = 75$ beats per minute (ecf for their period)(1)
(c) any one: conducting gel / paste used; skin cleaned / hair removed / dead skin rubbed off (ignore 'good contact' alone)(1)
Answer: (a)(i) A (P wave). (a)(ii) C (T wave). (b) 75 beats per minute. (c) Use conducting gel / remove hair and dead skin.
Examiner tip: Measure the period between matching points on consecutive beats (R to R) across several beats, not the width of one feature.

5Standard6 marks

Ultrasound is used to image a region where muscle lies on top of bone.

density of muscle = 1060 kg m$^{-3}$ speed of ultrasound in muscle = 1580 m s$^{-1}$
acoustic impedance of bone = $7.8 \times 10^{6}$ kg m$^{-2}$ s$^{-1}$
acoustic impedance of air = 430 kg m$^{-2}$ s$^{-1}$ acoustic impedance of skin = $1.6 \times 10^{6}$ kg m$^{-2}$ s$^{-1}$

(a) Calculate the acoustic impedance of muscle. Give an appropriate unit. [2]

(b) Calculate the fraction of the incident intensity that is reflected at the muscle–bone boundary. [2]

(a) $Z = \rho c = 1060 \times 1580 = 1.7 \times 10^{6}$ (allow 1.67 × 10⁶)(1)
unit kg m$^{-2}$ s$^{-1}$ (allow rayl)(1)
(b) $\frac{I_r}{I_i} = \left(\frac{7.8 - 1.67}{7.8 + 1.67}\right)^2$ (ecf Z from (a))(1)
$= 0.42$ (allow 0.41–0.42; allow 42%)(1)
(c) without gel there is air between transducer and skin: fraction reflected at air–skin $\approx 0.999$ / almost all ultrasound reflected (as $Z$ values very different)(1)
gel has impedance similar to skin so most ultrasound is transmitted into the body (allow 'acts as impedance matching / coupling medium')(1)
Answer: (a) $1.7 \times 10^{6}$ kg m$^{-2}$ s$^{-1}$. (b) 0.42. (c) Air–skin boundary reflects ~99.9%; gel matches impedance so ultrasound enters the body.
Examiner tip: In (c) 'removes air' is not enough: link it to the large impedance mismatch and almost total reflection.

6Hard8 marks

An ultrasound transducer is placed on a patient's abdomen and an A-scan is obtained. The diagram shows the echo pulses received. Pulse P is from the skin surface, pulse Q is from the boundary between a layer of fat and an organ, and pulse R is from the far side of the organ.

speed of ultrasound in fat = 1450 m s$^{-1}$ speed of ultrasound in the organ = 1580 m s$^{-1}$

(a) Determine the thickness of the fat layer. [2]

(b) Determine the thickness of the organ. [2]

(d) Describe how the piezoelectric transducer produces a pulse of ultrasound. [2]

(a) time in fat (one way) $= 10$ μs / uses total time 20 μs with $d = \frac{vt}{2}$(1)
thickness $= 1450 \times 10 \times 10^{-6} = 0.015$ m (allow 0.0145 m; allow 1.4–1.5 cm)(1)
(b) time between Q and R $= 80$ μs, so $d = \frac{1580 \times 80 \times 10^{-6}}{2}$ (ecf time read from graph)(1)
thickness $= 0.063$ m (allow 6.3 cm; allow 0.0632 m)(1)
(c) any two: ultrasound is attenuated / absorbed (and scattered) by the tissue as it travels further; some of the ultrasound is reflected at the fat–organ boundary (so less reaches the far boundary); the fraction reflected at the far boundary may be smaller (smaller impedance difference)(1)
second correct reason from the list above (list principle applies)(1)
(d) an alternating p.d. / voltage pulse is applied across the (piezoelectric) crystal(1)
making it change shape / vibrate (at its resonant frequency), producing ultrasound; crystal is damped so vibration stops quickly giving a short pulse (either)(1)
Answer: (a) 0.015 m. (b) 0.063 m. (c) Attenuation over the longer path; partial reflection at the earlier boundary. (d) Alternating p.d. makes crystal vibrate (damped to give short pulse).
Examiner tip: The time on an A-scan is for the pulse to go there and back: forgetting to halve it doubles every depth.

7Standard8 marks

A diagnostic X-ray tube operates with a tube voltage of 80 kV and a tube current of 4.0 mA.

(a) Calculate the maximum energy, in J, of an X-ray photon produced by the tube. [1]

(b) Calculate the minimum wavelength of the X-rays produced. [2]

(c) About 1% of the electrical power supplied to the tube is converted to X-rays. Calculate the rate at which energy is transferred to the anode as heat. [2]

(d) State why a rotating anode is used in this tube. [1]

(e) The tube voltage is increased to 100 kV with the tube current kept constant. Describe how the X-ray spectrum changes. [2]

(a) $E = eV = 1.6 \times 10^{-19} \times 80 \times 10^{3} = 1.3 \times 10^{-14}$ J (allow 1.28 × 10⁻¹⁴)(1)
(b) $\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{1.28 \times 10^{-14}}$ (ecf from (a))(1)
$\lambda = 1.6 \times 10^{-11}$ m (allow 1.55 × 10⁻¹¹)(1)
(c) power supplied $= IV = 4.0 \times 10^{-3} \times 80 \times 10^{3} = 320$ W(1)
heat transfer rate $= 0.99 \times 320 = 320$ W (allow 317 W; allow 316–320 W)(1)
(d) spreads the heating over a larger area (of the target) so the anode does not overheat / melt (allow allows higher tube current / shorter exposure)(1)
(e) maximum photon energy increases (to 100 keV) / minimum wavelength decreases; intensity increases at all energies (area under spectrum larger)(1)
characteristic lines remain at the same energies / wavelengths (allow line peaks become more intense)(1)
Answer: (a) $1.3 \times 10^{-14}$ J. (b) $1.6 \times 10^{-11}$ m. (c) 320 W (317 W). (d) Spreads heat over a larger area. (e) Spectrum extends to higher energy and is more intense; characteristic lines unchanged in position.
Examiner tip: Characteristic lines depend on the anode material, not the tube voltage, so their positions never move.

8Stretch8 marks

An X-ray beam of a single photon energy is used to image a patient's arm.

linear attenuation coefficient of muscle = 0.22 cm$^{-1}$
mass attenuation coefficient of bone = 0.42 cm$^{2}$ g$^{-1}$ density of bone = 1.9 g cm$^{-3}$

(a) Calculate the half-value thickness of muscle for this beam. [2]

(b) Show that the linear attenuation coefficient of bone for this beam is about 0.8 cm$^{-1}$. [1]

(c) Path 1 through the arm passes through 8.0 cm of muscle only. Path 2 passes through 5.5 cm of muscle and 2.5 cm of bone. Calculate the ratio

$$\frac{\text{intensity emerging from path 1}}{\text{intensity emerging from path 2}}$$ [3]

(d) A barium meal is used when the stomach is imaged with X-rays. Explain why. [2]

(a) $x_{1/2} = \frac{\ln 2}{\mu}$(1)
$= \frac{0.693}{0.22} = 3.2$ cm (allow 3.15 cm)(1)
(b) $\mu = \mu_m \rho = 0.42 \times 1.9 = 0.80$ cm$^{-1}$ (must see 0.798 or 0.80)(1)
(c) $I_1 = I_0 e^{-0.22 \times 8.0}$ / $I_1 = 0.17 I_0$(1)
$I_2 = I_0 e^{-(0.22 \times 5.5 + 0.80 \times 2.5)}$ / $I_2 = 0.041 I_0$ (allow ratio method $e^{(0.80-0.22)\times 2.5}$ for both of the first two marks)(1)
ratio $= 4.2$ or $4.3$ (allow 4.2–4.3)(1)
(d) stomach (soft tissue) has an attenuation coefficient similar to the surrounding tissue so gives little contrast(1)
barium has a high (proton number and so) attenuation coefficient so the stomach lining / outline absorbs strongly and shows up clearly / increases contrast(1)
Answer: (a) 3.2 cm. (b) 0.42 × 1.9 = 0.80 cm$^{-1}$. (c) 4.2–4.3. (d) Barium strongly absorbs X-rays, giving contrast where soft tissues are similar.
Examiner tip: In (c) the muscle along both paths matters: only the 2.5 cm that differs changes the ratio, so $e^{(\mu_b - \mu_m) \times 2.5}$ is the quick route.

9Hard7 marks

Technetium-99m is injected into a patient for a gamma camera scan. Its physical half-life is 6.0 hours and its biological half-life in the patient is 24 hours. The activity injected is 400 MBq.

(a) Calculate the effective half-life of technetium-99m in the patient. [2]

(b) Calculate the time after injection when the activity in the patient has fallen to 5.0% of its initial value. [2]

(c) Technetium-99m is obtained at the hospital from a generator containing molybdenum-99, which has a half-life of 66 hours. Explain why a generator is used. [2]

(d) State the purpose of the collimator in a gamma camera. [1]

(a) $\frac{1}{T_E} = \frac{1}{6.0} + \frac{1}{24}$(1)
$T_E = 4.8$ hours(1)
(b) $0.050 = \left(\frac{1}{2}\right)^{t/4.8}$ / $t = \frac{4.8 \ln 20}{\ln 2}$ (ecf $T_E$; allow use of $\lambda = \ln 2 / T_E$ and $A = A_0 e^{-\lambda t}$)(1)
$t = 21$ hours (allow 20.7 h)(1)
(c) Tc-99m has a short half-life so would decay significantly before it could be delivered / cannot be stored(1)
Mo-99 has a longer half-life so the generator can be transported and supplies Tc-99m (by decay) for several days (allow Tc-99m can be separated / eluted when needed)(1)
(d) only allows gamma rays travelling along the axis of its holes (perpendicular to the detector) to reach the crystal, so the position of the source can be located / image is sharp (ignore 'focuses the gamma rays')(1)
Answer: (a) 4.8 h. (b) 21 h. (c) Tc-99m half-life too short to transport; long-lived Mo-99 parent gives a supply on site. (d) Only lets through gamma rays travelling parallel to its holes, so the image shows where they came from.
Examiner tip: The effective half-life is always shorter than both the physical and biological half-lives; an answer between them signals the reciprocals were not used.

10Hard6 marks

A patient with a suspected soft-tissue injury in the knee may be imaged using either an MR scanner or a CT scanner.

Describe how an MR scanner produces an image and discuss the advantages and disadvantages of MR scanning compared with CT scanning for this patient. [6]

patient lies in a very strong, uniform magnetic field from a superconducting magnet(1)
hydrogen nuclei (protons) have spin and align with / against the field and precess about the field direction (at the Larmor frequency, which depends on field strength)(1)
gradient coils vary the field across the body so the precession frequency depends on position, allowing a slice / position to be selected(1)
RF pulse at the precession frequency excites the protons (flips them to the higher-energy state)(1)
when the pulse stops the protons de-excite / relax, emitting RF signals that are detected by coils(1)
relaxation times differ for different tissues; computer builds image from signal strength and timing(1)
MR advantage: no ionising radiation, so safer / can be repeated(1)
MR advantage: excellent soft-tissue contrast (good for ligament / cartilage injury); images in any plane(1)
MR disadvantages: more expensive; longer scan time; noisy / claustrophobic; patient must keep still for longer; unsuitable with some metal implants / pacemakers(1)
CT: uses X-rays (ionising, higher dose); quicker and cheaper; better for bone detail; resolution of both is similar (about 1 mm)(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered description of MR imaging that includes alignment and precession in a strong field, the role of gradient coils in locating position, RF excitation and detection of the emitted signal; plus a balanced comparison with CT covering at least one advantage and one disadvantage of MR, linked to this patient.
Level 2	3–4	A partial description of MR imaging with some correct steps (for example field, RF pulse and signal) but gaps in the sequence or no mention of how position is found; plus a comparison with CT that may be one-sided.
Level 1	1–2	Isolated relevant statements about MR or CT (for example 'uses a strong magnet' or 'CT uses X-rays') with little linking or reasoning.
Level 0	0	No relevant content.

Answer: MR: strong field aligns precessing protons; gradient coils locate position; RF pulse excites; relaxing protons emit RF signal detected and processed by computer. MR has no ionising radiation and better soft-tissue contrast but is costlier, slower and noisy, and is unsafe with some metal implants; CT is faster and cheaper but gives an ionising dose.
Examiner tip: Many answers describe the hardware but never explain how a position in the body is found: the gradient coils are the link that makes an image possible.

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