3.10 Medical physics (A-level only) — coverage pack

Answer

• Rod cells; the image has no colour or has low spatial resolution.

Method: Low light levels stimulate the more sensitive rod cells. Rods do not provide colour vision, and the convergence of several rods onto one nerve pathway gives low spatial resolution; either image property earns the second marking point.

Answer

• Several rod cells can share one nerve pathway, so signals from the two illuminated rods may be combined and the brain cannot distinguish their positions; sensitivity is increased but spatial resolution is reduced.

Method: Identify convergence: several rods connect to a common sensory pathway. Their signals are combined, which helps a small light stimulus reach the detection threshold. Because the same pathway does not identify which individual rod responded, the two source positions cannot be resolved.

Answer

• In bright light cones respond, so the grid is seen in colour and with high spatial resolution because cone pathways show little convergence. In dim light rods dominate, so colour is lost and fine lines are less easily resolved because many rods share a pathway, although this convergence makes the eye more sensitive to weak light.

Method: For bright light, award the linked chain cones, colour vision, and high resolution from separate or weakly convergent pathways. For dim light, award rods, no colour, and reduced resolution from convergence. Complete the comparison by linking that convergence to greater sensitivity because signals from several rods can combine.

Answer

• A diverging (concave) lens with negative power.

Method: A myopic eye is too powerful, so parallel rays would focus before the retina. A diverging lens reduces the convergence before the rays enter the eye; its focal length and power are negative.

Answer

• $v=+0.30\,\text{m}$ and $|m|=0.50$

Method: The focal length is $f=1/P=1/5.0=0.20\,\text{m}$. Hence $1/v=1/f-1/u=5.0-1/0.60=3.33\,\text{m}^{-1}$, so $v=+0.30\,\text{m}$. The magnification magnitude is $|m|=|v/u|=0.30/0.60=0.50$.

Answer

• $+2.8\,\text{D}$

Method: The object is real, so $u=+0.25\,\text{m}$. The required image is virtual, so $v=-0.80\,\text{m}$. Then $1/f=1/u+1/v=1/0.25+1/(-0.80)=4.00-1.25=2.75\,\text{m}^{-1}$. Therefore $P=1/f=+2.75\,\text{D}$, which is $+2.8\,\text{D}$ to $2$ significant figures. The positive sign confirms that a converging lens is required.

Answer

• The tympanic membrane (eardrum) and the ossicles.

Method: Sound pressure variations act on the tympanic membrane. Its vibration is transmitted through the malleus, incus and stapes, collectively called the ossicles.

Answer

• The ossicles provide a lever force gain, and this force acts on the oval window, whose area is much smaller than the tympanic membrane; since pressure is force divided by area, the pressure increases.

Method: First identify the lever action of the ossicles and link it to an increased force. Then compare areas: the oval window is smaller than the tympanic membrane. Using $p=F/A$, a larger force acting over a smaller area gives a greater pressure in the inner-ear fluid.

Answer

• $0.29\,\text{Pa}$

Method: The force on the tympanic membrane is $F_1=p_1A_1=0.012(5.5\times10^{-5})=6.6\times10^{-7}\,\text{N}$. The ossicle output force is $F_2=1.4F_1=9.24\times10^{-7}\,\text{N}$. Therefore $p_2=F_2/A_2=(9.24\times10^{-7})/(3.2\times10^{-6})=0.28875\,\text{Pa}=0.29\,\text{Pa}$ to $2$ significant figures.

Answer

• $40\,\text{dB}$

Method: Use $L=10\log_{10}(I/I_0)$. Thus $L=10\log_{10}((1.0\times10^{-8})/(1.0\times10^{-12}))=10\log_{10}(10^4)=40\,\text{dB}$.

Answer

• $63$

Method: The level difference is $\Delta L=76-58=18\,\text{dB}$. From $\Delta L=10\log_{10}(I_2/I_1)$, the ratio is $I_2/I_1=10^{18/10}=10^{1.8}=63.1$, so the larger intensity is about $63$ times the smaller.

Answer

• $14\,\text{m}$; normal hearing is less sensitive at $12\,\text{kHz}$ than near $3\,\text{kHz}$, so the higher-frequency tone has a greater equal-loudness threshold.

Method: First find the intensity: $I=I_0 10^{L/10}=(1.0\times10^{-12})10^{8.5}=3.16\times10^{-4}\,\text{W m}^{-2}$. For an isotropic source, $I=P/(4\pi r^2)$, so $r=\sqrt{P/(4\pi I)}=\sqrt{0.80/(4\pi(3.16\times10^{-4}))}=14.2\,\text{m}$, giving $14\,\text{m}$ to $2$ significant figures. Equal-loudness curves have their minimum near a few kilohertz; at $12\,\text{kHz}$ a larger level is required for the same perceived loudness.

Answer

• At high frequencies.

Method: Age-related hearing loss does not raise the threshold equally at all frequencies. The characteristic change is a larger loss towards the high-frequency end of an equal-loudness or threshold graph.

Answer

• $30\,\text{dB}$ and a factor of $1.0\times10^3$

Method: The hearing loss is the vertical threshold difference: $38-8=30\,\text{dB}$. Using $\Delta L=10\log_{10}(I_2/I_1)$ gives $I_2/I_1=10^{30/10}=10^3$. The impaired threshold intensity is therefore $1000$ times the normal value.

Answer

• A is consistent with age-related deterioration because the loss rises strongly at high frequency. B is consistent with excessive-noise exposure because its largest loss is near $4\,\text{kHz}$. Each equal-loudness curve is shifted upward most in its affected range, so greater intensity is needed there and sounds at those frequencies are perceived as quieter.

Method: Compare the frequency dependence rather than the total loss. A's largest value is at $10\,\text{kHz}$, matching age-related high-frequency loss. B has a pronounced maximum at $4\,\text{kHz}$, matching excessive-noise exposure. A positive hearing loss raises the required level on the equal-loudness curve; the larger the loss, the larger the upward displacement and the quieter an unchanged sound is perceived.

Answer

• The QRS complex is ventricular depolarisation and the T wave is ventricular repolarisation.

Method: Match each named feature to the electrical activity of the ventricles: rapid ventricular depolarisation produces QRS, and the later recovery or repolarisation produces T.

Answer

• $77\,\text{beats min}^{-1}$

Method: The five peaks span four complete cardiac periods. Their total time is $3.54-0.42=3.12\,\text{s}$, so $T=3.12/4=0.780\,\text{s}$. The rate is $60/T=60/0.780=76.9\,\text{beats min}^{-1}$, which rounds to $77\,\text{beats min}^{-1}$.

Answer

• Clean the skin and use low-resistance conductive gel, then secure non-reactive electrodes so movement does not change the contact. Use shielded leads kept away from mains sources and a differential low-noise amplifier. The amplifier needs high gain because the cardiac potential differences are only of order millivolts.

Method: Link each fault to a remedy. Cleaning and conductive gel lower and stabilise skin-electrode contact resistance. Securing the electrodes and keeping the patient still reduces movement artefact. Shielded leads and separation from mains wiring reduce induced electrical noise. A differential, low-noise amplifier rejects common interference, while high gain makes the millivolt ECG large enough to record.

Answer

• It is an A-scan; a B-scan displays a two-dimensional brightness image.

Method: An amplitude scan is one-dimensional, with echo size plotted against travel time or depth. A brightness scan places echoes spatially and represents their strength by brightness to form a two-dimensional image.

Answer

• $0.179\%$

Method: The impedances are $Z_A=\rho_Ac_A=1000(1500)=1.500\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$ and $Z_B=1060(1540)=1.6324\times10^6\,\text{kg m}^{-2}\text{s}^{-1}$. Thus $I_r/I_i=((Z_B-Z_A)/(Z_B+Z_A))^2=((1.6324-1.500)/(1.6324+1.500))^2=1.787\times10^{-3}$. Multiplying by $100$ gives $0.179\%$.

Answer

• $0.100\,\text{m}$ depth and $0.39\,\text{mm}$ resolution; the gel removes air and reduces the acoustic-impedance mismatch, so less intensity is reflected at the skin.

Method: The pulse makes a round trip, so $d=ct/2=1540(130\times10^{-6})/2=0.1001\,\text{m}$, giving $0.100\,\text{m}$. Its wavelength is $\lambda=c/f=1540/(4.0\times10^6)=3.85\times10^{-4}\,\text{m}=0.385\,\text{mm}$, so a one-wavelength resolution estimate is $0.39\,\text{mm}$. Air has a very different acoustic impedance from tissue and would reflect most incident ultrasound; gel displaces the air and provides a better impedance match, increasing transmission into the body.

Answer

• The coherent bundle carries the image, and the non-coherent (incoherent) bundle carries illumination.

Method: Image information requires the fibre positions to be preserved, so it uses a coherent bundle. Illumination only needs light delivery and therefore uses a non-coherent or incoherent bundle.

Answer

• $67.8^\circ$; total internal reflection occurs.

Method: For light travelling from core to cladding, $\sin c=n_{\text{cladding}}/n_{\text{core}}=1.50/1.62$. Hence $c=\sin^{-1}(1.50/1.62)=67.8^\circ$. The incidence angle $72.0^\circ$ is greater than $c$, and the light is travelling from higher to lower refractive index, so total internal reflection occurs.

Answer

• Rearranging the image fibres destroys the one-to-one spatial mapping, so light from each object point emerges at the wrong position. Missing cladding allows light to cross between neighbouring fibres or escape, reducing contrast and blurring the image. Random order does not prevent the illumination bundle working because it only delivers light into the body and does not encode spatial information.

Method: A coherent image bundle must preserve the relative position of every fibre; losing that order scrambles the image. Core-cladding total internal reflection confines each ray, so missing cladding permits leakage and cross-talk, producing dimmer or blurred regions. In the illumination bundle, only the total delivered light matters, so a non-coherent arrangement remains suitable.

Answer

• Hydrogen nuclei (protons); their spins align and precess about the magnetic field direction.

Method: The abundant hydrogen nuclei in body tissue supply the signal. In the strong static field, their spin states become aligned and the proton magnetic moments precess about the field lines.

Answer

• The pulse excites the protons and changes their spin state. When the pulse stops, they relax or de-excite and emit radio-frequency signals that can be detected.

Method: During the pulse, radio-frequency energy at the appropriate frequency is absorbed and selected protons change to an excited spin state. Once excitation ends, the protons return towards their original state. This relaxation or de-excitation releases an RF signal for the receiver coils.

Answer

• The main field aligns proton spins, which precess about the field. Gradient coils select and locate successive small regions of the cross-section. Short RF pulses excite protons in each selected region; as they relax, the protons emit RF signals. Receiver coils detect the signals and a computer uses their position and strength to construct the image.

Method: Give the process in causal order: alignment and precession in the static field; position-dependent gradient fields selecting a slice or small region; an RF pulse changing proton spin states; relaxation or de-excitation producing RF emissions; and detection followed by computer processing into a spatial image. No calculation of relaxation times is needed.

Answer

• $68\,\text{keV}$

Method: An electron accelerated through $68\,\text{kV}$ gains $68\,\text{keV}$. If it loses all this energy in one interaction, $E_{\max}=68\,\text{keV}$.

Answer

• $1.7\times10^{-11}\,\text{m}$

Method: Convert the voltage: $V=72\times10^3\,\text{V}$. The maximum photon energy is $E_{\max}=eV=(1.60\times10^{-19})(72\times10^3)=1.152\times10^{-14}\,\text{J}$. Hence $\lambda_{\min}=hc/E_{\max}=(6.63\times10^{-34})(3.00\times10^8)/(1.152\times10^{-14})=1.7266\times10^{-11}\,\text{m}$, which is $1.7\times10^{-11}\,\text{m}$ to two significant figures.

Answer

• $1.5\times10^{-14}\,\text{J}$ and $0.27\,\text{J}$; rotation spreads heating over the target

Method: The maximum photon energy is $E_{\max}=eV=(1.60\times10^{-19})(95\times10^3)=1.52\times10^{-14}\,\text{J}=1.5\times10^{-14}\,\text{J}$ to two significant figures. The electrical energy supplied is $VIt=(95\times10^3)(3.0\times10^{-3})(0.080)=22.8\,\text{J}$. Therefore the X-ray energy is $0.012\times22.8=0.2736\,\text{J}=0.27\,\text{J}$. Rotating the anode continually changes the impact area, spreading the large thermal energy and reducing local overheating or target damage.

Answer

• It converts incident X-ray photons into visible-light photons.

Method: The marking point is the energy conversion: the scintillator absorbs X-rays and emits visible light, which the photodiode pixels can detect.

Answer

• Barium absorbs X-rays strongly, so less radiation reaches the detector behind the digestive tract and its outline has greater contrast with surrounding soft tissue.

Method: Barium is X-ray opaque, so it has greater attenuation than the surrounding soft tissues. Regions behind the barium therefore give a different detector signal because fewer X-rays are transmitted. This increases image contrast and makes the shape of the digestive tract visible.

Answer

• Use the flat-panel detector: it is sensitive, gives an immediate digital image for enhancement and avoids the prolonged exposure of fluoroscopy; film needs processing and is less suitable for immediate enhancement.

Method: Photographic film records a still image but needs chemical processing and is not directly available for computer enhancement. An intensifying screen can reduce the film exposure, but the result is still not immediate. Fluoroscopic image intensification is valuable for live motion, but continuous viewing exposes the patient throughout the procedure and is unnecessary for one still image. A flat-panel detector is sensitive, so a lower exposure can produce a usable signal, and electronic scanning supplies an immediate digital image that can be enhanced. Therefore the flat-panel detector best meets both stated requirements.

Answer

• $2.5\,\text{cm}$

Method: Use $x_{1/2}=\ln 2/\mu=0.693/0.28=2.48\,\text{cm}$, which is $2.5\,\text{cm}$ to two significant figures.

Answer

• $1.8\,\text{W m}^{-2}$

Method: Use $I=I_0e^{-\mu x}$. The exponent is $-\mu x=-(0.18)(7.0)=-1.26$. Therefore $I=6.4e^{-1.26}=1.82\,\text{W m}^{-2}=1.8\,\text{W m}^{-2}$.

Answer

• $1.1\,\text{W m}^{-2}$, $3.3\,\text{W m}^{-2}$ and a ratio of $2.9$

Method: Attenuation exponents add for successive materials. For the bone path, $\mu x=(0.050)(18)+(0.32)(4.0)=0.90+1.28=2.18$, so $I_{\text{bone}}=10.0e^{-2.18}=1.130\,\text{W m}^{-2}=1.1\,\text{W m}^{-2}$. For soft tissue only, $\mu x=(0.050)(22)=1.10$, so $I_{\text{soft}}=10.0e^{-1.10}=3.329\,\text{W m}^{-2}=3.3\,\text{W m}^{-2}$. Using unrounded values, the ratio is $I_{\text{soft}}/I_{\text{bone}}=2.945=2.9$ to two significant figures.

Answer

• To obtain attenuation measurements through the body from many directions.

Method: Movement changes the beam direction, providing the many projections that the computer needs to reconstruct a cross-section.

Answer

• A narrow beam and detector array collect transmitted intensities at many angles as the tube moves, and a computer reconstructs these projection data into a cross-sectional image.

Method: The moving tube directs a narrow X-ray beam through the selected section from a succession of angles. The detector array measures the transmitted intensity for many paths. A computer processes this set of projections to calculate the attenuation distribution and display a cross-sectional image.

Answer

• Use CT to localise the lung lesion because cross-sectional images remove rib superposition and give better depth information; use a plain X-ray for the feeding tube because its radiopaque position is shown adequately at lower dose, lower cost and greater availability.

Method: CT reconstructs slices, so ribs at different depths do not remain superimposed on the lesion and its three-dimensional location is clearer. That extra information can justify CT's higher ionising dose, cost and lower availability for the lesion. A plain radiograph is quick, cheap, widely available and gives a lower dose. Because a radiopaque tube already has strong contrast and only its overall position must be checked, the projection image is sufficient. Therefore use CT for the obscured lesion and a plain X-ray for the tube check.

Answer

• Gamma radiation can pass out of the body and be detected externally.

Method: The required property is penetration: enough gamma photons must escape the tissue to reach the detector outside the patient.

Answer

• Its half-life is long enough for preparation and imaging but short enough to limit dose; $140\,\text{keV}$ gamma photons can leave the body for detection; and technetium-99m can be attached to compounds that concentrate in a chosen organ.

Method: Award one linked justification for each stated property. A $6\,\text{h}$ half-life provides usable activity during the scan but causes activity to fall soon afterwards. The $140\,\text{keV}$ gamma radiation is penetrating enough for external detection while avoiding an unnecessarily high photon energy. Labelling a compound with organ affinity makes the recorded count distribution show that organ's function or structure.

Answer

• The labelled glucose accumulates in active tissue; emitted positrons annihilate with electrons to produce two opposite $511\,\text{keV}$ photons; coincidence detectors identify a line of response, and many such events are reconstructed into an uptake map.

Method: The glucose analogue is carried into tissues, so regions with greater uptake contain more tracer. A radionuclide in the tracer emits a positron. After losing energy, the positron annihilates with an electron and produces two $511\,\text{keV}$ gamma photons travelling in approximately opposite directions. Detectors on opposite sides register near-simultaneous photons as a coincidence, locating the event somewhere along the joining line. A computer combines many lines of response to reconstruct the spatial distribution of uptake.

Answer

• The time for biological processes alone to remove half of the tracer from the organ.

Method: The definition must identify removal by biological processes and a reduction to one half; radioactive decay is excluded from this definition.

Answer

• $11\,\text{h}$

Method: Use $1/T_E=1/T_B+1/T_P=1/30+1/18=0.0889\,\text{h}^{-1}$. Therefore $T_E=1/0.0889=11.25\,\text{h}$, which is $11\,\text{h}$ to two significant figures.

Answer

• $18.0\,\text{h}$ and $12.5\%$

Method: Rearrange the reciprocal relation: $1/T_B=1/T_E-1/T_P=1/7.20-1/12.0=0.0556\,\text{h}^{-1}$, so $T_B=18.0\,\text{h}$. The elapsed time is $21.6/7.20=3.00$ effective half-lives. The remaining activity fraction is therefore $(1/2)^3=0.125$, giving $12.5\%$.

Answer

• It absorbs gamma photons arriving from unsuitable directions so that detected photons retain directional information.

Method: The collimator selects photon directions before the scintillator; it does not focus gamma rays with refraction.

Answer

• The photon produces a light flash in the scintillator, the photocathode emits photoelectrons, and successive dynodes multiply them to give an anode pulse.

Method: The gamma photon deposits energy in the scintillation crystal, which emits visible photons. Light reaching a photomultiplier photocathode releases electrons by the photoelectric effect. These electrons accelerate to successive dynodes and cause secondary emission at each stage, so a much larger electron pulse is collected at the anode.

Answer

• The collimator selects photon directions; relative signals from nearby photomultipliers locate each scintillation; pulse height estimates photon energy so scattered events are rejected; accumulating accepted events produces the image, with a resolution-sensitivity trade-off set by the collimator.

Method: The lead collimator accepts only a restricted range of directions, linking a detected photon to a line from the patient. A scintillation spreads light to several photomultipliers, and the relative sizes of their pulses allow the computer to estimate the interaction position. The total pulse height is related to deposited photon energy, so an energy window rejects lower-energy scattered photons that would be placed incorrectly. The computer accumulates the accepted positions to form the count image. Smaller collimator apertures sharpen directional selection and improve resolution, but fewer photons pass, so sensitivity falls.

Answer

• They are sufficiently penetrating to reach and ionise the deep tumour.

Method: The required link is high photon energy to penetration to the depth of the tumour.

Answer

• All beams overlap at the tumour so their doses add there, whereas each healthy region is crossed by fewer beams and receives a smaller dose.

Method: Aim each beam so that it passes through the tumour. At the tumour the energy deposited by all beam directions adds to the therapeutic dose. Away from the tumour, the entry and exit paths are spread through different healthy regions, so any one region receives only part of the total dose.

Answer

• $6.5\,\text{J}$; suitable methods include converging beams from different directions, collimation or shielding, and accurate imaging and planning

Method: Since $1\,\text{Gy}=1\,\text{J kg}^{-1}$, the energy absorbed per fraction is $E=Dm=(1.8)(0.18)=0.324\,\text{J}$. Over $20$ fractions, $E_{\text{total}}=20(0.324)=6.48\,\text{J}=6.5\,\text{J}$. Healthy-tissue exposure can be limited by using several beam directions that overlap at the tumour but spread entry dose, and by collimating or shielding the beam to exclude other tissue. Accurate imaging and computer planning improve targeting so that less healthy tissue lies within the treatment fields.

Answer

• Beta radiation has a short range in tissue, so its ionising dose is concentrated near the implant.

Method: Link the limited penetration of beta particles to a high local dose and reduced dose to distant healthy tissue.

Answer

• The implant is beside the tumour and beta particles have a short range, so most energy is deposited locally; an external X-ray beam must cross healthy tissue before and after reaching the tumour.

Method: The source is positioned within or adjacent to the target, avoiding a long entry path. Beta particles are strongly ionising over a limited range, so their energy is deposited close to the implant. High-energy X-rays are penetrating and an external beam necessarily irradiates some healthy tissue along its entry and exit paths.

Answer

• $1.7\,\text{MBq}$; placement in or beside the tumour and the short range of beta radiation localise the dose

Method: The elapsed number of half-lives is $120/48=2.5$. Hence $A=9.6(1/2)^{2.5}=1.70\,\text{MBq}=1.7\,\text{MBq}$. Placing the source within or close to the tumour avoids irradiating a long path of healthy tissue before the radiation reaches its target. Beta particles have a short range in tissue, so most ionisation occurs near the implant and relatively little dose reaches distant cells. The source can also be removed when the planned treatment time is complete.

Answer

• Ultrasound imaging

Method: Ultrasound is portable, produces real-time images and does not use ionising radiation, so it meets all three requirements.

Answer

• MR generally gives better soft-tissue contrast and no ionising dose, making it preferable for repeated scans, but CT is usually faster and more available while exposing the patient to ionising radiation.

Method: MR distinguishes soft tissues well, so tumour boundaries can have high contrast. It uses magnetic fields and RF signals rather than ionising radiation, which is advantageous for repeated imaging. CT is often quicker and more readily available, but each scan contributes an X-ray dose; the justified preference is therefore MR unless speed or access is the overriding constraint.

Answer

• Ultrasound is a reasonable first choice because it is available, convenient and non-ionising; CT is preferred if a more precise high-resolution location is needed despite its larger dose; MR is less suitable because stone-like calcified material gives poor contrast and the scan is costly and less available.

Method: Ultrasound is portable or widely accessible, quick and non-ionising, so it is suitable as an initial investigation, although its spatial resolution and image quality are lower and depend on the acoustic path. CT gives high-resolution cross-sectional localisation and strong contrast for dense stone material without superposition, but it uses a comparatively large ionising dose and is more costly. MR provides strong soft-tissue contrast without ionising radiation, yet calcified stone material is poorly shown, scans take longer and access is more limited. Therefore choose ultrasound first when it can answer the question, and use CT when the clinical need for precise localisation outweighs the radiation risk; MR is not the best choice for this target.