AQA A-level Physics (7408) · 3.10 Medical physics (option)
Mini-Lesson
Medical physics
This mini-lesson covers the AQA 3.10 Medical physics option: the physics of the eye (lens power, myopia and hypermetropia), the ear, ultrasound imaging and acoustic impedance, X-rays, attenuation and CT scanning, the ECG, and MRI.
Optional topic — examined in Paper 3 Section B.
Optional topic — AQA sections 3.9–3.13 are options: you study exactly ONE of the five (Astrophysics, Medical physics, Engineering physics, Turning points in physics, Electronics). It is examined in Paper 3 Section B. Check with your teacher which option your school teaches before revising this one.
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.10.1 · the eye
The eye: power, accommodation and defects of vision
The eye focuses light onto the retina. The cornea does most of the refraction; the lens fine-tunes it — a process called accommodation.
P = 1 ÷ fpower P in dioptres (D), f in metres · converging lens: P positive · diverging lens: P negative
1/u + 1/v = 1/f · magnification m = v ÷ ureal-is-positive convention
Myopia (short sight): distant objects focus in front of the retina (the eye is too powerful or too long). The far point is closer than infinity. Corrected with a diverging (concave) lens of negative power.
Hypermetropia (long sight): near objects focus behind the retina. The near point is further than 25 cm. Corrected with a converging (convex) lens of positive power.
Sensitivity = response per unit light intensity. Spatial resolution is limited by the spacing of the cones in the fovea. Rods are more sensitive (they are grouped and share a nerve) but give poorer resolution and no colour; cones give colour and fine detail.
Worked example — correcting myopia
A short-sighted eye has a far point of 0.40 m. The lens must make an object at infinity appear to come from 0.40 m in front, so f = −0.40 m.
P = 1 ÷ (−0.40) = −2.5 D — a diverging lens, as expected.
Calculate
Your turn — calculation 1
1A short-sighted person has a far point of 0.40 m. Calculate the power of the correcting lens, in dioptres. (Remember the sign!)
D
Hint: The lens has f = −0.40 m (it must form a virtual image at the far point). P = 1 ÷ f.
Quick check
Think it through
?Which lens corrects myopia (short sight), and why?
3.10.2 · the ear
The ear, intensity and the decibel scale
Sound is funnelled by the pinna, vibrates the eardrum, is amplified by the ossicles (a force-multiplying lever system, matching air to fluid), and stimulates hair cells in the fluid-filled cochlea, which convert vibration into nerve impulses.
The ear’s response is logarithmic in both intensity and frequency, which is why we use a logarithmic scale:
intensity level = 10 log(I ÷ I0) dBI0 = 1.0×10−12 W m−2, the threshold of hearing at 1 kHz
The dBA scale weights the measurement to match the frequency response of the human ear.
A ×10 increase in intensity is +10 dB; a ×100 increase is +20 dB.
Equal loudness curves show that the ear is most sensitive around 1–4 kHz.
Worked check: going from 10−12 to 10−6 W m−2 is a factor of 106, so the intensity level is 10 log(106) = 60 dB — normal conversation.
3.10.3 · ultrasound
Ultrasound imaging and acoustic impedance
Ultrasound is sound above 20 kHz, generated and detected by a piezoelectric crystal (an alternating pd makes it vibrate; a returning echo squeezes it and generates a pd — the same crystal does both jobs).
Z = ρcacoustic impedance, unit kg m−2 s−1 · ρ = density, c = speed of sound in that medium
Ir ÷ Ii = [(Z2 − Z1) ÷ (Z2 + Z1)]²the fraction of intensity reflected at a boundary between two media
A big impedance mismatch → almost total reflection. Air/tissue is a huge mismatch, so without coupling gel almost all the ultrasound would reflect at the skin and never enter the body. The gel has an impedance close to that of tissue — this is impedance matching.
A-scan: a single line of echoes vs time (used for eye depth measurement). B-scan: many A-scans combined into a 2D image.
Depth = ½ × c × t (halved, because the pulse travels there and back).
Higher frequency → shorter wavelength → better resolution, but less penetration. That is the fundamental trade-off.
Worked example
Soft tissue: ρ = 1060 kg m−3, c = 1540 m s−1.
Z = ρc = 1060 × 1540 = 1.63×106 kg m−2 s−1.
Calculate
Your turn — calculation 2
2Soft tissue has density 1060 kg m−3 and a speed of sound of 1540 m s−1. Calculate its acoustic impedance. Give the number in front of ×106 kg m−2 s−1.
× 10⁶ kg m⁻² s⁻¹
Hint: Z = ρc = 1060 × 1540 = 1 632 400.
Calculate
Your turn — calculation 3
3Ultrasound of speed 1540 m s−1 in tissue returns an echo 40 µs after the pulse is transmitted. Calculate the depth of the reflecting boundary, in cm.
cm
Hint: The pulse travels there AND back: depth = ½ × 1540 × 40×10−6 m. Convert m → cm.
Calculate
Your turn — calculation 4
4Ultrasound passes from air (Z1 = 430 kg m−2 s−1) into soft tissue (Z2 = 1.63×106 kg m−2 s−1). Calculate the percentage of the intensity reflected at the boundary.
%
Hint: Z2 − Z1 = 1 631 970 and Z2 + Z1 = 1 632 830. Ratio = (1 631 970 ÷ 1 632 830)² = 0.9989. Multiply by 100.
Quick check
Think it through
?Why must coupling gel be smeared on the skin before an ultrasound scan?
3.10.4 · X-rays & CT
X-rays, attenuation and CT scanning
In an X-ray tube, electrons are released by thermionic emission, accelerated through a large pd, and slammed into a rotating metal (tungsten) anode. Most energy becomes heat; a small fraction becomes X-rays — a continuous bremsstrahlung spectrum plus sharp characteristic lines.
I = I0e−µxµ = attenuation (absorption) coefficient in m−1 · x = thickness of absorber
half-value thickness x½ = ln2 ÷ µthe thickness that halves the intensity
Contrast media: barium meals or iodine solutions have high atomic number, so they absorb X-rays strongly and reveal soft tissue (the gut, blood vessels) that would otherwise be invisible.
CT scanning: an X-ray tube and detector array rotate around the patient, taking many projections. A computer reconstructs 2D slices, which stack into a 3D image.
CT gives far better soft-tissue contrast and full 3D information than a plain X-ray, but at a much higher radiation dose — a real risk, since X-rays are ionising.
6For the same tissue (µ = 0.25 cm−1), calculate the half-value thickness in cm.
cm
Hint: x½ = ln2 ÷ µ = 0.693 ÷ 0.25.
Sort it
Ultrasound, X-ray / CT, or MRI?
Tap a statement, then tap the imaging technique it belongs to.
🔊 Ultrasound
🩻 X-ray / CT
🧲 MRI
3.10.5 · ECG & MRI
The ECG and magnetic resonance imaging
ECG (electrocardiogram): electrodes on the skin (with conducting gel) pick up the tiny potential differences — typically about 1 mV — produced as heart muscle depolarises and repolarises. The signal needs a high-gain, high input impedance amplifier with good common-mode rejection to beat the noise.
P wave — atrial depolarisation (the atria contract).
QRS complex — ventricular depolarisation (the big spike; it swamps atrial repolarisation).
T wave — ventricular repolarisation (the ventricles recover).
MRI: a very strong superconducting magnet aligns the magnetic moments of hydrogen nuclei (protons), which precess at the Larmor frequency. A radio-frequency pulse at that frequency flips them (resonance). When the pulse stops they relax, re-emitting RF signals. Gradient coils make the field vary with position, so the resonant frequency encodes where the signal came from.
Advantages: no ionising radiation, superb soft-tissue contrast, and images in any plane.
Disadvantages: very expensive, noisy, claustrophobic, slow, and impossible for patients with pacemakers or ferromagnetic implants.
Match it
Match each item to the right physics
Tap a card on the left, then its partner on the right.
Statement
Answer
Quick check
Think it through
?What does the QRS complex on an ECG trace represent?
Quick check
Think it through
?Which is the single biggest advantage of MRI over CT scanning?
Recap
The big ideas to know
The eye: P = 1/f (dioptres) · myopia → diverging (−) lens · hypermetropia → converging (+) lens
The ear: intensity level = 10 log(I/I0) dB · I0 = 10−12 W m−2 · most sensitive at 1–4 kHz
Ultrasound: Z = ρc · reflection = [(Z2−Z1)/(Z2+Z1)]² · coupling gel matches impedance · depth = ½ct
X-rays & CT: I = I0e−µx · x½ = ln2/µ · CT = 3D but a high dose
ECG: P = atrial depolarisation · QRS = ventricular depolarisation · T = ventricular repolarisation