Notes · PhysicsUK · A-Levels

Medical physics (optional)

This is one of the five optional A-Level modules (AQA Paper 3, Section B); a student studies just one option. Medical physics applies core physics to the body and to diagnosis: the eye and ear as optical and acoustic detectors, the decibel scale, the electrical signals of the heart, and the principal imaging techniques - ultrasound, X-rays and CT, MRI and radionuclide imaging - compared by resolution, risk and suitability.

5 sections~14 min reading time3 competenciesLevel Standard 3 · Advanced 2

T·101010 / 13

Exam profile

AO1 · Describe the eye and ear as physical systems and the principal medical imaging techniques (optional module)AO2 · Calculate lens powers, intensity levels in decibels, acoustic impedance and reflection, and X-ray attenuationAO3 · Compare imaging techniques by resolution, risk and suitability and interpret attenuation and biological data

Operators:describecalculateexplaincomparedetermineevaluate

basic level

This is an optional A2 module examined in Paper 3 Section B; it is studied only if chosen as the school's option. It requires the eye and ear, the decibel scale and the outline of the imaging techniques.

higher level

The full option demands quantitative lens correction, decibel and acoustic-impedance calculations, X-ray attenuation and a reasoned comparison of imaging methods.

Depth

Text

Contents · 5 sections

Medical physics (optional)

Physics of the eye and its defects#

StandardAQA 7408 3.10.1DfE GCE Physics - medical physics option

The eye as a converging lens

Fig. 1The eye's lens system focuses light from an object to a real, inverted image on the retina; the three construction rays converge at the image tip.

The eye is a converging optical system. The cornea does most of the refraction and the flexible lens fine-tunes the focus, together forming a real, inverted, diminished image on the light-sensitive retina. The power of a lens is

P = \dfrac{1}{f}

, measured in dioptres (D) when

f

is in metres, and powers add when lenses are combined - so the eye's total power is the sum of the corneal and lens powers.

Accommodation is the eye's ability to change the shape of its lens to focus on objects at different distances: ciliary muscles thicken the lens (increasing its power) to focus on near objects, and relax it for distant ones. The near point (closest clear vision) is conventionally

25\,\text{cm}

for a normal eye, and the far point is infinity. The eye also adapts to a huge range of light intensities and is most sensitive in the green-yellow part of the spectrum.

The common defects are corrected with lenses that change the eye's total power. In short sight (myopia) the eye is too powerful (or too long) and focuses distant objects in front of the retina; a diverging (concave) lens of negative power corrects it. In long sight (hypermetropia) the eye is too weak and focuses near objects behind the retina; a converging (convex) lens of positive power corrects it. Astigmatism, an unevenly curved cornea, is corrected with a cylindrical lens.

Designing a correction is a power calculation. The corrective lens must move the image of the problem object onto the retina - equivalently, form a virtual image of the object at the eye's own far or near point, which the uncorrected eye can then focus. Using

\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

with the appropriate distances gives the required lens power in dioptres.

P = \dfrac{1}{f}

Lens power

In dioptres when the focal length is in metres; powers add.

\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

Thin-lens equation

Used to design a correcting lens.

Worked example

Power of a correcting lens

A short-sighted person cannot see clearly beyond a far point of $2.5\,\text{m}$ . Find the power of the spectacle lens needed for clear distant vision.

What the lens must do
It must take parallel light from infinity ( $u = \infty$ ) and form a virtual image at the far point, $v = -2.5\,\text{m}$ (virtual, on the same side).
Apply the lens equation
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{-2.5} - 0 = -0.40\,\text{m}^{-1}$ .
Power
$P = 1/f$ .
$P = -0.40\,\text{D}$

Result: A diverging lens of power $-0.40\,\text{D}$ corrects the short sight.

Typical mistakes

Forgetting that a diverging correcting lens has a negative power (and focal length).
Using centimetres instead of metres for the focal length when finding power in dioptres.

Active revision

A short-sighted eye has a far point of $2.0\,\text{m}$ . Calculate the power of the diverging lens needed to give clear distant vision.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Physics of the ear and the decibel scale#

StandardAQA 7408 3.10.2DfE GCE Physics - the ear

Threshold of hearing against frequency

Fig. 2The quietest sound the ear can detect varies with frequency: the ear is most sensitive (lowest threshold) around

2

4\,\text{kHz}

and far less sensitive at very low and very high frequencies.

The ear detects sound over an enormous range - the loudest tolerable sound carries about a million million times the intensity of the quietest audible one. Because the response is roughly logarithmic, loudness is measured on a logarithmic scale. The intensity level in decibels is

\text{IL} = 10\log\left(\dfrac{I}{I_0}\right)

, where

I_0 = 1.0\times10^{-12}\,\text{W m}^{-2}

is the threshold of hearing. A tenfold increase in intensity adds

10\,\text{dB}

; a doubling adds about

3\,\text{dB}

The ear's sensitivity depends on frequency. It responds to roughly

20\,\text{Hz}

20\,\text{kHz}

and is most sensitive around

2

4\,\text{kHz}

, where the ear canal resonates. A plot of the threshold of hearing against frequency is therefore U-shaped, dipping to its lowest (most sensitive) in the mid-range and rising steeply at very low and very high frequencies. The dBA scale weights measured sound levels to match this frequency response, so a dBA reading reflects perceived loudness.

Sound is transmitted through the ear in stages. The outer ear funnels sound to the eardrum, which vibrates; three small bones (the ossicles) in the middle ear amplify and transmit these vibrations, acting as a lever and matching the impedance from air to fluid; and in the inner ear the cochlea converts the vibrations into nerve impulses, with different positions along it responding to different frequencies.

Equal-loudness curves map, for each frequency, the intensity level that sounds equally loud as a reference tone, and they show why we hear a bass note as quieter than a mid-range note of the same physical intensity. Understanding the logarithmic scale and the frequency response is central to audiometry and hearing-protection design.

\text{IL} = 10\log\!\left(\dfrac{I}{I_0}\right)

Intensity level in decibels

$I_0 = 1.0\times10^{-12}\,\text{W m}^{-2}$ is the threshold of hearing.

Worked example

Intensity level of a sound

A sound has intensity $5.0\times10^{-5}\,\text{W m}^{-2}$ . Find its intensity level in decibels. Take $I_0 = 1.0\times10^{-12}\,\text{W m}^{-2}$ .

Intensity ratio
$\dfrac{I}{I_0} = \dfrac{5.0\times10^{-5}}{1.0\times10^{-12}} = 5.0\times10^{7}$ .
Apply the decibel formula
$\text{IL} = 10\log(5.0\times10^{7}) = 10 \times 7.70$ .
$\text{IL} = 77\,\text{dB}$

Result: The intensity level is about $77\,\text{dB}$ - roughly the level of busy traffic.

Typical mistakes

Adding decibel values directly - decibels are logarithmic, so intensities (not levels) combine.
Forgetting the reference intensity $I_0 = 1.0\times10^{-12}\,\text{W m}^{-2}$ in the decibel formula.

Active revision

A sound has intensity $2.0\times10^{-6}\,\text{W m}^{-2}$ . Calculate its intensity level in decibels, taking $I_0 = 1.0\times10^{-12}\,\text{W m}^{-2}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Biological measurement and the ECG#

StandardAQA 7408 3.10.3DfE GCE Physics - biological measurement

Many diagnoses rest on measuring the body's own electrical signals. Nerve and muscle cells maintain a potential difference across their membranes; when they are stimulated, a rapid change called an action potential sweeps along them, and the resulting tiny voltages can be picked up by electrodes on the skin. The heart, a large coordinated muscle, produces the strongest such signal.

The electrocardiogram (ECG) records the electrical activity of the heart from electrodes placed on the skin. Each heartbeat produces a characteristic trace: a small P wave (the atria contracting), a large sharp QRS complex (the ventricles contracting) and a T wave (the ventricles recovering). The shape, size and timing of these features reveal the health of the heart - an abnormal rhythm, an enlarged chamber or damaged muscle each alters the trace.

Because the signals are very small (of the order of millivolts) and easily swamped by electrical noise, the electrodes are attached with a conducting gel to lower the contact resistance, and the signal is amplified with a high-gain differential amplifier that rejects interference common to both electrodes. The heart rate is read directly from the interval between successive R peaks: rate

= 1/(\text{R-R interval})

The same principles - detect a small biological potential, reduce contact resistance, amplify while rejecting noise - underlie other measurements, from the electroencephalogram (brain) to the electromyogram (muscle). Physics thus provides the instrumentation on which much of clinical diagnosis depends.

\text{heart rate} = \dfrac{1}{\text{R-R interval}}

Heart rate from an ECG

The reciprocal of the time between successive R peaks.

Worked example

Heart rate from an ECG

On an ECG, successive R peaks are separated by $0.80\,\text{s}$ . Find the heart rate in beats per minute.

Beats per second
$\text{rate} = 1/(\text{R-R interval}) = 1/0.80 = 1.25\,\text{beats s}^{-1}$ .
Convert to per minute
Multiply by 60.
$1.25 \times 60 = 75\,\text{beats min}^{-1}$

Result: The heart rate is $75$ beats per minute - a normal resting rate.

Typical mistakes

Confusing the roles of the P wave (atria) and the QRS complex (ventricles).
Forgetting that the biological signals are very small and must be amplified while rejecting noise.

Active revision

An ECG trace shows successive R peaks $0.75\,\text{s}$ apart. Calculate the heart rate in beats per minute.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Ultrasound and non-ionising imaging#

AdvancedAQA 7408 3.10.4DfE GCE Physics - ultrasound imaging

Ultrasound A-scan

Fig. 3A piezoelectric transducer sends a pulse into the body; echoes reflected at tissue boundaries return to the transducer, and their timing gives the depth of each boundary.

Ultrasound is sound above the range of human hearing (above

20\,\text{kHz}

; medical ultrasound uses

1

15\,\text{MHz}

). It is generated and detected by a piezoelectric transducer: a crystal that vibrates when an alternating voltage is applied (producing the pulse) and generates a voltage when a returning echo makes it vibrate (detecting it). The same crystal thus both transmits and receives.

An image is built from echoes. When an ultrasound pulse meets a boundary between two tissues, part of it is reflected. The fraction reflected depends on the mismatch in acoustic impedance

Z = \rho c

(density times the speed of sound), through the reflection coefficient

\dfrac{I_r}{I_0} = \left(\dfrac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2

. A large mismatch (as at a tissue-air or tissue-bone boundary) reflects almost everything, which is why a coupling gel is used to exclude air between the transducer and the skin.

Two display modes are common. An A-scan (amplitude) shows the echoes from one line as a graph of amplitude against time; the time delay of each echo, with the known speed of sound in tissue, gives the depth of each boundary. A B-scan (brightness) sweeps the beam across the body and represents each echo as a bright dot, building a two-dimensional cross-sectional image - as used in obstetric scans.

Ultrasound's great advantage is that it is non-ionising and therefore safe, even in pregnancy, and it images soft tissue and movement in real time. The Doppler effect extends it to measuring blood flow: ultrasound reflected from moving red blood cells is shifted in frequency in proportion to their speed, so a Doppler scan reveals the direction and rate of blood flow. Its limitations are poorer resolution than X-rays and an inability to see through bone or gas.

Z = \rho c

Acoustic impedance

Density times the speed of sound in the medium.

\dfrac{I_r}{I_0} = \left(\dfrac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2

Reflection coefficient

Fraction of intensity reflected at a boundary.

Worked example

Reflection at a tissue boundary

Ultrasound passes from fat ( $Z_1 = 1.38\times10^{6}$ ) into muscle ( $Z_2 = 1.70\times10^{6}\,\text{kg m}^{-2}\,\text{s}^{-1}$ ). Find the fraction of intensity reflected.

Impedance difference and sum
$Z_2 - Z_1 = 0.32\times10^{6}$ ; $Z_2 + Z_1 = 3.08\times10^{6}$ .
Reflection coefficient
$\dfrac{I_r}{I_0} = \left(\dfrac{0.32}{3.08}\right)^2 = (0.1039)^2$ .
$\dfrac{I_r}{I_0} = 0.0108 \approx 1.1\%$

Result: About $1.1\%$ of the intensity is reflected - the small mismatch lets most of the beam pass, which is why gel and matching are essential at the skin.

Typical mistakes

Forgetting to square the reflection-coefficient bracket.
Omitting the coupling gel's purpose - to remove the strongly reflecting air gap (impedance matching).

Active revision

At a boundary, soft tissue ( $Z_1 = 1.63\times10^{6}$ ) meets bone ( $Z_2 = 6.40\times10^{6}\,\text{kg m}^{-2}\,\text{s}^{-1}$ ). Calculate the fraction of ultrasound intensity reflected.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

X-rays, CT, MRI and radionuclide imaging#

AdvancedAQA 7408 3.10.5DfE GCE Physics - X-ray and other imaging

Exponential attenuation of X-rays

Fig. 4The transmitted X-ray intensity falls exponentially with thickness; each half-value thickness halves the intensity.

X-rays are produced when high-energy electrons strike a metal target and decelerate. Passing through the body, the beam is attenuated - absorbed and scattered - according to

I = I_0\,e^{-\mu x}

, where

\mu

is the attenuation coefficient of the tissue. Dense tissue such as bone (high

\mu

) absorbs strongly and appears white on the image; soft tissue absorbs less. The half-value thickness,

x_{1/2} = \dfrac{\ln 2}{\mu}

, is the thickness that halves the intensity, and contrast media (barium, iodine) are used to make soft structures visible.

A CT (computed tomography) scanner takes many X-ray images from different angles around the patient and computes a three-dimensional image, giving far better soft-tissue contrast than a single X-ray but at the cost of a higher radiation dose. Because X-rays are ionising, their use is always balanced against the risk, especially for children and in pregnancy.

Magnetic resonance (MR) imaging uses no ionising radiation. A strong magnetic field aligns the magnetic axes of hydrogen nuclei (protons) in the body; a radio-frequency pulse tips them, and as they relax back they emit radio signals whose strength and timing depend on the tissue. MRI gives excellent soft-tissue contrast and is the method of choice for the brain and joints, though it is expensive and unsuitable for patients with certain metal implants.

Radionuclide imaging works the opposite way - the radiation comes from inside. A gamma-emitting tracer (such as technetium-99m) is introduced into the body and taken up by the organ of interest; a gamma camera detects the emitted gamma rays to form a functional image showing how the organ works, not just its structure. In PET (positron emission tomography), a positron-emitting tracer annihilates with electrons to produce back-to-back gamma photons, whose detection pinpoints activity such as tumour metabolism. Choosing between these techniques weighs resolution, functional versus structural information, cost and radiation risk.

I = I_0\,e^{-\mu x}

X-ray attenuation

Intensity decays exponentially with thickness.

x_{1/2} = \dfrac{\ln 2}{\mu}

Half-value thickness

The thickness that halves the intensity.

Worked example

Attenuation and half-value thickness

An X-ray beam passes through tissue of attenuation coefficient $\mu = 0.25\,\text{cm}^{-1}$ . Find the half-value thickness and the fraction transmitted through $4.0\,\text{cm}$ .

Half-value thickness
$x_{1/2} = \ln 2/\mu = 0.693/0.25 = 2.8\,\text{cm}$ .
Fraction transmitted
$I/I_0 = e^{-\mu x} = e^{-0.25 \times 4.0} = e^{-1.0}$ .
$\dfrac{I}{I_0} = e^{-1.0} = 0.37$

Result: The half-value thickness is $2.8\,\text{cm}$ and about $37\%$ of the beam is transmitted through $4.0\,\text{cm}$ .

Typical mistakes

Confusing the attenuation coefficient $\mu$ with the half-value thickness (they are inversely related).
Describing MRI or ultrasound as ionising - they are not; only X-rays, CT and radionuclide imaging use ionising radiation.

Active revision

An X-ray beam is attenuated by a tissue of attenuation coefficient $\mu = 0.20\,\text{cm}^{-1}$ . Find the half-value thickness and the fraction of the beam transmitted through $5.0\,\text{cm}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Medical physics (optional)

5 sections~14 min reading time3 competenciesLevel Standard 3 · Advanced 2

Medical physics (optional)

Key points

Power of a correcting lens

Key points

Intensity level of a sound

Key points

Heart rate from an ECG

Key points

Reflection at a tissue boundary

Key points

Attenuation and half-value thickness

Sources

Medical physics (optional)

Key points

Power of a correcting lens

Key points

Intensity level of a sound

Key points

Heart rate from an ECG

Key points

Reflection at a tissue boundary

Key points

Attenuation and half-value thickness

Sources