AQA A Level Physics Paper 3 Section B: Medical physics June 2025 question paper 7408/3BB

Figure 1 shows a simplified diagram of an eye that has no sight defects. The eye is viewing a distant point object.

Complete, on Figure 1, the paths of the two rays to the back of the eye.

Accommodation is the ability of the eye to change its optical power.

Which part of the eye enables accommodation?

Tick (✓) one box.

The accommodation amplitude A, in dioptres, is given by

where

d₁ is the distance, in m, from the eye to its near point d₂ is the distance, in m, from the eye to its far point.

Figure 2 shows how the value of A varies with age for a typical person.

An eye of a typical 30-year-old person has a far point at infinity. The focal length of the eye when viewing an object at its far point is 23 mm.

Calculate the focal length of the eye when viewing an object at its near point.

For a typical person, the far point remains at infinity as the person’s age increases. Some studies suggest that hypermetropia increases with age.

Explain how Figure 2 supports this suggestion.

Technetium-99_m (^99mTc) is produced from the decay of molybdenum-99 (⁹⁹Mo). In hospitals,^99mTc is extracted from a molybdenum-technetium generator.

⁹⁹Mo is put into the generator at time t = 0 As⁹⁹Mo decays, the amount of^99mTc increases to a maximum at t = 24 hours. ^99mTc is extracted from the generator at regular intervals. At each extraction, the activity of^99mTc in the generator falls to zero and then increases again.

Figure 3 shows how the activities of⁹⁹Mo and^99mTc in the generator vary with t when extractions are made at t = 24 hours and t = 48 hours.

Show that about 6 × 10¹⁵ nuclei of⁹⁹Mo decay between t = 0 and t = 24 hours.

decay constant for⁹⁹Mo = 2.8 × 10⁻⁶ s⁻¹

90% of the⁹⁹Mo nuclei decay to form^99mTc nuclei. The other 10% decay to form different nuclei.

Estimate the number of^99mTc nuclei that are extracted from the generator at t = 24 hours.

Information about heart function can be obtained from a multi-gated acquisition (MUGA) scan or from a transthoracic echocardiogram (TTE) scan. Heart specialists use these scans to diagnose a range of heart conditions.

In a MUGA scan, a patient is injected with a compound containing^99mTc which binds to red blood cells. Using images of the red blood cells, a precise reading can be made of blood flow rate through the heart.

A TTE is an ultrasound scan of the heart. Images of the heart are seen on a monitor as a transducer is moved across the surface of the patient’s chest.

Compare the MUGA method of investigating heart function with the TTE method of investigating heart function.

In your answer you should:

• outline how an image of the red blood cells is obtained during a MUGA scan

• explain the advantages that a MUGA scan and a TTE scan have in common

• compare the strengths and weaknesses of a MUGA scan and a TTE scan.

State what is meant by the threshold of hearing for a person with good hearing.

In a hearing test, a person is found to have good hearing. Figure 4 shows the equal loudness curve produced in the test.

The eardrum of the person being tested has a cross-sectional area of 60 mm².

Calculate the power that is incident on the eardrum when the frequency is 50 Hz.

Sketch, on Figure 4, an equal loudness curve that passes through 100 dB at 1 kHz.

Figure 5 shows a rotating-anode X-ray tube used to produce an image of a bone.

X-rays originate from a region on the rotating anode known as the focal spot.

Explain:

• one advantage of using a small focal spot

• why the focal spot should not be too small.

State and explain three methods used to limit exposure of the patient’s healthy cells to X-rays during a diagnostic procedure.

Figure 6 shows a scale diagram of a cross-section through a human bone called a femur.

A dashed line has been drawn from point A to point B through the centre of the femur. Along AB the femur has a circular cross-section with an external diameter of 27 mm. The femur has a hollow centre.

A beam of X-ray photons is incident on the femur at A in the direction AB. For these photons, the half-value thickness of bone is 15 mm.

Determine, using the scale diagram in Figure 6, the percentage of the X-ray photons incident on A that emerge at B.