Answers · The eye
The pupil reflex — answer key
Check meaning, not exact wording. These are teaching points, not official Cambridge marks.
1. PQ1 · Practice
Model answer. The light becomes brighter, so the pupil reflex makes the pupils narrower: their diameter decreases and less light enters each eye.
A full answer includes
- I chose: they get narrower, and their diameter decreases.
- I did not choose "the whole eyeball gets smaller".
Also accept
- narrower: diameter decreases
Watch for
- “They get wider: their diameter increases.” — The pupils widen when the light becomes dim. Brighter light makes them narrower.
- “They stay the same: only the lens changes.” — The lens changes shape to focus, not to control how much light enters. The pupils change size with the light.
- “The whole eyeball gets smaller instead.” — The eyeball keeps its size. The iris changes the size of the pupil, the hole in its middle.
2. PQ2 · Practice
Model answer. A narrower pupil lets less light into the eye. In bright light this protects the retina, whose receptors could be damaged by too much light.
A full answer includes
- I chose: less light enters, so the retina is protected from damage.
- I did not choose the lens: the light-sensitive part is the retina.
Also accept
- less light enters, protecting the retina
Watch for
- “More light enters, so the retina gets enough light to see.” — A narrower hole lets less light in, not more. Letting more in is what the wider pupil does in dim light.
- “Less light enters, so the lens is protected from damage.” — It is the retina, with its light receptors, that too much light could damage.
- “The pupil shuts, so no light at all can reach the retina.” — The pupil never shuts: it narrows, and some light still enters so that you can see.
3. PQ3 · Practice
Model answer. The diameter is the distance from one edge of the pupil to the other: 7.8 − 2.5 = 5.3 mm.
Checked automatically
- Pupil diameter: 5.3 mm (within ±0.05 mm)
- Common wrong answer 10.3 mm: That adds the two readings. The distance between two marks is their difference.
A full answer includes
- I subtracted the left-edge reading from the right-edge reading.
- I wrote 5.3 mm.
- I did not add the two readings together.
Also accept
- 5.3 mm
Watch for
- “10.3 mm.” — That adds the two readings. The distance between two marks on a scale is the difference between them.
4. PQ4 · Practice
Model answer. The diameter falls at every step as the light gets brighter, from 8.0 mm to 3.9 mm. The drops are 1.6, 1.2, 0.8 and 0.5 mm, so the biggest drop is at the dim end, and the change gets smaller as the light gets brighter.
A full answer includes
- I chose: the diameter falls as the light gets brighter, with the biggest drop at the dim end.
- I did not choose "by the same amount each time": the drops get smaller.
Also accept
- falls as the light gets brighter; the biggest drop is at the dim end
Watch for
- “Diameter rises as the light gets brighter; the biggest rise is at the dim end.” — Read the diameters downwards: 8.0, 6.4, 5.2, 4.4, 3.9 mm. They get smaller, not bigger.
- “Diameter falls as the light gets brighter; the biggest drop is at the bright end.” — The last drop, from 4.4 to 3.9 mm, is only 0.5 mm. The first, from 8.0 to 6.4 mm, is the biggest.
- “Diameter falls as the light gets brighter, by the same amount each time.” — Work out each drop: 1.6, 1.2, 0.8 and 0.5 mm. They are not equal.
5. PQ5 · Practice
Options as numbered on the pupil sheet: 1 — the muscles of the iris; 2 — a change in pupil diameter; 3 — a change in light intensity; 4 — light receptors in the retina.
Model answer. The stimulus is the change in light intensity. Light receptors in the retina detect it. The effector is the muscles of the iris, and the response is the change in pupil diameter.
A full answer includes
- I matched the stimulus to a change in light intensity.
- I matched the receptors to the light receptors in the retina.
- I matched the effector to the muscles of the iris.
- I matched the response to a change in pupil diameter.
- I did not match the effector to the retina.
Also accept
- stimulus light intensity; receptors in retina; effector iris muscles; response pupil diameter
Watch for
- “The effector is the retina.” — The retina contains the receptors, which detect the light. The effector, which acts, is the muscles of the iris.
- “The response is the muscles of the iris.” — The muscles are the effector: the part that acts. The response is the result of their action, a change in the pupil's size.
6. PQ6 · Practice
Model answer. Row C. In bright light the circular muscles contract and the radial muscles relax, so the pupil narrows. In dim light it is the other way round: the radial muscles contract and the circular muscles relax, so the pupil widens. The two sets always do opposite things.
A full answer includes
- I chose row C.
- I did not choose a row where both sets contract, or both relax, in the same light.
Also accept
- row C
Watch for
- “Row A: radial muscles contract in bright light.” — Row A has the muscles the wrong way round: contracting radial muscles widen the pupil, which happens in dim light, not bright.
- “Row B: both sets contract together in bright light.” — In row B both sets contract together, then both relax. The two sets are antagonistic: when one contracts, the other relaxes.
- “Row D: the radial muscles stay relaxed in dim light.” — Row D is right for bright light, but in dim light the pupil must widen, so the radial muscles contract; they do not stay relaxed.
7. PQ7 · Practice
Model answer. The change in pupil size is the pupil reflex. It is automatic and rapid: it happens without a decision, so it cannot be done on purpose.
A full answer includes
- I wrote that the change in pupil size is the pupil reflex.
- I wrote that a reflex is automatic: it happens without a decision.
- I did not write that the brain decides to change the pupils.
Also accept
- the pupil reflex: automatic, without a decision
Watch for
- “The brain decided to make the pupils smaller.” — The impulses pass through the brain, but no decision is made: it is a reflex.
8. PQ8 · Practice
Model answer. (a) 3.7 mm is the anomaly: the other three readings are all between 6.2 and 6.6 mm. (b) The mean of the rest is (6.2 + 6.6 + 6.3) ÷ 3 = 19.1 ÷ 3 = 6.37, which is 6.4 mm to 1 decimal place.
Checked automatically
- (a) Anomalous reading: 3.7 mm (exact value)
- Common wrong answer 6.6 mm: An anomaly does not fit the pattern of the others; 6.6 mm is close to 6.2 and 6.3 mm.
- (b) Mean diameter: 6.4 mm (2 significant figures)
- Common wrong answer 5.7 mm: That includes the anomalous reading. Leave it out and divide by 3.
A full answer includes
- (a) I chose the reading far from all the others.
- (b) I added the other three readings and divided by 3.
- (b) I wrote 6.4 mm.
- (b) I did not include the anomalous reading in the mean.
Also accept
- (a) 3.7 mm; (b) 6.4 mm
Watch for
- “(a) 6.6 mm, the largest.” — An anomaly is a reading that does not fit the pattern of the others. 6.6 mm is close to 6.2 and 6.3 mm.
- “(b) 5.7 mm.” — That includes the anomaly: 22.8 ÷ 4 = 5.7. Leave the anomalous reading out of the mean.
9. PQ9 · Practice
Model answer. A muscle can only pull, by contracting; it cannot push. Contracting circular muscles narrow the pupil, and contracting radial muscles widen it, so one set is needed for each direction. They are antagonistic: when one set contracts, the other relaxes.
A full answer includes
- I wrote that a muscle can only pull, by contracting.
- I wrote that the circular muscles narrow the pupil when they contract.
- I wrote that the radial muscles widen the pupil when they contract.
- I wrote that the two sets are antagonistic: when one contracts, the other relaxes.
- I did not write that one set of muscles pushes the pupil wider.
Also accept
- muscles only pull; circular narrow, radial widen; antagonistic
Watch for
- “One set contracts to narrow the pupil and pushes to widen it.” — Muscles cannot push. A second set, the radial muscles, contracts to widen the pupil.
10. PQ10 · Practice
Model answer. The decrease is 8.0 − 3.9 = 4.1 mm. As a percentage of the starting diameter: 4.1 ÷ 8.0 × 100 = 51.25, which is 51 % to 2 significant figures.
Checked automatically
- Percentage decrease: 51 % (2 significant figures)
- Common wrong answer 105 %: That divides by the final diameter. Divide the decrease by the starting diameter.
- Common wrong answer 49 %: That is the final diameter as a percentage of the first. Find the decrease first.
- Common wrong answer 4.1 : That is the decrease in millimetres. Turn it into a percentage of the starting diameter.
A full answer includes
- I found the decrease and divided it by the starting diameter, then multiplied by 100.
- I wrote 51 %.
- I did not divide by the final diameter.
Also accept
- 51 %
Watch for
- “105 %.” — That divides the decrease by the final diameter, 3.9 mm. A percentage change is worked out from the starting value.
- “49 %.” — That is the final diameter as a percentage of the first. The question asks for the decrease.
11. PT1 · Apply
Model answer. In the changing room the area is π × 3.6² = 40.7 mm²; in the sunshine it is π × 1.2² = 4.52 mm². 40.7 ÷ 4.52 = 9.0, so the area is 9 times larger. The diameter is only 3 times larger, but the area goes up with the square of the diameter, so about 9 times as much light can enter in the dark.
Checked automatically
- Times larger: 9.0 (within ±0.1)
- Common wrong answer 3 : That compares the diameters. Compare the areas: the area goes up with the square of the diameter.
- Common wrong answer 36.2 : That is the difference between the areas. Divide one area by the other.
- Common wrong answer 4.8 : That is the difference between the diameters. Work out both areas and divide.
A full answer includes
- I worked out both areas, about 40.7 mm² and 4.52 mm².
- I divided the larger area by the smaller one.
- I wrote about 9 times.
- I did not stop at the ratio of the diameters.
Also accept
- 9 times
Watch for
- “3 times.” — That compares the diameters. The area goes up with the square of the diameter: 3² = 9.
- “36.2 times.” — That is the difference between the two areas in mm². The question asks how many times larger: divide.
12. PT2 · Apply
Model answer. The forest is dimmer, so the pupils must widen to let more light in. (a) The circular muscles relax. (b) The radial muscles contract and pull the pupil wider. (c) The pupil diameter increases.
A full answer includes
- (a) I wrote that the circular muscles relax.
- (b) I wrote that the radial muscles contract.
- (c) I wrote that the pupil diameter increases.
- (a) I did not write that the circular muscles contract.
Also accept
- (a) relax; (b) contract; (c) increases
Watch for
- “(a) The circular muscles contract.” — Contracting circular muscles narrow the pupil, which happens in bright light. In the dim forest they relax.
- “(c) It decreases.” — Moving into dimmer light, the pupil widens so that more light enters.
13. PT3 · Apply
Model answer. The pupil widens and stays wide. The circular muscles, which narrow the pupil, are relaxed and cannot contract, so nothing can narrow it even in the bright room. The radial muscles, their antagonists, can still contract and pull the pupil wider.
A full answer includes
- I predicted that the pupil widens and stays wide.
- I wrote that the radial muscles can still contract and pull the pupil wider.
- I wrote that the circular muscles cannot contract to narrow it, even in bright light.
- I did not predict that the pupil narrows.
Also accept
- pupil widens; radial muscles still contract; circular cannot narrow it
Watch for
- “The pupil narrows, because the muscles are relaxed.” — Relaxing the circular muscles takes away the pull that narrows the pupil, so it widens.