Specification: AQA A-level Physics (7408) 3.10.1.2 Medical physics option | Paper: 3 Section B | Data and formulae booklet: P = 1/f is given in the exam; P = P1 + P2 for thin lenses in contact is not, so learn it
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Power of a lens: P = 1 ÷ focal length in metres. The unit is the dioptre (D), where 1 D = 1 m⁻¹. Converging lenses have positive power; diverging lenses have negative power.
Near point and far point: the near point is the closest point the unaided eye can focus on (about 25 cm for a normal eye); the far point is the furthest point the unaided eye can focus on (infinity for a normal eye).
Defects of vision: in myopia (short sight) the far point is closer than infinity, corrected with a diverging lens. In hypermetropia (long sight) the near point is further than 25 cm, corrected with a converging lens.
The equation
power = 1 ÷ focal length P = 1/f and, for thin lenses in contact, total power = sum of the powers P = P1 + P2 + …
| Quantity | Symbol | Unit |
|---|---|---|
| power of the lens (+ converging, − diverging) | P | dioptres (D), equivalent to m⁻¹ |
| focal length | f | metres (m) |
Rearranged: f = 1/P | Convert the focal length to metres before you divide, or the answer will not be in dioptres. To design a correcting lens, use 1/f = 1/u + 1/v with the virtual image placed at the person’s own near or far point (v negative).
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A converging lens has a focal length of 0.25 m. Calculate its power.
Show solution
F: P = 1/f
I: P = 1 ÷ 0.25
F: P is already the subject
A: +4.0 D
Q2 (Core) A spectacle lens has a power of −2.5 D. Calculate its focal length and state the type of lens.
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F: P = 1/f
I: −2.5 = 1/f
F: f = 1 ÷ (−2.5)
A: f = −0.40 m; the negative power means it is a diverging lens
Q3 (Core) A converging lens in a magnifier has a focal length of 40 mm. Calculate its power.
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Convert first: 40 mm = 0.040 m
F: P = 1/f
I: P = 1 ÷ 0.040
F: P is already the subject
A: +25 D
Watch out: dividing by 40 gives 0.025, which is in mm⁻¹, not dioptres. Power must use f in metres.
Q4 (Core) A thin converging lens of power +5.0 D is placed in contact with a thin diverging lens of focal length 50 cm. Calculate the power and focal length of the combination, to 2 significant figures.
Show solution
Convert first: f = −50 cm = −0.50 m, so P2 = 1 ÷ (−0.50) = −2.0 D
F: P = P1 + P2
I: P = 5.0 + (−2.0) = +3.0 D
F: f = 1/P = 1 ÷ 3.0 = 0.333… m
A: +3.0 D; f = 0.33 m (2 s.f.), a converging combination
Q5 (Exam) A short-sighted person has a far point 2.0 m from the eye. Calculate the power of the lens needed so they can see very distant objects clearly. Ignore the distance between the lens and the eye.
Show solution
The lens must take an object at infinity and form a virtual image at the far point, so u = ∞ and v = −2.0 m
F: 1/f = 1/u + 1/v and P = 1/f
I: P = 1/∞ + 1/(−2.0) = 0 − 0.50
F: P is already the subject
A: −0.50 D, a diverging lens
Watch out: giving +0.50 D. Myopia is corrected with a diverging lens, so the power must be negative.
Q6 (Exam) A long-sighted person has a near point 1.0 m from the eye. Calculate the power of the lens that lets them read a book held 25 cm from the eye. Ignore the distance between the lens and the eye.
Show solution
The lens must form a virtual image at the near point of an object 25 cm away, so u = 0.25 m and v = −1.0 m
F: P = 1/f = 1/u + 1/v
I: P = 1/0.25 + 1/(−1.0) = 4.0 − 1.0
F: P is already the subject
A: +3.0 D, a converging lens
Q7 (Exam) In a simple model of the eye, the lens–retina distance is 1.70 cm. A normal eye has a near point of 25 cm and a far point at infinity. Calculate the power of the eye when it views (a) a very distant object and (b) an object at the near point, each to 2 significant figures. (c) Calculate the power of accommodation (the increase in power from (a) to (b)).
Show solution
Convert first: v = 1.70 cm = 0.0170 m
Step 1 – (a) distant object
F: P = 1/u + 1/v
I: P = 0 + 1/0.0170 = 58.82… D
A: 59 D (2 s.f.)
Step 2 – (b) object at the near point
I: P = 1/0.25 + 1/0.0170 = 4.0 + 58.82… = 62.82… D
A: 63 D (2 s.f.)
Step 3 – (c) power of accommodation
62.82… − 58.82… = 4.0 D, the extra power the ciliary muscles can add
Q8 (Exam) A short-sighted student has a far point of 50 cm and a near point of 15 cm. She wears a −2.0 D lens. Calculate her new near point while wearing it, to 2 significant figures. Ignore the distance between the lens and the eye.
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Check the lens first: a −2.0 D lens has f = −0.50 m, so it forms the image of a distant object at her far point, as it should.
New near point: the closest object whose virtual image lies at her near point, so v = −0.15 m
F: P = 1/u + 1/v
I: −2.0 = 1/u + 1/(−0.15)
F: 1/u = −2.0 + 6.66… = 4.66… m⁻¹, so u = 1 ÷ 4.66… = 0.2142… m
A: 21 cm (2 s.f.), so the diverging lens moves her near point further away
Q9 (Stretch) An older person has a far point of 4.0 m and a near point of 50 cm. They need bifocal lenses. Ignore the lens–eye distance. (a) Calculate the power of the upper (distance) part. (b) Calculate the power of the lower (reading) part needed to read at 25 cm. (c) Calculate the furthest distance at which they can see clearly through the reading part, to 2 significant figures.
Show solution
Step 1 – (a) distance part
u = ∞, v = −4.0 m: P = 0 + 1/(−4.0)
A: −0.25 D (diverging)
Step 2 – (b) reading part
u = 0.25 m, v = −0.50 m: P = 1/0.25 + 1/(−0.50) = 4.0 − 2.0
A: +2.0 D (converging)
Step 3 – (c) furthest clear object through the reading part
the image must be no further than the far point, so v = −4.0 m
I: 2.0 = 1/u + 1/(−4.0)
F: 1/u = 2.0 + 0.25 = 2.25 m⁻¹, so u = 1 ÷ 2.25 = 0.444… m
A: 0.44 m (44 cm, 2 s.f.), which is why reading glasses blur the far side of a room
Q10 (Stretch) A person’s eye has a lens–retina distance of 1.70 cm. Its power can be varied from 59.3 D (relaxed) to 61.8 D (fully accommodated). Treat any spectacle lens as a thin lens in contact with the eye. (a) Find the far point and near point, to 2 significant figures. (b) Find the power of the lens that corrects the far point to infinity, to 2 significant figures. (c) Show that the near point while wearing this lens is 0.40 m.
Show solution
Step 1 – (a) far and near points
power needed to focus on the retina: P = 1/u + 1/0.0170, so 1/u = P − 58.82… D
relaxed: 1/u = 59.3 − 58.82… = 0.4764… m⁻¹, so far point u = 2.098… m = 2.1 m (2 s.f.)
accommodated: 1/u = 61.8 − 58.82… = 2.976… m⁻¹, so near point u = 0.3359… m = 0.34 m (2 s.f.)
Step 2 – (b) correcting lens
to focus parallel light, relaxed eye + lens must total 58.82… D
F: P = P1 + P2
I: 58.82… = 59.3 + Plens, so Plens = −0.4764… D
A: −0.48 D (2 s.f.), diverging, so the eye is myopic
Step 3 – (c) new near point
maximum total power = 61.8 − 0.4764… = 61.32… D
1/u = 61.32… − 58.82… = 2.5 m⁻¹, so u = 1 ÷ 2.5 = 0.40 m
Watch out: the power of accommodation (61.8 − 59.3 = 2.5 D) does not change when you add a lens. The lens shifts both the near and far points.
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