IGCSE Physics Thin Lenses Questions and Worked Solutions (3.2.3) | 0625

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IGCSE Physics 0625 · Topic 3.2.3

IGCSE Physics Thin Lenses Questions and Worked Solutions

50 topic-wise practice questions on Thin Lenses (Unit 3: Waves), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions26 Core14 Extended10 Challenge124 marksTextbook §3.2 · pp. 136–159
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Thin Lenses

Converging lenses bring parallel rays together at the principal focus; diverging lenses spread them out. Ray diagrams predict the position and type of image. Lenses are used in cameras, projectors, magnifying glasses and spectacles.

Key formulas

magnification = image height ÷ object heightno unitsCore
3 standard raysparallel → through F; through centre → straight; through F → parallelCore

Prerequisite knowledge

  • Refraction (3.2.2)
  • Accurate ray drawing with a ruler

Exam tips & common mistakes

  • Use at least two rays and a sharp pencil
  • Virtual rays are dashed
  • Learn the three image words in one phrase, e.g. 'real, inverted, diminished'

Key facts from the syllabus (3.2.3)

  • Principal axis, principal focus (focal point) and focal length must be defined
  • Object beyond 2F: real, inverted, diminished (camera, eye)
  • Object at 2F: real, inverted, same size
  • Object between F and 2F: real, inverted, enlarged (projector)
  • A virtual image is formed where diverging rays appear to come from; it cannot be shown on a screen
  • Describe images using: enlarged/same size/diminished, upright/inverted, real/virtual
  • Extended object inside F gives a virtual, upright, enlarged image (magnifying glass)
  • Extended short sight corrected with a diverging lens; long sight with a converging lens

Textbook: §3.2 Light, pp. 136–159

1
Core[2 marks]

Describe the effect of a thin converging lens on a parallel beam of light.

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Method: Rays meet at the focus.

The rays are refracted to meet (converge) at the principal focus
2
Core[2 marks]

Describe the effect of a thin diverging lens on a parallel beam of light.

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Method: Rays spread out.

The rays spread out as if they came from a principal focus on the same side as the incoming light
3
Core[1 mark]

Define the principal focus of a converging lens.

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Method: Point where parallel rays meet.

The point where rays parallel to the principal axis meet after passing through the lens
4
Core[1 mark]

Define the focal length.

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Method: Lens centre to focus.

The distance from the centre of the lens to the principal focus
5
Core[1 mark]

State what is meant by the principal axis.

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Method: Line through centre.

The line through the centre of the lens at right angles to it
6
Core[3 marks]

State the three standard rays used to draw a ray diagram for a converging lens.

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Method: Parallel, centre, through F.

A ray parallel to the axis refracts through F; a ray through the optical centre goes straight; a ray through F on the object side emerges parallel to the axis
7
Core[3 marks]

An object is placed at 3f from a converging lens. Describe the image.

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Method: Beyond 2f.

Real, inverted, diminished (between F and 2F on the other side)
8
Core[3 marks]

An object is placed at 2f from a converging lens. Describe the image.

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Method: At 2F.

Real, inverted, same size, at 2F on the other side
9
Core[3 marks]

An object is placed between f and 2f from a converging lens. Describe the image.

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Method: Beyond 2F.

Real, inverted, enlarged, beyond 2F on the other side
10
Core[2 marks]

Explain what is meant by a virtual image.

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Method: Rays appear to come from it.

An image formed where diverging rays appear to come from when extended backwards; it cannot be shown on a screen
11
Core[1 mark]

State one use of a converging lens forming a real, diminished image.

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Method: Camera.

Camera (or eye)
12
Core[1 mark]

State one use of a converging lens forming a real, enlarged image.

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Method: Projector.

Projector
13
Core[2 marks]

A ray diagram shows an object 4.0 cm tall and its image 12 cm tall. Calculate the magnification.

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Method: Magnification = image height ÷ object height.

m = 12 ÷ 4.0 = 3.00
14
Core[2 marks]

State the difference in shape between a converging and a diverging lens.

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Method: Thick middle vs thin middle.

Converging: thicker at the centre; diverging: thinner at the centre
15
Core[2 marks]

A converging lens focuses sunlight to a point 15 cm from the lens. State its focal length.

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Method: Sun's rays parallel.

15 cm
16
Core[2 marks]

Describe the image formed by a converging lens of an object very far away.

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Method: At F.

Real, inverted, very diminished, at the principal focus
17
Core[2 marks]

Explain why a converging lens can be used to start a fire with sunlight.

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Method: Concentrates energy.

It converges the Sun's rays to a small spot at the focus, concentrating energy
18
Core[3 marks]

An object 2.0 cm tall is placed 30 cm from a converging lens of focal length 10 cm. Describe the image.

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Method: Beyond 2F.

Real, inverted, diminished, between F and 2F on the other side
19
Core[3 marks]

An object is 15 cm from a converging lens of focal length 10 cm. Describe the image.

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Method: Between F and 2F.

Real, inverted, enlarged, beyond 2F
20
Core[2 marks]

State the meaning of a real image.

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Method: Rays meet.

An image formed where light rays actually meet; it can be projected on a screen
21
Core[2 marks]

State what is meant by an inverted image.

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Method: Upside down.

The image is upside down compared with the object
22
Core[2 marks]

An image is twice the height of the object. State whether it is enlarged, same size or diminished.

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Method: Ratio.

Enlarged
23
Core[2 marks]

Describe how to find the focal length of a converging lens quickly in the laboratory.

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Method: Distant object.

Focus a sharp image of a distant object (e.g. a window) onto a screen and measure the lens-to-screen distance
24
Core[2 marks]

Describe the path of a ray passing through the optical centre of a thin lens.

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Method: Undeviated.

It passes straight through without changing direction
25
Core[2 marks]

Describe the path of a ray passing through the principal focus before reaching a converging lens.

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Method: Emerges parallel.

It emerges parallel to the principal axis
26
Core[2 marks]

State whether a diverging lens can form a real image of a real object.

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Method: Always virtual.

No, it always forms a virtual image
27
Extended[3 marks]

An object is placed closer to a converging lens than the focal length. Describe the image.

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Method: Inside F.

Virtual, upright, enlarged, on the same side as the object
28
Extended[2 marks]

Describe how a single converging lens is used as a magnifying glass.

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Method: Object inside f.

The object is placed closer to the lens than the focal length; the eye sees a virtual, upright, enlarged image
29
Extended[2 marks]

State the type of lens used to correct short-sightedness and explain.

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Method: Diverging lens.

Diverging lens: it spreads the rays so the eye focuses them on the retina instead of in front of it
30
Extended[2 marks]

State the type of lens used to correct long-sightedness and explain.

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Method: Converging lens.

Converging lens: it converges the rays a little before the eye so they focus on the retina instead of behind it
31
Extended[2 marks]

A short-sighted person cannot focus on distant objects. State where the image forms without correction.

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Method: In front of the retina.

In front of the retina
32
Extended[3 marks]

A converging lens has focal length 10 cm. An object is 30 cm away. Using a ray diagram (or calculation), the image is 15 cm from the lens. Calculate the magnification and describe the image.

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Method: m = image distance ÷ object distance.

m = 15 ÷ 30 = 0.500; real, inverted, diminished
33
Extended[3 marks]

A converging lens of focal length 5.0 cm is used as a magnifying glass. State the range of object distances for which it acts as a magnifier.

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Method: Object must be inside f.

Between 0 and 5.0 cm from the lens
34
Extended[2 marks]

Describe the image formed by a magnifying glass.

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Method: Virtual, upright, enlarged.

Virtual, upright, enlarged, on the same side as the object
35
Extended[3 marks]

Explain how to draw a ray diagram for an object inside the focal length of a converging lens.

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Method: Extend rays back.

Draw a ray parallel to the axis refracting through F and a ray through the centre; they diverge, so extend them backwards with dashed lines; they meet at the virtual image
36
Extended[2 marks]

A long-sighted person's eye focuses light behind the retina. Explain how a converging lens helps.

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Method: Adds convergence.

It converges the light before it enters the eye so it focuses on the retina
37
Extended[2 marks]

Explain why a short-sighted person's eye cannot see distant objects clearly.

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Method: Focus in front.

The eye converges light too much, focusing it in front of the retina
38
Extended[2 marks]

State the type of lens used in spectacles for someone who cannot read small print close up.

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Method: Long sight.

Converging lens
39
Extended[3 marks]

A converging lens of focal length 8.0 cm is used as a magnifier. An object is placed 5.0 cm from it. State whether the image is real or virtual and explain.

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Method: Inside f.

Virtual, because the object is closer than the focal length so the refracted rays diverge
40
Extended[2 marks]

Explain why a magnifying glass must be held close to the object.

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Method: Object inside f.

The object must be within the focal length to form a virtual, enlarged image

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

Draw (describe) a ray diagram for an object 4.0 cm tall placed 24 cm from a converging lens of focal length 8.0 cm. State the position and nature of the image.

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Method: Object beyond 2F.

Image about 12 cm from the lens on the other side; real, inverted, diminished (about 2 cm tall)
42
ChallengeCore[4 marks]

An object is placed 12 cm from a converging lens of focal length 8.0 cm. Describe the image and estimate its distance using a ray diagram.

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Method: Between F and 2F.

Real, inverted, enlarged, about 24 cm from the lens on the other side
43
ChallengeCore[3 marks]

A projector lens has focal length 10 cm. Explain where the slide must be placed and why the slide is put in upside down.

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Method: Between F and 2F.

Between 10 cm and 20 cm from the lens to give an enlarged real image; the image is inverted, so the slide is inserted upside down
44
ChallengeCore[4 marks]

Explain how a camera focuses on objects at different distances using a converging lens.

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Method: Move lens.

For nearer objects the image forms further from the lens, so the lens is moved away from the sensor to focus
45
ChallengeCore[3 marks]

Explain why an image formed by a diverging lens cannot be shown on a screen.

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Method: Virtual.

Rays diverge after the lens and never meet; they only appear to come from a point (virtual image)
46
ChallengeExtended[4 marks]

An object is 4.0 cm from a converging lens of focal length 6.0 cm. Describe the image and how a ray diagram is used to locate it.

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Method: Inside f.

Virtual, upright, enlarged, on the same side as the object (about 12 cm from the lens); the refracted rays diverge so are extended backwards to meet
47
ChallengeExtended[4 marks]

Explain why a magnifying glass must have a short focal length and be held close to the object.

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Method: Object within f, large magnification.

The object must be within the focal length; a short focal length allows a large, virtual, upright image at a convenient distance
48
ChallengeExtended[4 marks]

A short-sighted person's far point is 2.0 m away. Explain the defect and the type of lens needed to correct it.

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Method: Diverging lens.

Light from distant objects focuses in front of the retina because the eye converges too strongly; a diverging lens spreads rays so they focus on the retina
49
ChallengeExtended[4 marks]

A long-sighted person cannot focus closer than 1.0 m. Explain the defect and how a converging lens corrects it.

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Method: Converging lens.

Light from near objects would focus behind the retina; a converging lens adds convergence so it focuses on the retina
50
ChallengeExtended[4 marks]

Compare the images formed by a converging lens of an object at 3f and at 0.5f.

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Method: Real vs virtual.

At 3f: real, inverted, diminished, between F and 2F on the far side. At 0.5f: virtual, upright, enlarged, on the same side
All IGCSE Physics topics← 3.2.2 Refraction of Light3.2.4 Dispersion of Light →

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