Free Thin lenses questions for IGCSE Physics (0625), part of Waves. Each question has the answer and the marking points an examiner would look for.
Generate a full Thin Lenses worksheet →All Waves questionsAn object is placed closer to a thin converging lens than its principal focus.
Which row describes the image?
| nature | orientation | size compared with object | |
|---|---|---|---|
| A | virtual | upright | larger |
| B | virtual | upright | smaller |
| C | real | inverted | larger |
| D | virtual | inverted | larger |
Answer: A
Where must an object be placed for a converging lens to act as a magnifying glass?
Answer: C
An object is placed 30 cm from a converging lens of focal length 10 cm.
On a ray diagram, describe the two rays you would draw to locate the image.
State three properties of the image.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | parallel ray through focus; ray through centre undeviated a ray parallel to the axis refracts through the principal focus B1 a ray through the centre of the lens goes straight on B1 | 2 |
| (b) | real, inverted, diminished real B1 inverted B1 diminished (smaller than the object) B1 | 3 |
An object is placed 36 cm from a thin converging lens of focal length 12 cm.
State what is meant by the focal length of a converging lens.
On the grid, draw a ray diagram to locate the image.
Describe the image formed.
Use your diagram to find the magnification of the image.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | distance from lens to principal focus, where parallel rays meet distance from the centre of the lens to the principal focus B1 where rays parallel to the principal axis are brought to a focus B1 | 2 |
| (b) | two correct rays; image at intersection ray parallel to axis refracted through F (allow ray through F emerging parallel) B1 ray through the centre of the lens undeviated B1 inverted image arrow drawn from the axis to where the rays cross, 18.0 ± 1 cm from the lens B1 | 3 |
| (c) | real, inverted, diminished real B1 inverted B1 diminished (smaller) B1 | 3 |
| (d) | 0.500 magnification = image height ÷ object height (or image distance ÷ object distance = 18.0/36) C1 0.500 (allow ± 0.1, ft their diagram) A1 | 2 |
A student finds the focal length of a converging lens.
Describe a simple method.
Define the principal focus.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | focus distant object onto screen; measure lens–screen distance focus a sharp image of a distant object (window) on a screen B1 measure the distance from the lens to the screen B1 this is the focal length B1 | 3 |
| (b) | point where parallel rays converge the point where rays parallel to the principal axis converge after passing through the lens B1 | 1 |
A converging lens forms an image of a distant tree on a screen.
Define the focal length of a converging lens.
State whether the image is real or virtual, and whether it is upright or inverted.
Describe how a converging lens can be used as a magnifying glass.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | distance from lens to principal focus the distance from the centre of the lens to the principal focus B1 | 1 |
| (b) | real, inverted real B1 inverted B1 | 2 |
| (c) | object closer than focal length gives virtual magnified image place the object closer to the lens than the principal focus B1 a virtual, upright, magnified image is formed B1 | 2 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in