Construct a virtual image with a converging lens
Place the object inside the focal length and extrapolate the diverging outgoing rays backward.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Follow the selected Light section; ray diagrams use an ideal thin-lens or plane-boundary model. Written work is self-checked. Human teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Draw parallel-to-focus and optical-centre rays.
- Explain virtual image formation from extensions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which placement gives the stated virtual converging-lens image?
What you will learn
- Draw a converging-lens virtual-image construction.
- Describe its object-side location, enlarged size and upright orientation.
Keep the object inside F
A converging lens with an object between its optical centre and near-side focus sends the relevant outgoing rays in diverging directions. From the object top, draw a parallel ray that refracts toward the far-side focus and a centre ray that continues straight. The two actual output rays do not meet on the far side.
Extend those outgoing directions backward with dashed lines. Their intersection is on the object side, farther from the lens than the object. The top image point remains above the axis for an upright object and is farther from the axis, giving an enlarged upright virtual image in this ideal construction.
Read a consistent scaled diagram
In the supplied diagram f is 15 cm and the object is 10 cm from the lens. The ray construction places the virtual image 30 cm from the lens on the object side. A 5 cm object becomes a 15 cm apparent image, a height ratio of 3. These are supplied original diagram values; construction, rather than a memorised position for every magnifier, determines the image.
A screen cannot directly project that apparent image at its dashed crossing. An observer on the output side receives the diverging actual rays and sees their apparent origin. Moving the object outside F changes the construction to a real-image case; the word 'converging' alone does not require every object placement to produce a real image.
Pause and explain
What intersects at the inside-F virtual image?
Worked example
A supplied scaled diagram has a 5 cm object inside F and a 15 cm upright image where backward extensions cross. Describe size and status.
Show the worked solution
- Image/object height ratio = 15/5 = 3.
- Both arrows point above the axis, so orientation is upright.
- Only backward extensions intersect, making the image virtual.
Answer Enlarged threefold, upright and virtual, on the object side.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe the two-ray construction for a converging lens with an object inside F.
Give me a hint
Use the usual physical ray rules and extend the outgoing lines backward.
Compare my reasoning
- Parallel input refracts toward the far focus.
- The optical-centre ray continues straight.
- Backward extensions meet on the object side above the axis.
Upright enlarged virtual image from backward extensions.
Look for these in your work
- Use the inside-F condition.
- Separate actual rays and extensions.
A supplied diagram has object height 4 cm and upright virtual image height 10 cm. Find the size ratio and image characteristics.
Give me a hint
Compare positive height magnitudes before describing status.
Compare my reasoning
- Height ratio = 10/4 = 2.5.
- The image is larger and upright.
- Its extension crossing makes it virtual.
2.5 times the height; enlarged, upright and virtual.
Look for these in your work
- Compare matching heights.
- State orientation and status independently.
An object is moved from inside F to between F and 2F. Explain the change in image type.
Give me a hint
Reconstruct the actual outgoing rays after the placement changes.
Compare my reasoning
- Inside F, output diverges and gives an upright virtual image.
- Beyond F, the actual rays can converge on the far side.
- Between F and 2F, the image is enlarged, inverted and real beyond 2F.
Upright virtual → inverted real, with different-side image locations.
Look for these in your work
- Use the changed object placement.
- Explain physical convergence rather than just naming types.
Common mistakes
- Place the object inside F.
- Extend actual outgoing directions, not arbitrary lines.
- A converging lens can give a virtual image for this placement.
What can you explain now?
Is a converging-lens virtual image inverted in the inside-F case?
Compare with the explanation
No.
The backward construction gives an upright enlarged image.
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Make it stick
Draw the inside-F construction and explain all characteristics from the ray paths.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.