Construct a real image with a converging lens
Use a parallel-to-focus ray and an optical-centre ray to locate an actual crossing.
Section 3.2.3 · Thin lenses
Exam scope and assessment notes
Core and Extended · 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
- Locate the principal axis and both focal points.
- Use the ideal optical-centre ray.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens after the lens to a ray initially parallel to the axis?
What you will learn
- Draw a real-image ray construction.
- Explain where an appropriate screen receives the image.
Start two rays at the same object point
Place an upright object beyond the incident-side focal point of a converging lens. From its top draw one ray parallel to the principal axis; after the lens it passes through the far-side focus. Draw a second ray from the same top through the optical centre, continuing straight in the ideal thin-lens model.
The actual outgoing rays intersect on the far side. Their intersection locates the image of the object's top. The object's base on the axis images onto the axis, so joining the points gives an inverted real image. Arrows on the outgoing paths show physical convergence rather than dashed apparent origin.
Check with another ray and the scale
A third check ray can approach the lens through the near-side focus and then leave parallel to the axis. It should meet the same image point. Use consistent focal distances on both sides and begin all top-point rays at the identical object top; mixing different source points creates a false crossing.
For an object at 2F, the ideal converging construction gives a same-size inverted image at 2F on the opposite side. A screen at that crossing can receive a sharp real image. Moving the screen to another position produces a blur; the existence of a screen does not move the geometrically determined image point.
Pause and explain
An object is at 2F before a converging lens. Where is its real image?
Worked example
A 4 cm object is placed at 2F before an ideal converging lens. Describe the constructed image.
Show the worked solution
- The parallel ray refracts through the far focus.
- The optical-centre ray continues straight.
- They meet at 2F on the other side below the axis.
Answer Real, inverted and 4 cm tall at 2F on the opposite side.
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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 two rays sufficient to locate a real image of an object beyond F.
Give me a hint
Begin at one object point and use two different principal-ray rules.
Compare my reasoning
- Draw a parallel ray refracting through the far focus.
- Draw an optical-centre ray continuing straight.
- Their actual intersection locates the image point.
Parallel-to-focus and straight centre rays meet at the real image.
Look for these in your work
- Use one source point.
- Identify actual convergence.
A 3 cm object is placed 30 cm from a converging lens with f = 15 cm. Describe its image.
Give me a hint
The object is exactly at twice the focal length.
Compare my reasoning
- u = 2f = 30 cm.
- The ray construction places the image 30 cm on the other side.
- Its height is 3 cm, inverted and real.
30 cm on the far side; 3 cm, inverted, real.
Look for these in your work
- Recognise the 2F case.
- Include size and status.
A screen is placed before the constructed crossing and shows a blur. What should be changed?
Give me a hint
The screen must intercept the rays where each point's rays converge.
Compare my reasoning
- The lens and object already define the image location.
- A misplaced screen intercepts spreading or converging bundles before they meet.
- Move the screen to the constructed image plane and adjust for sharp focus.
Move the screen to the real image plane; the screen does not set that plane.
Look for these in your work
- Explain the blur.
- Match screen position to ray geometry.
Common mistakes
- Start both rays at the same source point.
- Use the far-side focus for the parallel ray.
- A screen must be at the actual image plane for sharp focus.
What can you explain now?
Are dashed backward extensions needed to locate this real image?
Compare with the explanation
No.
The actual outgoing rays cross at the real image.
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Make it stick
Draw a two-ray real image, check with a third ray and explain its screen location.
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.