Distinguish virtual lens images from screen projections
Recognise an apparent image from diverging output and explain why it can be seen without direct projection.
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
- Distinguish physical rays from dashed extensions.
- Know actual convergence defines a real image.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which paths locate a virtual image?
What you will learn
- Explain virtual image formation by backward extrapolation.
- State why a virtual image does not directly project onto a screen.
Trace diverging light to an apparent origin
When outgoing rays from one object point diverge, extend those rays backward with dashed lines. If the extensions meet, that point is the virtual image of the source point. The transmitted rays physically move away from the lens; they are perceived as originating from the extension crossing rather than meeting there.
The same principle explains the apparent principal focus of a diverging lens and the image construction of a magnifying glass. The real/virtual distinction uses actual ray behaviour rather than whether the observer finds the view large, bright or convincing. A virtual image can be upright and visible without a physical crossing at its apparent point.
Separate direct projection and further optics
A screen at the apparent image location receives no converging bundle from the outgoing rays, so it cannot directly show that virtual image as a focused projection. An eye or camera can still receive the diverging rays and use its own lens to form a real image on a retina or sensor. That later real image is distinct from the virtual image made by the first optical system.
A diagram must show solid actual rays and dashed backward extensions clearly. A crossing of dashed lines alone is not a real light crossing. For a near object viewed through a converging magnifier, the apparent image is on the object side; do not put a physical screen behind that dashed crossing and call it a directly projected lens image.
Pause and explain
Why is direct projection at a virtual image point unavailable?
Worked example
A diagram shows diverging actual rays and intersecting dashed backward extensions. Can a screen directly capture the image at that dashed intersection?
Show the worked solution
- The dashed lines are extrapolations, not physical rays.
- Actual rays do not converge at their apparent meeting point.
- The virtual image can be seen by receiving the output, but it cannot directly project there.
Answer Visible virtual image; no direct screen projection at the extension crossing.
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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.
Explain how diverging output rays locate a virtual image.
Give me a hint
Keep the physical and extrapolated paths separate.
Compare my reasoning
- Follow the actual outgoing rays forward.
- Extend them backward with dashed lines.
- Their apparent common origin is the virtual image point.
Backward-extension intersection locates the apparent image.
Look for these in your work
- Identify actual divergence.
- Label construction extensions.
Compare screen formation at a real-image point and at a virtual-image point.
Give me a hint
Ask whether actual light converges at each location.
Compare my reasoning
- At a real point the actual rays meet.
- A screen there can receive a focused direct image.
- At a virtual point only extensions meet, so direct projection fails.
Real: direct focused projection; virtual: no direct projection at its apparent point.
Look for these in your work
- Use convergence as the criterion.
- Distinguish direct projection from observation.
A camera records a virtual magnifier image. Explain without claiming physical convergence at the magnifier's apparent point.
Give me a hint
The camera contributes its own optical system.
Compare my reasoning
- The magnifier sends diverging rays to the camera.
- Their extensions define the magnifier's virtual image.
- The camera lens forms a later real image on its sensor.
The camera makes a separate real sensor image from the received diverging light.
Look for these in your work
- Keep the two optical systems distinct.
- Retain the first image's virtual status.
Common mistakes
- Dashed crossings are apparent, not physical.
- Virtual images can be seen and photographed through further optics.
- Brightness and size do not determine real/virtual status.
What can you explain now?
Does a camera photograph make the original virtual image a real image of that first lens?
Compare with the explanation
No.
The camera forms its own later real sensor image from received rays.
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Make it stick
Explain virtual construction, visibility and failure of direct screen projection using actual ray directions.
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.