Use a single converging lens as a magnifying glass
Choose inside-F object placement and explain a visible enlarged view without direct screen projection.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Construct a converging-lens virtual image.
- Compare object position with focal length.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which lens is used for this single-lens magnifier?
What you will learn
- Describe magnifying-glass operation.
- Explain why the object must be inside the focal length for the intended upright view.
Use the upright virtual-image case
A simple magnifying glass uses one converging lens with the object between the lens and its principal focus. The observer receives transmitted diverging rays on the opposite side. Their backward extensions give an enlarged upright virtual image on the object side, allowing a small feature to be seen as a larger apparent image.
For a lens with f = 10 cm, an object 6 cm from the optical centre is inside F and supports this intended mode. An object 14 cm away is outside F, so the ideal lens instead forms a real inverted image for a suitable observer or screen arrangement. The lens type alone does not set the magnifier's upright operating case.
Separate apparent image size and viewing claims
A larger diagram image height describes geometric enlargement. The apparent angular size at an eye also depends on viewing distances and the observer's accommodation. The supplied syllabus task is to describe the single-lens use, not to invent an exact angular magnification or promise a particular medical or reading benefit from a focal-length label alone.
An eye or camera can see the virtual view by receiving the rays and making its own image. A screen placed at the virtual image point cannot directly project it. Keep the lens and object stable enough to identify the intended placement; glare, aberrations and finite lens size can affect a real viewing arrangement beyond the simple ray sketch.
Sources for this lesson
Pause and explain
For f = 12 cm, which object distance is inside F?
Worked example
A lens has f = 10 cm. Which object placement, 6 or 14 cm, supports an upright virtual magnifier view?
Show the worked solution
- Compare each object distance with f.
- 6 cm is inside F; 14 cm is outside.
- The inside-F placement gives the intended upright enlarged virtual image.
Answer Use 6 cm; the 14 cm placement is outside the intended virtual-image mode.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Explain the lens and object placement used by a magnifying glass.
Give me a hint
Name the lens action and compare object distance with f.
Compare my reasoning
- Use a single converging lens.
- Place the object inside its focal length.
- View the enlarged upright virtual image through the transmitted rays.
Converging lens, object inside F, upright enlarged virtual view.
Look for these in your work
- State the placement condition.
- Describe the image status.
For f = 15 cm compare object distances 9, 15 and 20 cm.
Give me a hint
Distinguish below, equal to and above the focal distance.
Compare my reasoning
- 9 cm gives the inside-F upright virtual case.
- 15 cm is exact F with no finite image crossing in the ideal model.
- 20 cm is outside F and gives an inverted real image.
9 cm virtual; 15 cm limiting parallel output; 20 cm real.
Look for these in your work
- Keep equality separate.
- Do not infer the same image for all placements.
A magnifier has an enlarged virtual image but no direct screen image at its apparent point. Is its operation inconsistent?
Give me a hint
Viewing and direct physical projection are different.
Compare my reasoning
- The transmitted light diverges toward the observer.
- Backward extensions locate the enlarged virtual view.
- No actual crossing at the apparent point is expected for direct projection.
No; visible virtual magnification and no direct screen projection are consistent.
Look for these in your work
- Explain the actual ray paths.
- Connect the screen result to virtual status.
Common mistakes
- Use a converging lens and inside-F placement.
- A visible virtual image is not a direct screen projection.
- Geometric image height alone does not establish an exact angular magnification.
What can you explain now?
Does putting the object farther than f preserve the usual upright virtual magnifier view?
Compare with the explanation
No.
It changes to the outside-F construction with an inverted real image.
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Make it stick
Explain magnifying-glass operation from lens type, inside-F placement and virtual ray construction.
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.