Compare real lens-image size and position
Read the object position relative to F and 2F before naming image size, orientation and status.
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 real image with two rays.
- Identify F and 2F on both sides.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which feature makes an image real?
What you will learn
- Describe images as enlarged, same size or diminished.
- Use upright/inverted and real/virtual as separate characteristics.
Read three real-image cases
For an ideal converging lens, an object beyond 2F forms a diminished inverted real image between F and 2F on the far side. At 2F the image is inverted, real and the same size at the other 2F. Between F and 2F the image is enlarged, inverted and real beyond the far-side 2F. These cases follow the principal-ray construction.
Use three independent descriptors: size compares image and object height; orientation compares their arrow directions; real or virtual distinguishes actual crossing from apparent origin. An enlarged image is not automatically upright or virtual. The between-F-and-2F case is an enlarged real inverted image.
Handle the limiting placement carefully
As an object approaches F from beyond it, the real image becomes more distant and enlarged in the ideal model. Exactly at F the principal outgoing rays are parallel, so they do not meet at a finite screen position. This is a limiting construction, not a claim that a finite image forms at the lens itself.
At a diagram scale, compare measured image and object heights without mixing centimetres on paper with physical lengths. A 6 cm object and a 3 cm image imply a size ratio 0.5. If the image points below the axis while the object points above, its orientation is inverted; the height magnitude remains positive when stating its size.
Sources for this lesson
Pause and explain
Where does an object beyond 2F form its ideal real image?
Worked example
A constructed image is 3 cm tall for a 6 cm object and the actual rays cross below the axis. Describe it.
Show the worked solution
- Size ratio = 3/6 = 0.5, so it is diminished.
- Opposite arrow direction means inverted.
- Actual ray crossing means real.
Answer Diminished to half size, inverted and real.
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.
Describe the image for an object beyond 2F.
Give me a hint
Use the far-side F-to-2F range and list three properties.
Compare my reasoning
- The image lies between F and 2F on the opposite side.
- It is smaller and inverted.
- Actual rays meet, so it is real.
Diminished, inverted, real, between far-side F and 2F.
Look for these in your work
- Give the position range.
- State all three characteristics.
Describe the image for an object between F and 2F.
Give me a hint
Follow the actual ray intersection beyond the far-side 2F.
Compare my reasoning
- The outgoing rays meet beyond 2F.
- The image is enlarged and inverted.
- The physical crossing makes it real.
Enlarged, inverted, real, beyond far-side 2F.
Look for these in your work
- Avoid assuming enlarged means virtual.
- Give the range and orientation.
An object is placed exactly at F. Explain why moving a screen a little farther away cannot find a finite sharp image in the ideal model.
Give me a hint
The outgoing principal rays no longer converge at a finite distance.
Compare my reasoning
- At F the outgoing rays from an object point are parallel.
- They do not cross at a finite image plane.
- A finite screen cannot capture a sharp point image from that ideal bundle.
Parallel outgoing rays give no finite real image plane at the exact focal placement.
Look for these in your work
- State the limiting geometry.
- Do not invent a finite crossing.
Common mistakes
- Size, orientation and real/virtual are separate properties.
- Enlarged does not always mean virtual.
- An object at F has no finite real screen image in the ideal model.
What can you explain now?
For an object between F and 2F, is its enlarged image upright?
Compare with the explanation
No.
Its actual outgoing rays give an inverted real image beyond 2F.
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Make it stick
Compare beyond 2F, at 2F and between F and 2F, then describe the exact-F limit.
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.