Describe the image formed by a plane mirror
Explain same size, equal perpendicular distance and a virtual image without putting light behind the mirror.
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
- Use the reflection law.
- Distinguish a travelling ray from a construction line.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Where do backward extensions of reflected rays meet for a plane mirror?
What you will learn
- State the required plane-mirror image characteristics.
- Explain virtual image location using backward ray extensions.
Locate an apparent source
Light from an object reflects at a plane mirror and can enter an observer's eye. The reflected rays reaching the eye diverge, but their backward extensions meet at an apparent point behind the mirror. The observer attributes the light to that point. The virtual image location is determined by these extensions, not by light actually travelling through an opaque mirror.
For a plane mirror, each image point is the same perpendicular distance behind the surface as its object point is in front. The image is the same size and upright. The usual left–right appearance follows the mirror reversal relative to the observer; the geometric construction reflects the coordinate perpendicular to the mirror, rather than swapping every coordinate in space.
Separate an image from a screen projection
A virtual image cannot be projected directly onto a screen placed at its apparent location, because the reflected rays do not physically converge there. It can still be seen and photographed by an eye or camera receiving those rays. 'Virtual' therefore does not mean invisible or imaginary light.
If an object is 0.40 m in front of the mirror, its image lies 0.40 m behind it and the object–image separation is 0.80 m along the perpendicular. Measure distances from the reflecting surface, not from the observer. The image's size does not depend on where an observer stands within a position from which the rays can be received.
Pause and explain
An object is 0.30 m in front of a plane mirror. Where is its image?
Worked example
A 6 cm tall object stands 40 cm in front of a plane mirror. Describe its image.
Show the worked solution
- Equal perpendicular distance places it 40 cm behind the surface.
- A plane mirror preserves the object's 6 cm image height.
- Backward extensions, not real convergence, give an upright virtual image.
Answer 6 cm tall, upright and virtual, 40 cm behind the mirror.
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 of a 4 cm object 25 cm in front of a plane mirror.
Give me a hint
List size, orientation, distance and real/virtual status separately.
Compare my reasoning
- Height remains 4 cm and orientation is upright.
- The equal distance puts the image 25 cm behind.
- Its location comes from backward ray extensions, so it is virtual.
4 cm, upright, virtual, 25 cm behind.
Look for these in your work
- Include all characteristics.
- Use equal perpendicular distances.
An object is 0.75 m from a plane mirror. Find the object–image separation.
Give me a hint
There is one equal distance on each side of the surface.
Compare my reasoning
- The image is 0.75 m behind the mirror.
- The object is 0.75 m in front.
- The total perpendicular separation is 1.50 m.
1.50 m.
Look for these in your work
- Add the two distances.
- Do not call this the image distance.
A student can see a mirror image but gets no direct screen projection at its apparent position. Explain.
Give me a hint
Seeing an image and physical ray convergence are different requirements.
Compare my reasoning
- The eye receives reflected rays.
- Their extensions appear to originate behind the mirror.
- The rays do not really meet there, so no direct image is formed on that screen.
A visible virtual image has apparent, not real, ray convergence.
Look for these in your work
- Explain visibility.
- Explain why direct projection fails.
Common mistakes
- Measure perpendicular distances from the surface.
- Virtual light paths are dashed extensions, not transmitted rays.
- Virtual images can be seen but not directly projected at their apparent location.
What can you explain now?
Is a virtual plane-mirror image twice the object's distance behind the mirror?
Compare with the explanation
No.
Image distance equals object distance; their separation is twice that distance.
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Make it stick
State plane-mirror image properties and explain virtual location using two backward extensions.
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.