Construct plane-mirror images from reflected rays
Trace two actual outgoing rays and extend them backward to a consistent image point.
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
- Use i = r.
- Know equal perpendicular object/image distances.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How should a virtual ray extension be distinguished on a diagram?
What you will learn
- Explain how a plane-mirror image forms from reflected rays.
- Distinguish construction extensions from actual light paths.
Use two rays to determine a point
Mark a point object O in front of a straight mirror. Choose two different contact points on the reflecting surface. At each contact draw a local normal and construct a reflected ray with i = r. Put arrows on the physical incident and reflected paths. Extend the two reflected rays backward behind the mirror with dashed lines.
The extensions intersect at image point I. A correct plane-mirror construction agrees with reflecting O perpendicularly across the mirror line: equal distance on the opposite side. Two distinct rays determine the apparent point; an arbitrary single extension on its own leaves many possible positions along that line.
Check an image of an extended object
Repeat the point construction for the top and bottom of an object. Their image points form an equally tall upright image. A horizontal mirror, sloping mirror or rotated page uses the same perpendicular-distance rule; the reference is the mirror surface rather than left–right screen coordinates.
A ray reaching an eye must physically leave a point on the actual mirror. A backward extension can locate a virtual image but does not prove that an observer can see it through a mirror of insufficient size. Check actual ray contact positions against the mirror ends when a finite mirror is drawn.
Sources for this lesson
Pause and explain
Across a vertical mirror x = 0, where is the image of (−4, 1) cm?
Worked example
A vertical mirror is x = 0. An object point is at (−3, 2) cm. Locate its plane-mirror image and explain the construction.
Show the worked solution
- Reflect the perpendicular x coordinate across the mirror.
- Keep the coordinate parallel to the mirror unchanged.
- Two backward reflected-ray extensions meet at (3, 2) cm.
Answer Virtual image point (3, 2) cm; actual light stays on the object side.
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.
Locate the image of (−2, 3) cm in a vertical mirror x = 0.
Give me a hint
Reverse the coordinate perpendicular to this mirror.
Compare my reasoning
- The perpendicular coordinate is x.
- Replace −2 by +2 and retain y = 3.
- The virtual image is at (2, 3) cm.
(2, 3) cm.
Look for these in your work
- Reflect the correct coordinate.
- Retain the parallel position.
For a horizontal mirror y = 0, locate the image of (3, 2) cm.
Give me a hint
Now y, rather than x, measures perpendicular position.
Compare my reasoning
- The mirror is horizontal.
- Reflect y = 2 to y = −2.
- Retain x = 3.
(3, −2) cm.
Look for these in your work
- Identify the surface orientation.
- Use equal perpendicular distances.
A correct dashed extension is drawn, but its physical reflection point lies beyond a short mirror's end. Can that drawn ray reach the observer?
Give me a hint
A construction cannot supply reflection where no mirror exists.
Compare my reasoning
- Identify the required physical contact point.
- It lies outside the stated reflecting surface.
- That particular light path is unavailable, even if its virtual extension is mathematically drawn.
No; the physical ray requires a reflection point on the actual mirror.
Look for these in your work
- Separate image location from visibility.
- Check the apparatus boundary.
Common mistakes
- Use at least two distinct rays to locate a point.
- Draw virtual extensions dashed.
- Check finite mirror contact points for visibility.
What can you explain now?
Does one backward line alone determine a unique image point?
Compare with the explanation
No.
Two different constructed rays or an additional distance rule locate the point.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain a two-ray image construction and test its location and physical mirror contacts.
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.