PhysicsICSE Class 9 · Chapter 8

Reflection of Light by Plane Mirrors ICSE Class 9: Laws, Images and Parallel Mirrors

Updated 15 October 2026 · Fact-checked

When light is reflected, the angle of incidence equals the angle of reflection, and the incident ray, normal and reflected ray lie in one plane. A plane mirror forms a virtual, upright, same-size, laterally inverted image as far behind it as the object is in front. ICSE Class 9 adds verifying the laws, uses and parallel-mirror images.

What Reflection of Light by Plane Mirrors is about

Reflection is the bouncing back of light from a surface into the same medium. A plane mirror is a flat, polished reflecting surface.

Terms: the incident ray strikes the mirror at the point of incidence; the normal is the line perpendicular to the mirror at that point; the reflected ray leaves the mirror. The angle of incidence $i$ and angle of reflection $r$ are measured from the normal, not from the mirror.

The CISCE 2028 Class IX syllabus covers: the laws of reflection and their experimental verification; the characteristics of images formed by plane mirrors; uses of plane mirrors; and the formation of multiple images when two plane mirrors are placed parallel to each other. (Images in perpendicular mirrors are not in the 2028 syllabus.)

Expect ray diagrams for the image of a point or an object, 2-mark angle questions, 3-mark descriptions of the verification experiment, and questions on positions of images in parallel mirrors.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Reflection of Light by Plane Mirrors is in the 6. Light (i) part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

Laws of reflection

First law: the angle of incidence is equal to the angle of reflection: $∠ i = ∠ r$ .

Second law: the incident ray, the normal at the point of incidence and the reflected ray all lie in the same plane.

Consequences:

a ray striking the mirror along the normal ( $i = 0°$ ) is reflected straight back along its path;
the angle between the incident and reflected rays is $2 i$ ;
if a ray makes angle $θ$ with the mirror surface, then $i = 90° - θ$ .

Worked example

A ray strikes a plane mirror making an angle of 55° with the mirror surface. Find the angle of incidence, the angle of reflection and the angle between the incident and reflected rays.

Try it first, then show the solutionSolution

$i = 90° - 55° = 35°$ .
$r = i = 35°$ .
Angle between the rays $= 35° + 35° = 70°$ .

Answer: 35°, 35°, 70°

Experimental verification (pin method)

1. Fix a sheet of paper on a drawing board and draw a line $M M^{'}$ for the mirror. Draw a normal $O N$ at point $O$ and a line $P O$ making an angle (say 30°) with $O N$ .
2. Stand the plane mirror vertically on $M M^{'}$ .
3. Fix two pins $P_{1}$ and $P_{2}$ upright on line $P O$ .
4. Looking into the mirror from the other side of the normal, fix pins $P_{3}$ and $P_{4}$ so that they appear in a straight line with the images of $P_{1}$ and $P_{2}$ .
5. Remove the mirror and pins, join $P_{3} P_{4}$ to meet $O$ , and measure the angle it makes with $O N$ .

Result: angle of reflection equals angle of incidence, for several angles. Since all lines are drawn on the flat sheet, the incident ray, normal and reflected ray lie in one plane.

Worked example

In a pin experiment the angles of incidence used are 20°, 35° and 50°, and the angles of reflection measured are 20°, 36° and 50°. What do the results show, and why is one value slightly off?

Try it first, then show the solutionSolution

Within experimental error, $∠ r = ∠ i$ in each case.
The 1° difference comes from errors in placing the pins or reading the protractor.

Answer: They verify the first law; small differences are experimental errors.

Image formed by a plane mirror

To locate the image of a point object $O$ : draw two rays from $O$ to the mirror, reflect each with $i = r$ , and extend the reflected rays backwards behind the mirror; they meet at the image $I$ .

Characteristics of the image:

virtual (cannot be caught on a screen; rays only appear to come from it);
erect (upright);
same size as the object;
as far behind the mirror as the object is in front ( $O M = M I$ , and the line $O I$ is perpendicular to the mirror);
laterally inverted (left and right interchanged).

Worked example

A girl stands 1.5 m in front of a plane mirror. How far is her image from her? If she walks 0.5 m towards the mirror, what is the new distance between her and her image?

Try it first, then show the solutionSolution

Image is 1.5 m behind the mirror: distance from her $= 3$ m.
New object distance $= 1.0$ m; image 1.0 m behind; distance $= 2$ m.

Answer: 3 m; 2 m

Lateral inversion

In a plane mirror, the left side of the object appears as the right side of the image and vice versa. This is lateral inversion: your right hand appears as the image's left hand, and printed words appear back to front.

That is why AMBULANCE is painted in mirror writing (reversed letters) on the front of ambulances: drivers ahead see it the right way round in their rear-view mirrors.

Letters with a vertical line of symmetry (A, H, I, M, O, T, U, V, W, X, Y) look the same after lateral inversion.

Worked example

Which of these words would read the same in a plane mirror: TOMATO, MOM, WOW, HOT?

Try it first, then show the solutionSolution

All their letters are symmetric about a vertical line, but the order of letters is also reversed.
MOM and WOW read the same backwards; TOMATO becomes OTAMOT; HOT becomes TOH.

Answer: MOM and WOW

Uses of plane mirrors

Looking mirrors (dressing tables), because the image is upright and the same size.
Periscopes: two plane mirrors, parallel to each other and at 45° to the tube, let you see over obstacles (submarines, crowds).
Kaleidoscopes: mirrors arranged to produce many images of coloured pieces.
Behind the scale of instruments to avoid parallax: you read the pointer when it hides its own image.
In barber shops and lifts, to give multiple images or a feeling of space.
To reflect sunlight into solar cookers and to signal over distances.

Worked example

A periscope has two plane mirrors set at 45° to its vertical tube. A horizontal ray enters the top. Trace its path.

Try it first, then show the solutionSolution

At the top mirror, $i = 45°$ , so the ray turns through 90° and travels down the tube.
At the bottom mirror, $i = 45°$ again, so the ray turns through 90° and leaves horizontally into the eye.

Answer: Turned down by 90°, then turned horizontal again by 90°.

Multiple images in two parallel mirrors

When an object is placed between two plane mirrors facing each other and parallel, light reflects back and forth between them. Each image acts as an object for the other mirror, so an infinite number of images is formed, all in a straight line perpendicular to the mirrors.

The images become fainter further away, because some light is absorbed at every reflection, so only a limited number are seen.

Positions: if the mirrors are $d$ apart and the object is $x$ from mirror $M_{1}$ , the images behind $M_{1}$ are at distances $x$ , $2 d - x$ , $2 d + x$ , $4 d - x$ , … from $M_{1}$ .

Worked example

Two parallel plane mirrors are 20 cm apart. An object is 5 cm from mirror $M_{1}$ . Find the distances of the first three images behind $M_{1}$ .

Try it first, then show the solutionSolution

First image in $M_{1}$ : 5 cm behind it.
Image in $M_{2}$ is 15 cm behind $M_{2}$ , i.e. 35 cm from $M_{1}$ ; its image in $M_{1}$ is 35 cm behind $M_{1}$ ( $2 d - x$ ).
The image of the first $M_{1}$ image in $M_{2}$ is 25 cm behind $M_{2}$ (45 cm from $M_{1}$ ); its image in $M_{1}$ is 45 cm behind ( $2 d + x$ ).

Answer: 5 cm, 35 cm, 45 cm

Physics in real life: Reflection of Light by Plane Mirrors

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Image distance

Suppose a barber's chair is 1.2 m from a mirror on the wall, and the customer sits in it.

How far does the customer see his image from himself?

Try it first, then show the solutionSolution

Image is 1.2 m behind the mirror.
Distance from customer to image $= 1.2 + 1.2 = 2.4$ m.

Answer: 2.4 m.

Memory hook: Image distance doubles the gap to the mirror.

Lateral inversion

Imagine a driver looks in the rear-view mirror at an ambulance behind with 'AMBULANCE' painted in mirror writing on its bonnet.

How does the word appear to the driver, and why?

Try it first, then show the solutionSolution

The plane mirror laterally inverts the reversed writing.
Reversing it twice gives the normal word.

Answer: It reads AMBULANCE normally.

Memory hook: Two reversals make a right.

Parallel mirrors

Suppose a lift has mirrors on two opposite walls, parallel to each other.

Why do you see a long row of your images, and why are the far ones dim?

Try it first, then show the solutionSolution

Each image in one mirror acts as an object for the other, so images form again and again.
Some light is absorbed at each reflection, so distant images (more reflections) are fainter.

Answer: Repeated reflections; light lost at each one.

Memory hook: More bounces, dimmer image.

Topics in this chapter

Reflection of Light by Plane Mirrors formula sheet

First law: $∠ i = ∠ r$; Both measured from the normal.
Angle with the mirror: $i = 9 0^{\circ} - θ$; θ = angle between ray and mirror.
Deviation: $angle between incident and reflected rays = 2 i$; For normal incidence, 0° (ray returns).
Image distance: $image distance = object distance$; Measured perpendicular to the mirror.
Parallel mirrors: $x, 2 d - x, 2 d + x, 4 d - x, \dots$; Image distances behind M₁.

Reflection of Light by Plane Mirrors: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Image of a point object in a plane mirror

Draw a ray diagram to locate the image of a point object; 2 to 3 marks.

Draw

Draw the mirror as a line with short hatching on the back (non-reflecting) side.
Mark the point object O in front of the mirror.
Draw two rays from O to the mirror; at each point draw the normal (dotted) and the reflected ray with r = i.
Extend the reflected rays backwards with dotted lines behind the mirror until they meet at I.
Put arrows on all real rays.

Label

Mirror (hatched back)
O (object)
I (image)
normals
i and r
dotted virtual rays

Watch out: Real rays are solid with arrows; virtual (extended) rays are dotted. Show OM = MI.

Pin method to verify the laws of reflection

Describe an experiment to verify the laws of reflection; 3 to 4 marks.

Draw

A line MM′ for the mirror with a normal ON at point O.
The incident line through pins P₁ and P₂ meeting O.
The reflected line through pins P₃ and P₄ from O on the other side of the normal.
Mark angles i and r.

Label

Mirror MM′
Normal ON
P₁, P₂ (incident ray)
P₃, P₄ (reflected ray)
∠i
∠r

Watch out: State the precaution: look at the feet of the pins (or their images) to avoid errors, and keep the mirror vertical.

How the school exam asks Reflection of Light by Plane Mirrors

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Laws and angles	State the laws; find i, r or the angle between rays from the angle with the mirror.	Measure angles from the normal: i = 90° − θ.
Verification	Describe the pin (or ray-box) experiment, with a diagram and precautions.	Pins on the incident line; align images; measure r; repeat for several i.
Image characteristics	List them; find image distance; explain lateral inversion and mirror writing.	Virtual, erect, same size, same distance behind, laterally inverted.
Ray diagram	Locate the image of a point or an object with two rays.	Normals, equal angles, dotted extensions.
Uses	Give uses; explain a periscope; explain the mirror behind an ammeter scale.	Link each use to an image property.
Parallel mirrors	Number of images; why they get fainter; positions of the first few images.	Infinite in theory; absorption at each reflection; x, 2d − x, 2d + x …

Tricks to remember Reflection of Light by Plane Mirrors

Short memory aids. Say each one aloud once, then use it on a question the same day.

Angles from the normal
Never measure i or r from the mirror; always from the normal.
Use it when: Angle questions.
VESSL
Plane-mirror image: Virtual, Erect, Same size, Same distance, Laterally inverted.
Use it when: Image characteristics.
Mirror walks, image runs
Move towards the mirror by x: you and your image get 2x closer.
Use it when: Distance questions.
Parallel = endless
Parallel mirrors bounce light for ever, so images go on (getting fainter).
Use it when: Multiple images.

Reflection of Light by Plane Mirrors: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Normal incidence · 1 mark · MCQExam level

A ray of light falls normally on a plane mirror. The angle of reflection is:
(a) 0°
(b) 45°
(c) 90°
(d) 180°

Try it first, then show the solutionSolution

Along the normal, $i = 0°$ , so $r = 0°$ ; the ray returns along its path.

Answer: (a)

Example 2: Angle between the rays · 2 marks · Short answerExam level

The angle between an incident ray and the reflected ray is 80°. Find the angle of incidence and the angle the incident ray makes with the mirror.

Try it first, then show the solutionSolution

$2 i = 80°$ , so $i = 40°$ .
Angle with the mirror $= 90° - 40° = 50°$ .

Answer: 40°; 50°

Example 3: Image characteristics · 3 marks · Short answerExam level

State four characteristics of the image formed by a plane mirror, and explain why the image is called virtual.

Try it first, then show the solutionSolution

Virtual, erect, same size as the object, as far behind as the object is in front (also laterally inverted).
It is virtual because the reflected rays do not actually meet; they only appear to come from behind the mirror, so the image cannot be caught on a screen.

Answer: Virtual, erect, same size, same distance (laterally inverted).

Example 4: Moving the mirror · 3 marks · Short answerHarder exam level

An object is 20 cm in front of a plane mirror. The mirror is moved 5 cm away from the object, parallel to itself. By how much does the image move, and what is the new distance between object and image?

Try it first, then show the solutionSolution

Original object–image distance $= 40$ cm.
New object distance $= 25$ cm, so the image is 25 cm behind: object–image distance $= 50$ cm.
The image moves $50 - 40 = 10$ cm, twice the mirror's movement.

Answer: 10 cm; 50 cm

Example 5: Rotating the mirror · 3 marks · Short answerHarder exam level

A ray falls on a plane mirror with $i = 30°$ . The mirror is rotated by 10° so that $i$ becomes 40°, the incident ray staying fixed. Through what angle does the reflected ray turn?

Try it first, then show the solutionSolution

Before: angle between incident and reflected rays $= 60°$ .
After: $= 80°$ .
The reflected ray turns through $80° - 60° = 20°$ , twice the mirror's rotation.

Answer: 20°

Example 6: Full-length mirror · 4 marks · Long answerHarder exam level

A man 1.8 m tall, with his eyes 10 cm below the top of his head, wants to see his full image in a vertical plane mirror. Find the least length of mirror needed and how high its lower edge must be above the floor.

Try it first, then show the solutionSolution

Ray from the top of the head reflects into the eye from a point half-way between head-top and eye level: 5 cm below head-top height.
Ray from the feet reflects into the eye from a point half-way between floor and eye level: eyes at 1.7 m, so at 0.85 m.
Mirror from 0.85 m to 1.75 m: length $= 0.90$ m, half his height.
Lower edge 0.85 m above the floor.

Answer: 0.9 m; lower edge 0.85 m above the floor

Example 7: Images between parallel mirrors · 4 marks · Long answerHOTS / competency

Two parallel plane mirrors $M_{1}$ and $M_{2}$ are 30 cm apart. A candle is 10 cm from $M_{1}$ . Find the distances behind $M_{2}$ of the first two images formed in $M_{2}$ , and the distance between them.

Try it first, then show the solutionSolution

The candle is 20 cm from $M_{2}$ : first image 20 cm behind $M_{2}$ .
The first image in $M_{1}$ is 10 cm behind $M_{1}$ , i.e. 40 cm from $M_{2}$ ; its image in $M_{2}$ is 40 cm behind $M_{2}$ .
Distance between these two images $= 40 - 20 = 20$ cm.

Answer: 20 cm and 40 cm behind M₂; 20 cm apart

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 9-mark question split into 2 + 3 + 4 marks.

Designing a dressing room

A dressing room has a large plane mirror on one wall. Riya, 1.6 m tall with eyes 0.1 m below the top of her head, stands 2 m from the mirror. A second mirror is fixed on the opposite wall, parallel to the first, 4 m away from it.

(i) · 2 marks
How far is Riya's image (in the first mirror) from her? If she walks 0.5 m towards the mirror, how far is it now?
Answer
Answer: 4 m; 3 m
Image 2 m behind the mirror: 4 m from her. At 1.5 m from the mirror: 3 m.
(ii) · 3 marks
What is the least height of the first mirror for her to see her full image, and where should its bottom edge be?
Answer
Answer: 0.8 m; 0.75 m above the floor
Least length $= \frac{1}{2} \times 1.6 = 0.8$ m. Eyes at 1.5 m; bottom edge at half of 1.5 m $= 0.75$ m.
(iii) · 4 marks
Standing between the two parallel mirrors, she sees many images. Explain why, and find how far behind the first mirror the second image in that mirror lies.
Answer
Answer: Repeated reflections; 6 m
Each image in one mirror is an object for the other, so images repeat (getting fainter as light is absorbed). With $d = 4$ m and $x = 2$ m: images behind the first mirror at $x = 2$ m, then $2 d - x = 6$ m.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

State the two laws of reflection. (2 marks)

Step	Marks
The angle of incidence equals the angle of reflection	1
The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane	1

Draw a ray diagram to show the formation of the image of a point object by a plane mirror, and show that the image is as far behind the mirror as the object is in front. (3 marks)

Step	Marks
Two rays from O reflected with i = r at the mirror, normals drawn	1
Reflected rays extended backwards (dotted) to meet at I	1
OI perpendicular to the mirror at M, with OM = MI shown (by congruent triangles or measurement)	1

Describe an experiment to verify the laws of reflection using pins, with one precaution. (4 marks)

Step	Marks
Mirror line, normal at O and incident line drawn on paper; mirror placed vertically	1
Pins P₁, P₂ on the incident line; P₃, P₄ fixed in line with their images	1
Join P₃P₄ to O; measure r and compare with i for several angles; all lines on one sheet (same plane)	1
Precaution: pins vertical and mirror vertical; align the feet of the pins	1

Where students lose marks in Reflection of Light by Plane Mirrors

The slips examiners see most often, and what to write instead.

Mistake: Measuring angles from the mirror
i and r are measured from the normal; using the mirror gives the complement.
Wrong working
Ray at 55° to the mirror: $i = 55°$
Right working
$i = 90° - 55° = 35°$
Do this: Draw the normal first and mark angles from it.
Usually costs: 1 mark.
Mistake: Solid lines behind the mirror
Rays do not actually pass behind a mirror; extensions must be dotted.
Do this: Dotted lines for virtual rays; arrows only on real rays.
Usually costs: 1 mark.
Mistake: No normals in ray diagrams
Without normals, the equal angles cannot be shown.
Do this: Draw a dotted normal at each point of incidence.
Usually costs: 1 mark.
Mistake: Giving a finite number of images for parallel mirrors
In theory the number is infinite; only the visible number is limited by fading.
Do this: Say 'infinite in principle; fainter with each reflection'.
Usually costs: 1 mark.
Mistake: Using the perpendicular-mirror formula
The 2028 Class IX syllabus covers parallel mirrors only; inclined or perpendicular mirror formulas are not needed.
Do this: Answer parallel-mirror questions by tracing images.
Usually costs: Time and possible errors.

Practice questions on Reflection of Light by Plane Mirrors

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The image formed by a plane mirror is:
(a) real and inverted
(b) virtual and erect
(c) real and erect
(d) virtual and inverted
Hint
Cannot be caught on a screen.
Answer
(b)
Q2 · 1 mark · MCQExam level
A ray makes an angle of 30° with the surface of a plane mirror. The angle of reflection is:
(a) 30°
(b) 60°
(c) 90°
(d) 120°
Hint
i = 90° − 30°.
Answer
(b)
Q3 · 2 marks · Short answerExam level
Why is AMBULANCE written in mirror writing on the front of an ambulance?
Hint
Lateral inversion.
Answer
Drivers ahead see it in their rear-view mirrors, which laterally invert it, so they read AMBULANCE the right way round and give way.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): When you walk 2 m towards a plane mirror, the distance between you and your image decreases by 4 m.
Reason (R): The image is always as far behind the mirror as the object is in front.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
You move 2 m closer to the mirror; where does the image go?
Answer
(c): the image also moves 2 m towards the mirror, because image distance always equals object distance, so the gap shrinks by 2 + 2 = 4 m.
Q5 · 2 marks · Short answerExam level
The angle between the reflected ray and the mirror surface is 25°. Find the angle of incidence and the angle between incident and reflected rays.
Hint
r = 90° − 25°.
Answer
$r = 65°$ , so $i = 65°$ ; angle between rays $= 130°$ .
Q6 · 2 marks · Short answerHarder exam level
A boy stands 3 m in front of a plane mirror. A lamp hangs 1 m behind him (further from the mirror). How far is the lamp's image from the boy?
Hint
Lamp is 4 m from the mirror.
Answer
Lamp's image is 4 m behind the mirror, so $3 + 4 = 7$ m from the boy.
Q7 · 3 marks · Short answerHarder exam level
Two plane mirrors are parallel, facing each other, 50 cm apart. An object is midway between them. Find the positions of the first three images behind one mirror.
Hint
x = 25, d = 50.
Answer
$x = 25$ cm; $2 d - x = 75$ cm; $2 d + x = 125$ cm behind that mirror.
Q8 · 3 marks · Short answerHarder exam level
A ray is incident on a plane mirror at 40°. Keeping the ray fixed, the mirror is turned so the angle of incidence becomes 25°. Through what angle has the mirror turned and through what angle has the reflected ray turned?
Hint
Reflected ray turns twice as much.
Answer
Mirror turned 15°; angle between rays changes from 80° to 50°, so the reflected ray turns 30°.
Q9 · 2 marks · Short answerHarder exam level
Explain how a mirror strip behind the scale of an ammeter helps you take an accurate reading.
Hint
Parallax.
Answer
When your eye is directly above the pointer, the pointer hides its own image in the mirror. Reading in that position avoids parallax error.
Q10 · 3 marks · Short answerHarder exam level
Draw and describe how a periscope made with two plane mirrors works.
Hint
Mirrors parallel, at 45° to the tube.
Answer
Two parallel plane mirrors are fixed at 45° to a vertical tube. Light from an object enters the top, reflects at 45° incidence down the tube, reflects again at the lower mirror at 45° and reaches the eye horizontally. The image is erect.
Q11 · 4 marks · Long answerHOTS / competency
Prove, using the laws of reflection and congruent triangles, that the image of a point object in a plane mirror is as far behind the mirror as the object is in front.
Hint
Use one ray along the normal and one oblique ray.
Answer
Ray OM along the normal returns along itself, so I lies on OM produced. Oblique ray OA reflects with $i = r$ ; extended back it meets OM produced at I. Angles OAM and IAM are equal (from $i = r$ and vertically opposite angles), right angles at M, AM common: $△ O A M ≅ △ I A M$ (AAS), so OM = MI.
Q12 · 3 marks · Short answerHOTS / competency
A person facing a plane mirror lifts his right hand. Explain what his image appears to do, and whether a camera photo of the image would show the same as a photo of the person.
Hint
Lateral inversion.
Answer
The image appears to lift its left hand (lateral inversion). A photo of the image shows this inverted version, which differs from a photo of the person, where the right hand is raised.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the first law of reflection.Show answer
The angle of incidence equals the angle of reflection.
2.From which line are angles of incidence and reflection measured?Show answer
The normal.
3.Is the image in a plane mirror real or virtual?Show answer
Virtual.
4.What is lateral inversion?Show answer
The interchange of left and right in the image.
5.How many images form between two parallel mirrors?Show answer
Infinitely many (fainter further away).
6.If a mirror moves 3 cm away from an object, how far does the image move?Show answer
6 cm.
7.Least mirror length to see your full image?Show answer
Half your height.
8.Name an instrument that uses two parallel plane mirrors.Show answer
The periscope.

Best order to study it

Learn the terms and the two laws of reflection.
Learn the pin experiment for verification.
Practise ray diagrams and image characteristics.
Learn lateral inversion and uses of plane mirrors.
Practise parallel-mirror image positions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Do you measure i and r from the normal?
Can you state both laws exactly?
Can you describe the pin experiment and a precaution?
Can you draw the image of a point object with dotted extensions?
Can you list VESSL characteristics?
Can you explain the periscope?
Can you find image positions in parallel mirrors?

Reflection of Light by Plane Mirrors: frequently asked questions

Are images in perpendicular or inclined mirrors in the 2028 syllabus?

No. The 2028 Class IX syllabus covers multiple images in two parallel mirrors; images in a pair of perpendicular mirrors were removed.

Why can't a virtual image be caught on a screen?

The rays only appear to come from it; they do not actually meet there, so no light reaches a screen at that place.

Why are the images in parallel mirrors fainter further away?

Some light is absorbed (and some scattered) at every reflection, and far images involve many reflections.

Does the second law matter in experiments?

Yes. Because the incident ray, normal and reflected ray are drawn on one sheet of paper, the experiment also shows they lie in one plane.

Is a plane mirror image always the same size?

Yes. However far the object is, its plane-mirror image is the same size; it only looks smaller because it is further from your eye.

Based on: CISCE ICSE 2028 Science (52) Physics syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028