PhysicsICSE Class 9 · Reflection of Light by Plane Mirrors

Laws of Reflection and Image in a Plane Mirror: Angles, Rotation and Relative Motion (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

The angle of incidence equals the angle of reflection, and the incident ray, the normal and the reflected ray lie in one plane; both angles are measured from the normal. A plane mirror forms a virtual, erect, same-size, laterally inverted image as far behind the mirror as the object is in front. Turning the mirror through θ turns the reflected ray through 2θ.

Reflected rays only appear to meet behind the mirror, so the image is virtual, and its distance behind equals the object's distance in front.

Illustration drawn for QuestionBuddy.

Understand Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion

Laws of reflection.
1. The angle of incidence equals the angle of reflection: $∠ i = ∠ r$ .
2. The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane.

Angles are measured from the normal, not from the mirror. If a ray makes an angle $θ$ with the mirror surface, then $i = 9 0^{\circ} - θ$ . The angle between the incident and reflected rays is $2 i$ . A ray along the normal ( $i = 0^{\circ}$ ) returns along its own path.

Image in a plane mirror. To locate the image of a point object, draw two incident rays, reflect each with $i = r$ , and extend the reflected rays backward; they meet at the image. The image is

virtual (cannot be taken on a screen),
erect,
the same size as the object,
as far behind the mirror as the object is in front: $O M = M I$ , with $O I$ perpendicular to the mirror,
laterally inverted.

Distances and speeds. Object at $u$ from the mirror, image at $u$ behind; separation $= 2 u$ . If the object moves towards a fixed mirror with speed $v$ , the image moves towards the mirror (from behind) with speed $v$ , so the object and image approach each other at $2 v$ . If the object is fixed and the mirror moves away at speed $v$ , the image moves at $2 v$ in the same direction.

Rotating the mirror. Keep the incident ray fixed and turn the mirror through an angle $θ$ about the point of incidence. The normal also turns through $θ$ , so $i$ changes by $θ$ and the angle between incident and reflected rays changes by $2 θ$ : the **reflected ray turns through $2 θ$ **. This is the idea of a mirror galvanometer or light-pointer: a small turn of the mirror produces a large movement of a light spot on a distant scale (spot shift $\approx L tan 2 θ$ at distance $L$ ).

Periscope. Two parallel plane mirrors, each at $4 5^{\circ}$ to the tube, face each other. A horizontal ray falls at $i = 4 5^{\circ}$ on the first mirror, is turned through $9 0^{\circ}$ , falls at $4 5^{\circ}$ on the second and is turned through $9 0^{\circ}$ again: the emerging ray is parallel to the original but displaced sideways. The observer sees an erect image over an obstacle.

Method for ray questions. (1) Draw the normal at the point of incidence. (2) Find $i$ from the given angle (to the surface or the normal). (3) Mark $r = i$ on the other side. (4) Use geometry for the angle between the rays or the shifts.

Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion: formulas and key facts

Laws of reflection: $∠ i = ∠ r; i = 9 0^{\circ} - θ$; θ = angle with the mirror surface.
Angle between the rays: $∠ (incident, reflected) = 2 i$; Both angles from the normal.
Mirror turned by θ: $reflected ray turns through 2 θ$; Incident ray kept fixed.
Image position and speeds: $v = u; v_{ima g e} = v_{o bj ec t} (fixed mirror); v_{ima g e} = 2 v_{mi r r or}$; Image distance equals object distance.

How to solve Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion questions

Always measure from the normal and keep track of which part is fixed.

1Draw the normal and mark i and r on either side of it.
2If the angle is given with the mirror surface, subtract from 90°.
3Angle between the rays = 2i (= i + r).
4For rotation, keep the incident ray fixed: the reflected ray turns by 2θ.
5For image motion use image distance = object distance, then differentiate in your head: speeds follow the distances.

Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion in real life

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.

A light pointer

Suppose a small plane mirror on a delicate instrument turns through 3° when a current passes, and a narrow light beam reflected from it falls on a scale 2 m away.

By how much does the light spot move on the scale?

Try it first, then show the solutionSolution

The reflected ray turns through $2 \times 3^{\circ} = 6^{\circ}$ .
Shift $\approx 2 \times tan 6^{\circ} = 2 \times 0.1051 \approx 0.21$ m.

Answer: About 21 cm: a tiny rotation is magnified.

Memory hook: Mirror turns θ, ray turns 2θ.

Tricks to remember Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion

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

Normal first
Draw the normal before any angle: i and r are from the normal.
Use it when: Every reflection diagram.
Ninety minus the surface angle
Angle with the mirror θ means i = 90° − θ.
Use it when: Angle given with the surface.
Twice the turn
Mirror turns θ, ray turns 2θ; image of a moving mirror moves 2v.
Use it when: Rotation and motion.

Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Angle with the surface · 1 mark · MCQBoard level

A ray makes an angle of 35° with a plane mirror. The angle between the incident and the reflected rays is:
(a) 35°
(b) 55°
(c) 70°
(d) 110°

Try it first, then show the solutionSolution

$i = 9 0^{\circ} - 3 5^{\circ} = 5 5^{\circ}$ ; angle between rays $= 2 i = 11 0^{\circ}$ .

Answer: (d) 110°

Example 2: Object and image separation · 2 marks · Short answerBoard level

An object is 40 cm in front of a plane mirror. Find the distance of the image from the mirror and its distance from the object. If the object is moved 15 cm towards the mirror, what is the new separation of object and image?

Try it first, then show the solutionSolution

Image distance $= 40$ cm behind the mirror; separation $= 80$ cm. (1)
New object distance $= 25$ cm, separation $= 2 \times 25 = 50$ cm. (1)

Answer: 40 cm; 80 cm; 50 cm

Example 3: Rotating the mirror · 3 marks · Short answerHarder board level

A ray falls on a plane mirror at $i = 2 0^{\circ}$ . Keeping the ray fixed, the mirror is turned through 15° so that the angle of incidence increases. Find the new angle of incidence, the angle between incident and reflected rays before and after, and the angle through which the reflected ray turned.

Try it first, then show the solutionSolution

New $i = 2 0^{\circ} + 1 5^{\circ} = 3 5^{\circ}$ . (1)
Angle between rays: before $4 0^{\circ}$ , after $7 0^{\circ}$ . (1)
Reflected ray turned through $7 0^{\circ} - 4 0^{\circ} = 3 0^{\circ} = 2 \times 1 5^{\circ}$ . (1)

Answer: 35°; 40° then 70°; 30°

Example 4: Approaching speed · 3 marks · Short answerHarder board level

A girl walks towards a fixed plane mirror at 2 m s⁻¹, starting 3 m from it. (a) Find the speed of her image. (b) Find the speed with which she approaches the image. (c) Find the time taken for the gap between her and the image to shrink from 6 m to 2 m. (d) If she stands still and the mirror moves away at 1.5 m s⁻¹, find the speed of the image.

Try it first, then show the solutionSolution

(a) 2 m s⁻¹ towards the mirror (from behind). (½)
(b) $2 + 2 = 4$ m s⁻¹. (½)
(c) Gap reduces by 4 m at 4 m s⁻¹, so $t = 1$ s. (1)
(d) Image distance $=$ object distance, so the image moves at $2 \times 1.5 = 3$ m s⁻¹. (1)

Answer: (a) 2 m s⁻¹ (b) 4 m s⁻¹ (c) 1 s (d) 3 m s⁻¹

Example 5: Periscope · 4 marks · Case-basedHOTS / competency

In a simple periscope two plane mirrors M₁ and M₂ are placed parallel to each other at 45° to the tube. A horizontal ray enters the tube.
(a) Find the angle of incidence at M₁.
(b) Find the total deviation of the ray after both mirrors.
(c) Describe the direction of the emergent ray.
(d) M₂ is turned through 10°. Through what angle is the emergent ray turned?

Try it first, then show the solutionSolution

(a) Ray makes $4 5^{\circ}$ with the mirror, so $i = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$ . (1)
(b) The first mirror turns the ray through $2 i = 9 0^{\circ}$ ; the second, parallel to it, turns it back through $9 0^{\circ}$ in the opposite sense, so the total deviation is $0^{\circ}$ . (1)
(c) Parallel to the incident ray, displaced sideways by the distance between the mirrors. (1)
(d) Turning M₂ through 10° turns the emergent ray through $2 \times 1 0^{\circ} = 2 0^{\circ}$ . (1)

Answer: (a) 45° (b) 0° (c) parallel to the incident ray (d) 20°

Where students lose marks in Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion

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

Mistake: Angle with the mirror used as the angle of incidence
i is from the normal.
Wrong working
Ray at 35° to the mirror, i = 35°.
Right working
i = 90° − 35° = 55°.
Do this: Draw the normal and label it.
Usually costs: The answer.
Mistake: Image called real because it is "seen"
Real images can be received on a screen.
Wrong working
Real and erect.
Right working
Virtual and erect.
Do this: Ask: do the rays actually meet?
Usually costs: 1 mark.
Mistake: Rotation angle not doubled
Reflected ray turns by 2θ.
Wrong working
Mirror rotated by 10°, reflected ray by 10°.
Right working
Reflected ray by 20°.
Do this: Remember the normal turns with the mirror.
Usually costs: The answer.

Exam tips

Draw the normal as a dashed line and mark i and r on the figure.
State "virtual, erect, same size, laterally inverted, as far behind as the object is in front".
For rotation questions write "the normal also turns through θ".
In the periscope answer, give the angle (45°) and the direction of the final ray.

Practice questions on Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion

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 · MCQBoard level
A plane mirror forms the image of an object placed 30 cm in front of it. The image is:
(a) 30 cm in front of the mirror
(b) 30 cm behind the mirror
(c) 15 cm behind the mirror
(d) on the mirror surface
Hint
Image distance.
Answer
(b) 30 cm behind the mirror
Q2 · 1 mark · Very short answerBoard level
What is the angle of reflection for a ray that falls along the normal?
Hint
Normal incidence.
Answer
0°.
Q3 · 2 marks · Short answerHarder board level
The angle of incidence of a ray is 25°. The mirror is turned so that the angle between the ray and the mirror surface becomes 40°. Find the new angle of reflection and the new angle between incident and reflected rays.
Hint
i = 90° − 40°.
Answer
New $i = 5 0^{\circ}$ , so $r = 5 0^{\circ}$ ; angle between rays $= 10 0^{\circ}$ .
Q4 · 3 marks · Short answerHarder board level
A plane mirror reflects a narrow beam of light onto a scale 2 m away. The mirror is turned through 3°. Find the shift of the spot on the scale (take $tan 6^{\circ} = 0.105$ ).
Hint
Reflected ray turns by 2θ.
Answer
Reflected ray turns through $6^{\circ}$ ; shift $= 2 \times 0.105 = 0.21$ m $= 21$ cm.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): The image of an object in a plane mirror cannot be obtained on a screen.
Reason (R): The reflected rays do not actually meet; they only appear to come from the image.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Virtual image.
Answer
(a)
Q6 · 3 marks · Long answerHOTS / competency
Prove that if a plane mirror is rotated through an angle θ about the point of incidence, keeping the incident ray fixed, the reflected ray turns through 2θ.
Hint
Normal turns by θ too.
Answer
Initially angle between rays $= 2 i$ . The normal turns through $θ$ , so the angle of incidence becomes $i + θ$ (or $i - θ$ ) and the angle between the rays $2 (i \pm θ)$ . The change is $2 θ$ , and the incident ray is fixed, so the reflected ray turns through $2 θ$ .

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.Laws of reflection?Show answer
i = r; incident ray, normal and reflected ray in one plane.
2.Image distance in a plane mirror?Show answer
Equal to the object distance (behind the mirror).
3.Mirror turned by θ?Show answer
Reflected ray turns by 2θ.
4.Nature of a plane mirror image?Show answer
Virtual, erect, same size, laterally inverted.
5.Angle of incidence for a ray 30° to the surface?Show answer
60°.

Laws of Reflection and Image Formation in a Plane Mirror: Rotation and Relative Motion: frequently asked questions

What are the laws of reflection of light?

The angle of incidence equals the angle of reflection, and the incident ray, the reflected ray and the normal lie in the same plane.

What happens when a mirror is rotated?

If the incident ray is fixed and the mirror turns through θ, the reflected ray turns through 2θ.

How does a periscope work?

Two parallel plane mirrors at 45° to the tube turn the ray through 90° twice, so the observer sees an erect image of objects above an obstacle.