IGCSE Physics Reflection of Light Questions and Worked Solutions (3.2.1) | 0625

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IGCSE Physics 0625 · Topic 3.2.1

IGCSE Physics Reflection of Light Questions and Worked Solutions

50 topic-wise practice questions on Reflection of Light (Unit 3: Waves), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions25 Core15 Extended10 Challenge118 marksTextbook §3.2 · pp. 136–159
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Reflection of Light

Light reflects from a plane mirror so that the angle of incidence equals the angle of reflection. The image in a plane mirror is virtual, upright, the same size and the same distance behind the mirror as the object is in front.

Key formulas

i = rangle of incidence = angle of reflection (both measured from the normal)Core

Prerequisite knowledge

  • Measuring angles with a protractor
  • Straight-line propagation of light

Exam tips & common mistakes

  • If an angle to the mirror surface is given, subtract from 90°
  • Draw virtual rays behind the mirror as dashed lines
  • Image distance = object distance

Key facts from the syllabus (3.2.1)

  • The normal is a line at 90° to the mirror at the point of incidence
  • Angles of incidence and reflection are measured from the normal, not the mirror
  • Plane mirror image: same size, same distance behind the mirror, virtual, upright, laterally inverted
  • Experiment: ray box, mirror and protractor to show i = r
  • Extended use constructions (two reflected rays extended behind the mirror) to locate images and solve problems

Textbook: §3.2 Light, pp. 136–159

1
Core[1 mark]

Define the normal in reflection.

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Method: Line at 90° to surface.

A line at right angles to the mirror surface at the point where the ray strikes it
2
Core[1 mark]

State the law of reflection.

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Method: i = r.

Angle of incidence = angle of reflection
3
Core[1 mark]

A ray strikes a plane mirror with an angle of incidence of 35°. State the angle of reflection.

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Method: i = r.

35°
4
Core[2 marks]

A ray makes an angle of 25° with the surface of a plane mirror. Calculate the angle of reflection.

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Method: Angles are measured from the normal.

i = 90 − 25 = 65°, so angle of reflection = 65°
5
Core[3 marks]

State three characteristics of an image in a plane mirror.

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Method: Same size, same distance, virtual.

Same size as the object, the same distance behind the mirror as the object is in front, virtual (and laterally inverted, upright)
6
Core[2 marks]

A person stands 2.5 m in front of a plane mirror. Calculate the distance between the person and their image.

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Method: Image is 2.5 m behind the mirror.

Distance = 2.5 + 2.5 = 5.00 m
7
Core[2 marks]

A person walks towards a mirror at 1.0 m/s. State how fast they approach their image.

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Method: Image moves at the same speed in the opposite direction.

2.0 m/s
8
Core[1 mark]

Explain why the image in a plane mirror is described as virtual.

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Method: Rays only appear to come from it.

The light rays do not actually pass through it, so it cannot be formed on a screen
9
Core[3 marks]

Describe an experiment to show that angle of incidence equals angle of reflection.

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Method: Ray box, mirror, protractor.

Draw a normal on paper; place a mirror on a line; shine a ray from a ray box at the mirror; mark incident and reflected rays; measure both angles with a protractor for several angles
10
Core[2 marks]

The angle between an incident ray and the reflected ray is 80°. Calculate the angle of incidence.

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Method: i + r = 80°, i = r.

i = 80 ÷ 2 = 40°
11
Core[2 marks]

A ray hits a mirror along the normal. State the angle of incidence and the direction of the reflected ray.

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Method: i = 0.

0°; it reflects straight back along the same path
12
Core[2 marks]

Explain why the word AMBULANCE is written in reverse on the front of an ambulance.

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Method: Lateral inversion in mirrors.

A driver sees it in their rear-view mirror, which laterally inverts it, so it reads correctly
13
Core[1 mark]

State whether an image in a plane mirror is upright or inverted.

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Method: Upright.

Upright
14
Core[1 mark]

Define the angle of incidence.

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Method: Between ray and normal.

The angle between the incident ray and the normal
15
Core[1 mark]

Define the angle of reflection.

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Method: Between reflected ray and normal.

The angle between the reflected ray and the normal
16
Core[2 marks]

A ray strikes a mirror at an angle of incidence of 52°. State the angle between the reflected ray and the mirror surface.

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Method: 90 − 52.

38°
17
Core[2 marks]

The angle between the incident ray and the mirror surface is 30°. Calculate the angle between the incident and reflected rays.

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Method: i = 60°.

120°
18
Core[2 marks]

A girl stands 1.5 m from a plane mirror. She moves 0.5 m closer. Calculate the new distance from her to her image.

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Method: Image 1.0 m behind.

2.00 m
19
Core[2 marks]

Explain why you can read words in a mirror more easily if they are written backwards.

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Method: Lateral inversion.

The mirror laterally inverts (swaps left and right) the image
20
Core[2 marks]

State why the image in a plane mirror cannot be projected onto a screen.

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Method: Virtual.

It is a virtual image; light does not actually pass through it
21
Core[2 marks]

A clock seen in a mirror appears to show 3:00. State the actual time.

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Method: Lateral inversion.

9:00
22
Core[2 marks]

Describe the size of the image of a 1.7 m tall person in a plane mirror.

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Method: Same size.

1.7 m tall (same size)
23
Core[2 marks]

Explain why a smooth surface gives a clear image but a rough surface does not.

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Method: Regular vs diffuse.

On a smooth surface parallel rays reflect in the same direction; on a rough surface they scatter in many directions
24
Core[1 mark]

State the angle of reflection when the angle of incidence is 0°.

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Method: Along the normal.

0°
25
Core[2 marks]

A ray of light reflects from a mirror and the angle between the incident and reflected rays is 110°. Calculate the angle of incidence.

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Method: Half.

55°
26
Extended[3 marks]

Describe how to locate the image of a pin in a plane mirror using ray construction.

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Method: Draw two reflected rays and extend them back.

Draw two rays from the pin reflecting off the mirror (i = r), then extend the reflected rays backwards behind the mirror; the image is where they meet
27
Extended[2 marks]

An object is 30 cm in front of a mirror. The mirror is moved 10 cm further away from the object. Calculate the new distance between the object and its image.

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Method: Object now 40 cm from mirror.

Distance = 2 × 40 = 80 cm
28
Extended[3 marks]

Two plane mirrors are at 90° to each other. A ray hits the first mirror at an angle of incidence of 30°. Calculate the angle of incidence on the second mirror.

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Method: In a right-angled corner, the angles of incidence add to 90°.

Angle of incidence on second mirror = 90 − 30 = 60°
29
Extended[3 marks]

A periscope uses two plane mirrors at 45°. Describe the path of light through it.

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Method: Each mirror turns the ray through 90°.

Light hits the top mirror at 45° and turns 90° down the tube, hits the lower mirror at 45° and turns 90° again into the eye, parallel to the original direction
30
Extended[2 marks]

A ray strikes a mirror with an angle of incidence of 20°. The mirror is rotated by 10° so that the angle of incidence increases. Calculate the new angle between incident and reflected rays.

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Method: New i = 30°.

Angle between rays = 2 × 30 = 60°
31
Extended[2 marks]

Explain why a ray diagram can be used to show that the image in a plane mirror is the same distance behind as the object is in front.

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Method: Geometry of equal angles.

Congruent triangles formed by the reflected rays and normals show the image distance equals the object distance
32
Extended[2 marks]

A small object is placed 12 cm in front of a plane mirror. The observer is 30 cm in front of the mirror on the same normal. Calculate the distance from the observer to the image.

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Method: Image 12 cm behind the mirror.

Distance = 30 + 12 = 42 cm
33
Extended[3 marks]

An object is 25 cm in front of a mirror. Describe how to locate its image using two rays and a ruler and protractor.

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Method: Construction.

Draw two rays from the object to the mirror; at each, draw a normal and the reflected ray with i = r; extend the reflected rays behind the mirror with dashed lines; their intersection is the image, 25 cm behind
34
Extended[3 marks]

A ray hits mirror 1 at 40° incidence and then mirror 2, which is at 90° to mirror 1. Show that the final ray is parallel to the original ray.

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Method: Angles sum.

Angle of incidence on mirror 2 = 50°; the total deviation is 180°, so the final ray travels back parallel to the incident ray
35
Extended[2 marks]

A plane mirror is rotated by 15° while the incident ray stays fixed. State the angle through which the reflected ray turns.

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Method: Twice.

30°
36
Extended[3 marks]

A person 1.8 m tall wants to see their full height in a vertical mirror. State the minimum mirror height and explain.

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Method: Half height.

0.90 m: rays from head and feet reflect from points halfway between the eyes and those points
37
Extended[2 marks]

Two people stand in front of a mirror. Person A can see person B's eyes. Explain why B can see A's eyes.

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Method: Reversibility.

Light paths are reversible: a ray from A to B via the mirror can travel back along the same path
38
Extended[3 marks]

A point object is 10 cm in front of a plane mirror. The observer's eye is 30 cm in front of the mirror and 20 cm to one side. Calculate the straight-line distance from the eye to the image.

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Method: Image 10 cm behind; horizontal 40 cm, sideways 20 cm.

d = √(40² + 20²) = 44.7 cm
39
Extended[2 marks]

Explain why a ray diagram for a plane mirror uses dashed lines behind the mirror.

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Method: Virtual rays.

Dashed lines show where rays appear to come from; no light actually travels there
40
Extended[2 marks]

An object is moved from 40 cm to 15 cm in front of a plane mirror. Calculate how far its image moves.

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Method: Image distance = object distance.

25 cm (from 40 cm to 15 cm behind the mirror)

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

Two plane mirrors are at 60° to each other. A ray hits the first mirror at an angle of incidence of 50°. Calculate the angle of incidence on the second mirror.

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Method: The reflected ray makes 90 − 50 = 40° with mirror 1. In the triangle formed with the 60° corner, the angle between the ray and mirror 2 is 180 − 60 − 40 = 80°.

Angle of incidence = 90 − 80 = 10°
42
ChallengeCore[4 marks]

A student stands 3.0 m from a plane mirror and walks towards it at 0.50 m/s for 2.0 s. Calculate the new distance between the student and the image, and the speed at which the image approaches the student.

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Method: Image same distance behind.

Distance from mirror = 2.0 m, so separation = 4.0 m; the image approaches at 1.0 m/s
43
ChallengeCore[3 marks]

Explain why, when you raise your right hand in front of a mirror, your image appears to raise its left hand.

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Method: Lateral inversion.

The image is laterally inverted: each point is the same distance behind the mirror, which reverses front-to-back, appearing as left-right reversal
44
ChallengeCore[4 marks]

A periscope uses two plane mirrors at 45°. Draw a description of the path of a ray and state the characteristics of the final image.

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Method: Two reflections.

The ray turns 90° at each mirror, emerging parallel to the original; the image is upright, same size, virtual
45
ChallengeCore[3 marks]

Explain why you can see an image in a still pond but not in a windy, rippled pond.

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Method: Regular vs diffuse reflection.

A flat surface reflects parallel rays in one direction (regular reflection); ripples scatter rays in many directions (diffuse)
46
ChallengeExtended[4 marks]

A ray strikes a mirror at an angle of incidence of 30°. The mirror is rotated by 12° so the angle of incidence increases. Calculate the angle through which the reflected ray turns.

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Method: Deviation changes by twice the mirror rotation.

24°
47
ChallengeExtended[4 marks]

An object is placed between two parallel mirrors A and B, 50 cm apart, 10 cm from mirror A. Calculate the positions of the first two images seen in mirror A.

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Method: First image: 10 cm behind A. B forms an image 40 cm behind B (90 cm from A); A then forms an image of that.

10 cm and 90 cm behind mirror A
48
ChallengeExtended[4 marks]

A girl 1.60 m tall, eyes 1.50 m above the floor, stands in front of a vertical mirror. Calculate the minimum length of mirror and the height of its bottom edge above the floor for her to see her whole body.

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Method: Half-height rule.

Length = 0.80 m; bottom edge at 0.75 m (half eye height), top at 1.55 m
49
ChallengeExtended[4 marks]

Describe how to locate the image of a pin in a plane mirror experimentally using ray-tracing with pins.

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Method: Sight lines.

Place an object pin; view its image from two positions and place two sighting pins in line with the image each time; draw lines through each pair and extend them behind the mirror; their intersection is the image
50
ChallengeExtended[3 marks]

Explain why a ray diagram for a plane mirror shows the image the same distance behind the mirror as the object is in front.

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Method: Congruent triangles.

Using i = r, the triangles formed by the object, the image and the normals are congruent, so the distances are equal
All IGCSE Physics topics← 3.1 General Properties of Waves3.2.2 Refraction of Light →

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