IGCSE Physics | Worked Answers | 25 Questions
Define the term normal as used in reflection diagrams.
The normal is an imaginary line drawn perpendicular (at 90°) to the mirror surface at the point where the light ray hits it. All angles in reflection are measured from the normal.
Define the angle of incidence.
The angle of incidence is the angle between the incoming (incident) ray and the normal, measured at the point where the ray meets the mirror.
Define the angle of reflection.
The angle of reflection is the angle between the reflected ray and the normal, measured at the same point on the mirror.
State the law of reflection.
The angle of incidence is equal to the angle of reflection: $i = r$.
State the three characteristics of the image formed by a plane mirror.
- Same size as the object
- Same distance from the mirror as the object
- Virtual
What is meant by a virtual image?
A virtual image is one where light rays only appear to come from a point — they do not actually meet there. A virtual image cannot be projected onto a screen.
From which line are the angle of incidence and the angle of reflection always measured?
Both angles are always measured from the normal — the imaginary line perpendicular to the mirror surface at the point of incidence.
A ray of light hits a plane mirror at an angle of incidence of 55°. State the angle of reflection.
By the law of reflection, $i = r$, so the angle of reflection = 55°.
Is the image in a plane mirror larger than, smaller than, or the same size as the object?
The image is the same size as the object. A plane mirror does not magnify or shrink.
An object is placed 8 cm in front of a plane mirror. State how far behind the mirror the image appears.
The image appears 8 cm behind the mirror. For a plane mirror, image distance = object distance.
A ray of light hits a plane mirror at an angle of incidence of 35°. (a) State the angle of reflection. (b) Calculate the angle between the incident ray and the reflected ray.
By the law of reflection, $i = r$, so the angle of reflection = 35°.
Both rays are on opposite sides of the normal, each at 35° from it.
$$\text{Angle between rays} = i + r = 35° + 35° = \mathbf{70°}$$A ray of light hits a plane mirror. The angle between the ray and the mirror surface is 25°. (a) Find the angle of incidence. (b) State the angle of reflection.
The angle between the ray and the mirror surface is 25°. Since the normal is at 90° to the surface:
$$i = 90° – 25° = \mathbf{65°}$$By the law of reflection, $r = i$ = 65°.
The angle between an incident ray and its reflected ray is 64°. Find the angle of incidence.
The incident and reflected rays are on opposite sides of the normal, each making angle $i$ with it. So the total angle between them is $i + r = 2i$.
$$2i = 64°$$ $$i = 64° \div 2 = \mathbf{32°}$$An object is placed 15 cm in front of a plane mirror. (a) State how far behind the mirror the image appears. (b) Calculate the total distance from the object to its image.
Image distance = object distance = 15 cm behind the mirror.
Explain why the image formed by a plane mirror cannot be projected onto a screen.
The image is virtual. The reflected light rays do not actually pass through or come from the image position — they only appear to come from behind the mirror. Because no real light comes from that point, there is nothing to form a spot on a screen, no matter how bright the source.
A student writes: “The angle of reflection is the angle between the reflected ray and the mirror surface.” Correct this statement and explain what the angle of reflection is actually measured from.
Corrected statement: “The angle of reflection is the angle between the reflected ray and the normal.”
The normal is an imaginary line at 90° to the mirror surface at the point where the ray hits. Angles in reflection are always measured from the normal, not from the mirror surface itself.
A ray of light hits a plane mirror and reflects back along exactly the same path as the incident ray. State the angle of incidence and explain why this happens.
The angle of incidence = 0°.
The ray is travelling along the normal — it hits the mirror at exactly 90° to its surface. By the law of reflection, $r = i = 0°$, so the reflected ray also travels along the normal, back in exactly the opposite direction.
An object is moved from 6 cm to 18 cm in front of a plane mirror. Describe what happens to: (a) the distance of the image from the mirror, (b) the size of the image.
The image distance equals the object distance. As the object moves from 6 cm to 18 cm, the image moves from 6 cm to 18 cm behind the mirror — it moves further away from the mirror by the same amount.
The size of the image stays the same. A plane mirror always produces an image the same size as the object, regardless of the object’s distance from the mirror.
An object is 20 cm in front of a plane mirror. The object then moves 5 cm closer to the mirror. (a) State where the image now appears. (b) How far has the image moved compared to its original position?
New object distance = 20 − 5 = 15 cm. Image distance = object distance, so the image now appears 15 cm behind the mirror.
Original image: 20 cm behind mirror. New image: 15 cm behind mirror.
$$\text{Image moved} = 20 – 15 = \mathbf{5 \text{ cm}}$$The image moved 5 cm closer to the mirror — the same distance the object moved.
The angle between an incident ray and the mirror surface is 50°. Show that the angle between the incident ray and the reflected ray is 80°.
- The angle between the ray and the mirror surface is 50°. Convert to angle from the normal: $$i = 90° – 50° = 40°$$
- By the law of reflection: $r = i = 40°$
- The incident ray is at 40° on one side of the normal; the reflected ray is at 40° on the other side: $$\text{Angle between rays} = i + r = 40° + 40° = \mathbf{80°} \checkmark$$
Describe the image formed by a plane mirror. For each of the three characteristics, give a brief explanation of what it means. Then explain why the image is described as “virtual” rather than “real”.
The image is exactly the same size as the object. A plane mirror does not magnify or shrink — a 10 cm object produces a 10 cm image.
The image appears as far behind the mirror as the object is in front. If the object is 12 cm from the mirror, the image appears 12 cm behind it.
The reflected rays spread out after hitting the mirror. When you trace them backwards (behind the mirror), they appear to come from the image position — but no actual light rays pass through that point.
A real image is formed where light rays actually converge. You can project a real image onto a screen because real light arrives there.
A virtual image is formed where light rays only appear to come from. No real light arrives at that point, so the image cannot be projected onto a screen — no matter how bright the light source.
An object is placed 7 cm in front of a plane mirror. Supplement
(a) Describe step by step how to draw and correctly reflect two rays from the object. (b) Explain how the reflected rays are used to find the image. (c) State the expected distance of the image behind the mirror.
- From the object, draw a ray to a chosen point on the mirror surface.
- At that point, draw the normal — a dashed line at 90° to the mirror.
- Measure the angle of incidence ($i$) between the ray and the normal.
- Draw the reflected ray on the other side of the normal at the same angle: $r = i$.
- Repeat steps 1–4 for a second ray hitting a different point on the mirror.
Extend both reflected rays behind the mirror using dashed lines. The image is located at the point where the two dashed lines meet. This is the virtual image — it appears to be behind the mirror at the point where the extended rays intersect.
Image distance = object distance = 7 cm behind the mirror.
A student makes three statements about the image in a plane mirror. For each statement, say whether it is correct or incorrect and give a full explanation.
(a) “The image is behind the mirror at the same distance as the object is in front.”
(b) “The image is the same size as the object.”
(c) “I can project the image onto a screen if I use a bright enough light source.”
Correct. For a plane mirror, the image distance always equals the object distance. If the object is, say, 10 cm in front of the mirror, the image appears exactly 10 cm behind it. This is a fixed property of plane mirror reflection.
Correct. A plane mirror always produces an image the same size as the object. This does not change as the object moves closer or further away. There is no magnification.
Incorrect. The image in a plane mirror is virtual. The reflected light rays do not actually pass through the image position — they only appear to come from there when extended behind the mirror. Because no real light exists at that point, it cannot form an image on a screen regardless of how bright the light source is. Brightness is not the issue; the type of image is.
A ray of light hits a plane mirror. The angle between the incident ray and the mirror surface is 30°. (a) Find the angle of incidence. (b) Find the angle of reflection. (c) Find the angle between the incident ray and the reflected ray. (d) A second ray hits the same mirror. Its reflected ray makes an angle of 20° with the normal. Find the angle this incident ray makes with the mirror surface.
By the law of reflection, $r = i$ = 60°.
The reflected ray makes 20° with the normal, so by $i = r$, the angle of incidence is also 20°.
$$\text{Angle with mirror surface} = 90° – 20° = \mathbf{70°}$$A student investigates reflection using a plane mirror and a ruler. Supplement
(a) A pin (object) is placed 9 cm in front of the mirror. State where the image appears and give the total distance from the pin to its image. (b) The student draws a ray from the pin that hits the mirror at 40° to the mirror surface. Find the angle of reflection. (c) Describe step by step how to use two reflected rays to locate the position of the image. (d) The student moves the pin to 14 cm from the mirror. State the new image position and the new total distance from the pin to its image.
Image distance = object distance = 9 cm behind the mirror.
$$\text{Total distance} = 9 + 9 = \mathbf{18 \text{ cm}}$$The ray hits the mirror at 40° to the mirror surface, so:
$$i = 90° – 40° = 50°$$By the law of reflection, $r = i$ = 50°.
- Draw a ray from the object to a point on the mirror. At that point, draw the normal (at 90° to the mirror). Measure the angle of incidence and draw the reflected ray at the same angle on the other side of the normal ($r = i$).
- Repeat for a second ray hitting a different point on the mirror.
- Extend both reflected rays behind the mirror using dashed lines. The point where the two dashed lines meet is the position of the virtual image.
New object distance = 14 cm. Image distance = object distance = 14 cm behind the mirror.
$$\text{Total distance} = 14 + 14 = \mathbf{28 \text{ cm}}$$
