These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: all mirrors are plane and flat over the region the light hits; angles of incidence (θᵢ) and reflection (θᵣ) are measured from the normal; light travels in straight lines between surfaces. A scientific calculator in degree mode is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A ray of light strikes a plane mirror. The ray makes an angle of 35° with the surface of the mirror. What is the angle of reflection?
- (A) 35°
- (B) 55°
- (C) 70°
- (D) 110°
Answer and explanation
Answer: (B). The normal is at 90° to the surface, so θᵢ = 90° − 35° = 55°. By the law of reflection, θᵣ = θᵢ = 55°.
- (A) uses the angle with the surface as if it were the angle of incidence.
- (C) is 2 × 35°, the angle through which the ray’s direction of travel changes. It is not an angle of reflection.
- (D) is the angle between the incident and reflected rays, 2 × 55° = 110°.
Question 2 (multiple choice · foundation)
Which observation cannot be explained using the ray model of light alone?
- (A) A laser beam crossing a dusty room follows a straight line.
- (B) The shadow of a door on a wall has fairly sharp edges.
- (C) Light passing through a slit about as wide as its wavelength spreads out over a wide area of a screen.
- (D) A beam reflects from a mirror with the angle of reflection equal to the angle of incidence.
Answer and explanation
Answer: (C). Spreading of light through a very narrow opening is diffraction. Straight rays would give a narrow bright strip the width of the slit. Explaining the spreading needs the wave model (Unit 14).
- (A) is exactly what the ray model predicts: light travels in straight lines in a uniform material.
- (B) Sharp shadows of large objects follow from straight-line rays that are blocked by the door.
- (D) is the law of reflection, which is part of the ray model.
Question 3 (multiple choice · core)
A small lamp produces circular wavefronts centred on the lamp. Point P is 2 m from the lamp, up and to the right of it. Which describes the light ray at P?
- (A) It points along the wavefront through P, at a tangent to the circle.
- (B) It points along the straight line from the lamp to P, away from the lamp.
- (C) It points along the straight line from P back towards the lamp.
- (D) It points vertically upward, whatever the position of P.
Answer and explanation
Answer: (B). A ray is perpendicular to the wavefronts and points in the direction the wave travels. For circular wavefronts, the line perpendicular to the circle at P is the radius through P. The wave moves outward from the lamp, so the ray points away from it.
- (A) runs along the wavefront rather than perpendicular to it.
- (C) has the right line but the wrong direction: light is travelling away from its source.
- (D) ignores where P is. Only a point directly above the lamp would have a vertical ray.
Question 4 (multiple choice · core)
A narrow beam strikes a plane mirror with an angle of incidence of 30°. The beam is kept fixed and the mirror is turned through 10° so that the angle of incidence becomes 40°. Through what angle does the reflected beam turn?
- (A) 5°
- (B) 10°
- (C) 20°
- (D) 40°
Answer and explanation
Answer: (C). Turning the mirror turns the normal by 10°. Before the turn, the incident and reflected beams were 2 × 30° = 60° apart. After it, they are 2 × 40° = 80° apart. The incident beam has not moved, so the reflected beam has turned through 80° − 60° = 20°, which is twice the mirror’s rotation.
- (A) halves the rotation instead of doubling it.
- (B) assumes the reflected beam turns with the mirror. It forgets that θᵣ changes as well as θᵢ.
- (D) is the new angle of incidence, not a change in direction.
Question 5 (calculation · core)
A laser is fixed 1.20 m above a level floor and points down at a small flat mirror lying on the floor. The beam must reflect and hit a target on a vertical wall, 0.80 m above the floor. The wall is 3.00 m from the laser, measured horizontally.
(a) Draw a ray diagram showing the laser, mirror, normal and target. (b) Calculate the horizontal distance from the point directly below the laser to the mirror. (c) Calculate the angle of incidence at the mirror.
Worked solution and suggested Marlbridge rubric
(a) A straight ray from the laser down to a point M on the floor, a dashed vertical normal at M, and a straight ray from M up to the target, with equal angles on each side of the normal and arrows showing the direction of travel.
(b) Let x be the horizontal distance from below the laser to M. The wall is then 3.00 − x from M. Because θᵢ = θᵣ, the two right-angled triangles (laser side and target side) have equal angles, so they are similar:
x ÷ 1.20 = (3.00 − x) ÷ 0.80
0.80x = 1.20(3.00 − x) → 0.80x + 1.20x = 3.60 → x = 1.80 m. The mirror is 1.20 m from the wall.
(c) tan θᵢ = 1.80 ÷ 1.20 = 1.5, so θᵢ = 56.3°. Check from the target side: tan θᵣ = 1.20 ÷ 0.80 = 1.5, so θᵣ = 56.3° as well.
| Point | What earns it |
|---|---|
| 1 | Ray diagram with a normal at the mirror, equal angles either side of it, and arrows |
| 1 | Uses equal angles to set up similar triangles (or equal tangents) linking x, 1.20 m, 0.80 m and 3.00 − x |
| 1 | x = 1.80 m |
| 1 | θᵢ = 56.3°, measured from the vertical normal (accept 56°) |
Common error: putting the mirror half-way (1.50 m) “by symmetry”. That only works if the laser and target are at the same height. At 1.50 m the angle of incidence would be 51.3° but the angle needed to reach the target would be 61.9°, so the law of reflection is broken. Another common error: giving 33.7°, which is the angle between the beam and the floor.
Question 6 (constructed response · core)
At night, a driver notices that on a dry road the white lines ahead are clearly lit by her headlights. After heavy rain, with a layer of water on the road, the lines and the road ahead look much darker, while oncoming drivers complain of glare.
(a) Sketch two ray diagrams, one for the dry road and one for the wet road, each showing three parallel headlight rays striking the road and reflecting. (b) Using your diagrams, explain why the road looks darker to the driver when it is wet. (c) Explain why the law of reflection applies to both surfaces even though they reflect light so differently.
Worked solution and suggested Marlbridge rubric
(a) Dry road: a rough surface drawn as many small tilted patches, each with its own normal; the three reflected rays leave in different directions, some of them back towards the car. Wet road: a smooth flat surface; the three reflected rays leave parallel to each other, forward and away from the car, at the same angle to the normal as they arrived.
(b) The dry road gives diffuse reflection. Because the normal changes from patch to patch, some light from each lit point is reflected back towards the driver’s eyes, so she sees the lines. The water gives a smooth surface and specular reflection. The headlight beams strike at a large angle of incidence and all reflect forward at the same large angle. Almost none comes back toward the driver, so the road looks dark to her, and the forward-reflected light reaches oncoming drivers as glare.
(c) At every small patch of either surface, the angle of reflection equals the angle of incidence measured from that patch’s normal. The difference is that the normal is the same everywhere on smooth water but varies on the rough road.
| Point | What earns it |
|---|---|
| 1 | Dry-road sketch: parallel incident rays leave in different directions, with normals at different angles |
| 1 | Wet-road sketch: parallel incident rays leave parallel, forward, with equal angles to a single normal direction |
| 1 | Links darkness of the wet road to specular reflection sending light forward, away from the driver |
| 1 | Links visibility on the dry road to diffuse reflection sending some light back towards the driver |
| 1 | States that θᵢ = θᵣ holds at each patch and the difference comes from the direction of the normal |
Do not award the third point for “the water makes the road dark” with no link to the direction in which the light is reflected.
Question 7 (constructed response · stretch)
Two plane mirrors meet at exactly 90°. A ray strikes mirror 1 with an angle of incidence of 35°, at a point 0.40 m from the corner, and then reflects from mirror 2.
(a) Calculate the angle of incidence at mirror 2. (b) Show that, for any angle of incidence θ on mirror 1 (as long as the ray reaches mirror 2), the final ray travels in exactly the opposite direction to the original ray. (c) A student says: “So the final ray goes back along exactly the same line as the incoming ray.” Evaluate this claim. Use the data in the question.
Worked solution and suggested Marlbridge rubric
(a) After mirror 1, the ray makes 90° − 35° = 55° with mirror 1. In the triangle formed by the corner, the two reflection points and the ray, the angles are 90° (corner) and 55°, so the ray makes 180° − 90° − 55° = 35° with mirror 2. The angle of incidence at mirror 2 is 90° − 35° = 55°.
(b) By the same triangle argument, if θᵢ = θ on mirror 1, then θᵢ = 90° − θ on mirror 2. Each reflection turns the direction of travel by 180° − 2θᵢ, and both turns are in the same sense. Total turn = (180° − 2θ) + (180° − 2(90° − θ)) = 180° − 2θ + 2θ = 180°. A 180° turn means the final ray is antiparallel to the incoming ray, whatever θ is.
(c) The claim is not correct in general. The ray hits mirror 2 at a distance 0.40 m × tan 55° = 0.571 m from the corner. The outgoing ray starts from this point, which is not on the incoming line. The two rays are parallel (opposite directions) but offset: here the lines are about 0.66 m apart. They would only coincide for a ray aimed exactly at the corner. In a small corner reflector the offset is tiny, so the light returns very close to its source, which is the idea behind the corner-shaped reflectors on bicycles.
| Point | What earns it |
|---|---|
| 1 | Uses θᵣ = 35° and the angle sum of the triangle to find the ray’s angle with mirror 2 |
| 1 | θᵢ at mirror 2 = 55° |
| 1 | Shows that the angles of incidence on the two mirrors add to 90° |
| 1 | Adds the two turns (or uses components) to reach a total turn of 180° for any θ |
| 1 | Rejects the claim: the outgoing ray is parallel but displaced, with evidence such as the second reflection point (0.571 m from the corner) not lying on the incoming line |
Accept a component argument for (b): reflection at mirror 1 reverses the component of the direction perpendicular to mirror 1; reflection at mirror 2 reverses the other component; both components reversed means the direction is reversed.
How did you do?
- Q1 or Q4 wrong: re-read Figure 2 and “Turning a mirror” in the study guide. Check the angle is from the normal.
- Q2 or Q3 wrong: revisit “Light as a ray” and “When the ray model works”.
- Q5 wrong: redo it with a large, careful ray diagram first.
- Q6 incomplete: compare the two panels of Figure 3.
- Q7 incomplete: work through Worked example 2 again, then repeat with 90°.
Then tick off the topic checklist.