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HKDSE Physics Recap: Reflection and Refraction of Light | Vanilla (2026)

HKDSE Physics — Reflection and Refraction of Light

Subject: Physics

Vanilla (2026)

Issue date: Friday, 2 October 2026

Total: 40 marks  |  Suggested time: 50–60 minutes  |  Speed of light in vacuum: c = 3.00 × 108 m s−1; take the refractive index of air as 1.00

01

Section A — Reflection and Refraction of Light

Multiple Choice

Each question has one correct answer. Choose the best option. (1 mark each)

  1. A boy of height 1.80 m stands in front of a vertical plane mirror fixed on a wall, as shown in Fig. 1. His eyes are 1.70 m above the floor. The lower and upper edges of the mirror are 0.95 m and 1.80 m above the floor respectively. The mirror is wide enough for him. What is the height above the floor of the lowest point of his body that he can see in the mirror?
    floor wall plane mirror eyes: 1.70 m 1.80 m 0.95 m 1.80 m
    Fig. 1 — Side view (not drawn to scale)
    1. 0 m
    2. 0.10 m
    3. 0.20 m
    4. 0.95 m
  2. Two plane mirrors M1 and M2 are placed perpendicular to each other. A ray of light in the plane of the paper is incident on M1 at an angle of incidence of 25° and is then reflected to M2, as shown in Fig. 2. What is the angle of reflection at M2?
    M1 M2 25° ?
    Fig. 2 — Plane of the paper (not drawn to scale)
    1. 25°
    2. 65°
    3. 50°
    4. 115°
  3. A boy walks towards a vertical plane mirror at 1.2 m s−1. At the same time, the mirror moves at 0.4 m s−1 in the same direction as the boy, as shown in Fig. 3. Both speeds are measured relative to the ground. What is the speed of the boy’s image relative to the boy?
    boy 1.2 m s−1 plane mirror 0.4 m s−1
    Fig. 3 — Both arrows show velocities relative to the ground (not drawn to scale)
    1. 0.4 m s−1
    2. 0.8 m s−1
    3. 2.4 m s−1
    4. 1.6 m s−1
  4. A point source S is 1.0 m above a horizontal plane mirror. Light from S is reflected by the mirror and reaches a point P on a vertical wall, 2.0 m above the mirror. The horizontal distance between S and the wall is 3.0 m, as shown in Fig. 4. What is the horizontal distance between the point of reflection on the mirror and the wall?
    horizontal plane mirror wall P S 1.0 m 2.0 m 3.0 m
    Fig. 4 — Side view (not drawn to scale)
    1. 2.0 m
    2. 1.0 m
    3. 1.5 m
    4. 2.5 m
  5. A ray of light in air is incident on a glass surface of refractive index 1.50 at an angle of incidence θ, as shown in Fig. 5. The reflected ray is perpendicular to the refracted ray. What is θ?
    air glass (n = 1.50) θ 90°
    Fig. 5 — Reflected and refracted rays at a glass surface (not drawn to scale)
    1. 33.7°
    2. 41.8°
    3. 56.3°
    4. 48.6°
  6. A layer of water of refractive index 1.33 lies on top of a layer of glass of refractive index 1.50. The boundaries are parallel. A ray of light in air is incident on the water at an angle of incidence of 45°, as shown in Fig. 6. What is the angle between the ray in the glass and the normal?
    air water (n = 1.33) glass (n = 1.50) 45° ?
    Fig. 6 — Ray passing through water into glass (not drawn to scale)
    1. 28°
    2. 32°
    3. 39°
    4. 45°
  7. Fig. 7 shows a ray of light passing from medium X into medium Y. The angle of incidence is 30° and the angle of refraction is 45°. The speed of light in X is 1.40 × 108 m s−1. What is the refractive index of Y?
    medium X medium Y 30° 45°
    Fig. 7 — Refraction at the boundary between X and Y (not drawn to scale)
    1. 1.07
    2. 2.14
    3. 3.03
    4. 1.52
  8. In Fig. 8, a ray of light in air is incident at an angle of incidence of 45° on a block of transparent plastic of refractive index √2 (≈ 1.41). What is the angle of refraction?
    air plastic (n = √2) 45° ?
    Fig. 8 — Refraction at an air–plastic boundary (not drawn to scale)
    1. 32°
    2. 30°
    3. 60°
    4. 45°
  9. A ray of light in air enters a glass prism through face AB along the normal, as shown in Fig. 9. The cross-section of the prism is a right-angled isosceles triangle ABC, with the right angle at B. The refractive index of the glass is 1.50. What happens to the ray when it reaches face AC?
    A B C glass (n = 1.50) air incident ray ?
    Fig. 9 — Right-angled isosceles glass prism (not drawn to scale)
    1. It is refracted into the air, bending away from the normal.
    2. It passes straight through into the air without changing direction.
    3. It is not refracted across the face AC.
    4. It travels along face AC.
  10. In Fig. 10, a narrow laser beam leaves a hole O in a vertical screen and strikes a plane mirror M at normal incidence, so the reflected beam returns to O. M is 2.00 m from the screen. M is then rotated through 5.0° about a vertical axis through its centre. The reflected beam now hits the screen at P. What is the distance OP?
    screen plane mirror M 5.0° incident beam reflected beam O P 2.00 m
    Fig. 10 — Top view (not drawn to scale)
    1. 0.17 m
    2. 0.35 m
    3. 0.70 m
    4. 1.00 m
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10 marks
02

Section B — Reflection and Refraction of Light

True or False

For each statement, state whether it is True or False. If it is false, rewrite the statement correctly. (1 mark each)

  1. The image formed by a plane mirror is virtual, upright and laterally inverted. The image is as far behind the mirror as the object is in front of it.
  2. A person can see the whole of his image in a vertical plane mirror only if the mirror is at least half his height. This minimum height becomes smaller if he stands farther away from the mirror.
  3. A ray of light is incident on a plane mirror and the incident ray is kept fixed. When the mirror is rotated through an angle θ about an axis which lies in the mirror surface and is perpendicular to the plane of incidence, the reflected ray turns through 2θ.
  4. When light passes from air into glass, both its speed and its frequency decrease.
  5. A ray of light passes from medium 1 into medium 2 and bends towards the normal. Light travels more slowly in medium 2 than in medium 1, and medium 2 has the greater refractive index.
  6. When light passes from one transparent medium into another, the angle of refraction is always smaller than the angle of incidence.
  7. A ray of light passes through a parallel-sided glass plate. The emergent ray is always parallel to the incident ray, whatever the media on the two sides of the plate (for example, air on one side and water on the other).
  8. A point object is fixed in front of a plane mirror. The mirror is moved a distance d directly away from the object. The image then moves a distance 2d away from the object.
  9. A girl walks towards a fixed plane mirror at speed v. Her image moves towards the mirror at speed 2v.
  10. A ray of light is incident normally on an air–glass boundary. It enters the glass without any change in direction, but its speed decreases.
10 marks
03

Section C — Comprehensive Structured Question 1

Reflection of Light: Where Can the Image Be Seen?

Fig. 11 shows the top view of a room. A plane mirror of width 0.80 m is fixed on a wall. A small lamp L is placed 1.0 m in front of the mirror, directly opposite the centre of the mirror. An observer walks along a straight path which is parallel to the wall and 3.0 m from the wall. L may be treated as a point source.

Wall plane mirror (width 0.80 m) lamp L observer Path of the observer (parallel to the wall) N Q 2.0 m 1.0 m 3.0 m
Fig. 11 — Top view: L is directly opposite the centre of the mirror (drawn to scale)

(a) Find the distance between the image of L in the mirror and the path of the observer. (1 mark)

(b) By considering similar triangles, show that the observer can see the image of L along a length of 3.2 m of his path. You may draw on Fig. 11. (3 marks)

(c) L is now moved along the perpendicular from the centre of the mirror until it is 2.0 m from the mirror. Calculate the length of the path along which the observer can now see the image of L. Is this length longer or shorter than that in (b)? (2 marks)

(d) L is returned to its original position, 1.0 m from the mirror. Consider the ray from L which strikes one edge of the mirror and is reflected towards one end of the 3.2 m length in (b). Calculate the angle of incidence of this ray. (2 marks)

(e) L is at its original position, 1.0 m from the mirror. In Fig. 11, N is the point on the path directly opposite L, and the observer now stands at point Q, which is 2.0 m to the right of N. Can the observer at Q see the image of L in the mirror? Support your answer with a calculation (or by drawing on Fig. 11). (2 marks)

10 marks
04

Section C — Comprehensive Structured Question 2

Refraction of Light: An Unknown Liquid Under a Glass Plate

Fig. 12 shows a transparent tank containing an unknown liquid of depth 12.0 cm. A plane mirror is placed at the bottom of the tank. A glass plate of thickness 2.0 cm and refractive index 1.50 lies on the surface of the liquid.

A narrow ray of light in air is incident on the top surface of the glass plate at an angle of incidence of 53.1°. The ray passes through the glass and the liquid, is reflected by the mirror and leaves through the glass plate. The angle between the ray in the liquid and the normal is 36.9°.

53.1° 36.9° air glass plate (n = 1.50, 2.0 cm thick) unknown liquid (depth 12.0 cm) plane mirror
Fig. 12 — Path of a light ray through a glass plate and a liquid (not drawn to scale)

(a) Calculate the speed of light in the glass. (1 mark)

(b) Calculate the angle between the ray in the glass and the normal. (2 marks)

(c)

  1. Show that the refractive index of the liquid is 1.33. (2 marks)
  2. Calculate the speed of light in the liquid. (1 mark)

(d) The ray leaves the top surface of the glass plate and travels in air.

  1. State the angle between the emergent ray and the normal. Explain your answer. (1 mark)
  2. Calculate the distance between the point where the ray enters the glass plate and the point where it leaves the glass plate, measured along the top surface of the plate. (3 marks)
10 marks

Question 1 — Multiple Choice

1. C
2. B
3. D
4. A
5. C
6. A
7. D
8. B
9. C
10. B
  1. C — 0.20 m. Let the eye be at height 1.70 m and let a point on the body be at height h. By the law of reflection and similar triangles, the ray from the point to the eye meets the mirror at the height halfway between them: (h + 1.70) ÷ 2. This does not depend on the boy’s distance from the mirror. The lowest point visible is seen through the lower edge of the mirror: (h + 1.70) ÷ 2 = 0.95, so h = 0.20 m. (The top of his head, 1.80 m, is seen at a mirror height of 1.75 m, which is within the mirror.)
  2. B — 65°. After reflection at M1, the ray makes 25° with the normal to M1. Since M1 ⊥ M2, this normal is parallel to the surface of M2. The ray therefore makes 25° with the surface of M2, so the angle of incidence (and reflection) at M2 is 90° − 25° = 65°. (115° = 180° − 65°.)
  3. D — 1.6 m s−1. The boy–mirror distance decreases at 1.2 − 0.4 = 0.8 m s−1. The image is as far behind the mirror as the boy is in front, so the boy–image distance is twice the boy–mirror distance and decreases at 2 × 0.8 = 1.6 m s−1. (2.4 m s−1 would apply to a stationary mirror.)
  4. A — 2.0 m. The image S′ of S is 1.0 m below the mirror. The reflected ray appears to come straight from S′ to P, so S′, the point of reflection and P lie on one straight line. P is 3.0 m above S′ and 3.0 m away horizontally. The line reaches the mirror level after rising 1.0 m, at a horizontal distance x from S where 1.0x = 3.03.0, so x = 1.0 m. The distance from the wall is 3.0 − 1.0 = 2.0 m.
  5. C — 56.3°. The angle between the reflected and refracted rays is 180° − θ − r = 90°, so r = 90° − θ. Then sin θ = 1.50 sin r = 1.50 cos θ, so tan θ = 1.50 and θ = 56.3°. (33.7° is the angle of refraction.)
  6. A — 28°. At each parallel boundary, n sin θ is constant, so the water layer does not affect the final angle: 1.00 × sin 45° = 1.50 × sin r, sin r = 0.471 and r = 28.1°. (The angle in the water is 32°.)
  7. D — 1.52. nX = c ÷ vX = 3.00 × 108 ÷ 1.40 × 108 = 2.14. From nX sin 30° = nY sin 45°, nY = 2.14 × 0.5 ÷ 0.707 = 1.52.
  8. B — 30°. sin r = sin 45° ÷ √2 = 0.707 ÷ 1.414 = 0.5, so r = 30°. (32° comes from dividing the angles, 45° ÷ 1.41. 60° is 90° − 30°.)
  9. C — It is totally internally reflected towards face BC. The ray enters face AB along the normal, so it is not deviated. Since the prism is right-angled isosceles, face AC makes 45° with the ray, so the angle of incidence at AC is 45°. The critical angle of the glass is given by sin c = 1 ÷ 1.50 = 0.667, so c = 41.8°. Since 45° > 41.8°, total internal reflection occurs at AC. (A and B would need an angle of incidence smaller than 41.8°, and D would need an angle of incidence of exactly 41.8°.) The reflected ray then meets face BC along the normal and leaves the prism without changing direction.
  10. B — 0.35 m. Rotating the mirror by 5.0° rotates the reflected beam by 2 × 5.0° = 10.0°. OP = 2.00 × tan 10.0° = 0.353 m. (0.17 m comes from using 5.0° instead of 10.0°.)

Question 2 — True or False

  1. True.
  2. False. The minimum height is half the person’s height, and it does not depend on how far the person stands from the mirror (similar triangles).
  3. True.
  4. False. The speed decreases, but the frequency stays the same, because it is fixed by the source.
  5. True. Bending towards the normal means sin r < sin i, so n2 > n1 and v2 < v1.
  6. False. The angle of refraction is smaller than the angle of incidence only when the light enters a medium with a greater refractive index (a slower medium). When it enters a medium with a smaller refractive index, the angle of refraction is larger. (At normal incidence the two angles are equal.)
  7. False. The emergent ray is parallel to the incident ray only when the media on both sides of the plate are the same (have the same refractive index).
  8. True. The object–mirror distance increases by d, so the image–mirror distance also increases by d. The object–image distance therefore increases by 2d, and with the object fixed, the image moves 2d away from the object.
  9. False. The image moves towards the mirror at speed v. It approaches the girl at a relative speed of 2v.
  10. True. For i = 0°, sin r = 0 and r = 0°, so the direction is unchanged, but v = c ÷ n decreases.

Question 3 — Reflection of Light: Where Can the Image Be Seen?

(a) Distance from the image to the observer’s path

The image L′ is 1.0 m behind the mirror (same distance as the object is in front).
Distance = 1.0 + 3.0 = 4.0 m (1)

(b) Length of path (similar triangles)

The observer sees the image only when a ray from L′ passes through the mirror, so the limits are given by the lines from L′ through the two edges of the mirror. (1)
The triangle formed by L′ and the mirror is similar to the triangle formed by L′ and the ends of the visible length ℓ:
0.80ℓ = 1.04.0 (1)
ℓ = 0.80 × 4.0 = 3.2 m (1)

(c) Lamp 2.0 m from the mirror

The image is 2.0 m behind the mirror and 2.0 + 3.0 = 5.0 m from the path.
0.80ℓ = 2.05.0 (1)
ℓ = 2.0 m, which is shorter than 3.2 m, because the mirror subtends a smaller angle at the farther image. (1)

(d) Angle of incidence at the edge of the mirror

The edge of the mirror is 0.40 m from the centre line and 1.0 m from L along the normal direction:
tan i = 0.401.0 (1)
i = 21.8° (1)
Check: The end of the visible length is 1.60 m from the centre line, so the reflected ray from the edge travels 1.60 − 0.40 = 1.20 m sideways over 3.0 m: tan r = 1.20 ÷ 3.0 = 0.40. This agrees with the law of reflection (r = i).

(e) Can the observer at Q see the image of L?

The observer at Q sees the image L′ only if the straight line from Q to L′ passes through the mirror. L′ is 4.0 m from the path (from (a)) and 1.0 m from the mirror, and Q is 2.0 m to one side of the centre line.
By similar triangles, the line from Q to L′ crosses the mirror at a sideways distance from the centre line of
2.0 × 1.04.0 = 0.50 m (1)
The mirror extends only 0.80 ÷ 2 = 0.40 m on each side of the centre line, and 0.50 m > 0.40 m. The line misses the mirror, so the observer at Q cannot see the image of L. (1)
Check: From (b), the image can be seen only within 3.2 ÷ 2 = 1.6 m on either side of N. Q is 2.0 m from N, which is outside this range, so the answer is consistent. Accept a correct ray diagram drawn on Fig. 11 (line from L′ to Q meeting the wall outside the mirror) for both marks.
Marking note: In (b), accept a labelled ray diagram with the image L′ and the two limiting rays in place of a written explanation for the first mark. Award the last mark in (b) only if the final answer is consistent with the ratio 0.80 : ℓ = 1.0 : 4.0.

Question 4 — Refraction of Light: An Unknown Liquid Under a Glass Plate

(a) Speed of light in glass

v = cn = 3.00 × 1081.50 = 2.00 × 108 m s−1 (1)

(b) Angle in the glass

sin 53.1°sin rg = 1.50 (1)
sin rg = 0.7997 ÷ 1.50 = 0.533
rg = 32.2° (1)

(c) The liquid

  1. Applying n sin θ = constant from the air to the liquid (the glass plate does not affect the result):
    1.00 × sin 53.1° = nL × sin 36.9° (1)
    nL = 0.79970.6004 = 1.33 (1)

    Alternative: 1.50 sin 32.2° = nL sin 36.9°, which gives the same result.

  2. v = cnL = 3.00 × 1081.33 = 2.25 × 108 m s−1 (1)

(d) Emergent ray

  1. 53.1°. The mirror reflects the ray with the angle of reflection equal to the angle of incidence, so the angles in the liquid and in the glass on the way up are the same as on the way down (32.2° in the glass). At the glass–air boundary, sin θ = 1.50 sin 32.2°, so θ = 53.1° (the path is reversible). (1)

  2. Horizontal distance for one journey down (or up):
    2.0 tan 32.2° + 12.0 tan 36.9° (1)
    = 2.0(0.630) + 12.0(0.751) = 1.26 + 9.01 = 10.27 cm (1)
    Total distance = 2 × 10.27 = 20.5 cm (1)
Common errors: (1) using the angle of incidence in air (53.1°) in the glass or in the liquid; (2) forgetting that the ray crosses both layers twice; (3) using sin instead of tan for the horizontal displacement.
Mark weighting: Section A (10) + Section B (10) + Question 3 (10) + Question 4 (10) = 40 marks.
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