By /
HKDSE Physics Recap: Mirror Images, Reflection Calculations and Refraction | Vanilla (2026)

HKDSE Physics — Mirror Images, Reflection Calculations and Refraction of Light

Subject: Physics

Vanilla (2026)

Issue date: Thursday, 8 October 2026

Total: 70 marks  |  Suggested time: 90–100 minutes  |  Speed of light in vacuum: c = 3.00 × 108 m s−1; take the refractive index of air as 1.00. Use a sharp pencil and a ruler for ray diagrams: real rays are solid lines with arrows; virtual rays and images are dashed lines. Total internal reflection is not involved in this paper. Give numerical answers to 3 significant figures unless stated otherwise.

01

Section A — Images, Reflection and Refraction of Light

Multiple Choice

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

  1. Which of the following statements about the image formed by a plane mirror are correct?
    (1) The image is virtual because the reflected rays only appear to diverge from the image position, so the image cannot be formed on a screen.
    (2) If half of the mirror is covered by black paper, the image of an object is still formed at the same position, but the region from which it can be seen becomes smaller.
    (3) The image is laterally inverted, which means that the upper and the lower parts of the object are interchanged in the image.
    1. (1) and (2) only
    2. (1) and (3) only
    3. (2) and (3) only
    4. (1), (2) and (3)
  2. Two plane mirrors M1 and M2 are placed parallel to each other, facing each other and 4.0 m apart. A small object O is placed between them, 1.0 m from M1. Let I1 be the image of O formed by M1 alone and I2 be the image of O formed by M2 alone. What is the distance between I1 and I2?
    1. 4.0 m
    2. 6.0 m
    3. 8.0 m
    4. 10.0 m
  3. A man of height 1.76 m has his eyes 1.66 m above the floor. He wants to see his whole body at once in a vertical plane mirror fixed on a wall. What are the minimum length of the mirror and the maximum height of the lower edge of the mirror above the floor?
    1. Length 0.83 m; lower edge at most 0.88 m above the floor
    2. Length 0.88 m; lower edge at most 1.66 m above the floor
    3. Length 1.76 m; lower edge at most 0.83 m above the floor
    4. Length 0.88 m; lower edge at most 0.83 m above the floor
  4. A boy walks towards a vertical plane mirror at 0.50 m s−1. At the same time the mirror moves in the same direction as the boy (that is, away from him) at 0.30 m s−1. Both speeds are measured relative to the ground. What is the speed of the boy’s image relative to the boy?
    1. 0.20 m s−1
    2. 0.40 m s−1
    3. 1.10 m s−1
    4. 1.60 m s−1
  5. A simple periscope uses two parallel plane mirrors, each inclined at 45° to a vertical tube, so that an observer can look over a wall. Which of the following describes the final image seen by the observer?
    1. Virtual, upright and laterally inverted
    2. Real, upright and not laterally inverted
    3. Virtual, upright and not laterally inverted
    4. Real, inverted and laterally inverted
  6. A small lamp L is 0.60 m in front of the centre of a plane mirror of width 0.30 m fixed on a wall. An observer walks along a straight line parallel to the wall and 2.40 m from the wall, as shown in Fig. 1. What is the length of this line along which the observer can see the image of L in the mirror?
    wall plane mirror (width 0.30 m) L path of the observer (parallel to the wall) 0.60 m 2.40 m
    Fig. 1 — Top view (drawn to scale)
    1. 0.30 m
    2. 0.90 m
    3. 1.20 m
    4. 1.50 m
  7. A narrow beam of light strikes a plane mirror at normal incidence and is reflected back along its own path. The mirror is then rotated about a vertical axis in its surface, while the incident beam is kept fixed. The reflected beam falls on a straight scale placed 2.50 m from the mirror, perpendicular to the original beam, and the spot of light moves by 0.50 m from its original position. Through what angle was the mirror rotated?
    1. 2.8°
    2. 5.7°
    3. 11.3°
    4. 22.6°
  8. A girl places a small plane mirror M flat on the level ground and moves until she can just see the top T of a tree in the mirror, as shown in Fig. 2. Her eyes are 1.60 m above the ground. M is 2.4 m from her feet and 36 m from the foot of the tree. What is the height of the tree?
    M eye T 1.60 m 2.4 m 36 m
    Fig. 2 — Side view (not drawn to scale)
    1. 1.5 m
    2. 16 m
    3. 24 m
    4. 54 m
  9. An arrow AB points to the right, as shown in Fig. 3 (the squares are 1 cm by 1 cm). A plane mirror is inclined at 45° to the arrow, and the reflecting side of the mirror faces the arrow (the hatched side is the back). Which of the following describes the image A′B′ of the arrow, as seen on the page?
    plane mirror B A
    Fig. 3 — Arrow AB in front of a plane mirror inclined at 45°
    1. A vertical arrow, pointing upwards
    2. A horizontal arrow, pointing to the left
    3. A horizontal arrow, pointing to the right
    4. A vertical arrow, pointing downwards
  10. Two plane mirrors M1 and M2 are placed at right angles to each other, as shown in Fig. 4. A small object O is 3.0 m from M1 and 2.0 m from M2. Light from O is reflected first by one mirror and then by the other, forming an image I. What is the distance between O and I?
    M1 M2 O 3.0 m 2.0 m
    Fig. 4 — Top view: the mirrors are perpendicular to the page and their reflecting sides face O
    1. 5.0 m
    2. 7.2 m
    3. 10.0 m
    4. 14.4 m
  11. A ray of monochromatic light travels from air into glass. Which of the following statements are correct?
    (1) The frequency of the light decreases.
    (2) The wavelength of the light decreases.
    (3) If the angle of incidence is not zero, the angle of refraction is smaller than the angle of incidence.
    1. (1) and (2) only
    2. (1) and (3) only
    3. (2) and (3) only
    4. (1), (2) and (3)
  12. Monochromatic light has a wavelength of 480 nm in water of refractive index 1.33. The light then passes from the water into glass of refractive index 1.52. What is the wavelength of the light in the glass?
    1. 420 nm
    2. 480 nm
    3. 549 nm
    4. 638 nm
  13. Fig. 5 shows a ray of light passing from air into a transparent medium. In air the ray makes an angle of 40° with the boundary, and in the medium the ray makes an angle of 60° with the boundary. What is the speed of light in the medium?
    air transparent medium 40° 60°
    Fig. 5 — The angles are measured from the boundary (not drawn to scale)
    1. 1.96 × 108 m s−1
    2. 2.23 × 108 m s−1
    3. 2.33 × 108 m s−1
    4. 4.04 × 108 m s−1
  14. A ray of light in air is incident at 60° on a rectangular glass block of thickness 5.0 cm and refractive index 1.50. The ray passes through the block and emerges from the opposite parallel face into air. What is the lateral displacement, i.e. the perpendicular distance between the emergent ray and the extension of the incident ray?
    1. 2.6 cm
    2. 3.5 cm
    3. 5.1 cm
    4. 6.1 cm
  15. A ray of light travels in a transparent plastic of refractive index 1.60 and is incident on a boundary with water of refractive index 1.33. The angle of incidence is 30°. What is the angle of refraction in the water?
    1. 24.6°
    2. 30.0°
    3. 37.0°
    4. 53.1°
  16. A student shines a ray of light from air into a block of glass at various angles of incidence i and measures the angle of refraction r each time. Fig. 6 shows the graph of sin r against sin i. What is the speed of light in the glass?
    0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 sin i sin r 0.75 0.48
    Fig. 6 — Graph of sin r against sin i for light passing from air into glass
    1. 1.44 × 108 m s−1
    2. 2.25 × 108 m s−1
    3. 4.69 × 108 m s−1
    4. 1.92 × 108 m s−1
  17. A ray of light in air is incident on the surface of water of refractive index 1.33 in a tank, at an angle of incidence of 45°, as shown in Fig. 7. The water is 8.0 cm deep. What is the horizontal distance x between the point where the ray enters the water and the point where it reaches the bottom of the tank?
    air water (n = 1.33) 45° 8.0 cm x
    Fig. 7 — Side view of a tank of water
    1. 4.3 cm
    2. 5.0 cm
    3. 6.0 cm
    4. 8.0 cm
  18. Monochromatic light of frequency 5.00 × 1014 Hz in air is incident at an angle of incidence of 40° on the surface of a liquid. The wavelength of the light in the liquid is 450 nm. What is the angle of refraction?
    1. 28.8°
    2. 30.0°
    3. 40.0°
    4. 59.0°
  19. A thin layer of water of refractive index 1.33 lies between a block of glass of refractive index 1.50 and the air above it. The boundaries are parallel. A ray of light in the glass is incident on the glass–water boundary at an angle of incidence of 30°, passes through the water and emerges into the air. What is the angle between the emergent ray in the air and the normal?
    1. 30.0°
    2. 34.3°
    3. 45.0°
    4. 48.6°
  20. A ray of light in air is incident on the plane surface of a transparent block at an angle of incidence of 30°, as shown in Fig. 8. The angle between the reflected ray and the refracted ray is 130°. What is the refractive index of the block?
    air transparent block 30° 130°
    Fig. 8 — Reflection and refraction at the surface of a block (not drawn to scale)
    1. 0.68
    2. 1.18
    3. 1.46
    4. 1.93
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
20 marks
02

Section B — Comprehensive Structured Question 1

Images Formed by Plane Mirrors: Ray-Diagram Drawing

Fig. 9 shows four situations (a) to (d). In each, an object AB (an arrow with its head at A) is placed in front of a plane mirror; the hatched side of the mirror is its back. The eye of an observer is also shown. The side of each large square represents 1.0 cm. Use a sharp pencil and a ruler.

Fig. 9 — Objects, mirrors and eyes (1 large square = 1.0 cm)

(a)–(d) For each of the panels (a), (b), (c) and (d) of Fig. 9:

  1. Draw the image A′B′ of the object AB, using a dashed line with the arrowhead at A′. (1 mark)
  2. Draw a ray from A which is reflected by the mirror and enters the eye. Mark the point of reflection P, and use dashed lines for any virtual ray (construction line). (1 mark)

[8 marks in total for the four panels]

(e)

  1. In panel (c), find the distance between B and its image B′. (1 mark)
  2. In panel (d), state the angle between the object AB and its image A′B′. (1 mark)
10 marks
03

Section B — Comprehensive Structured Question 2

Reflection of Light: What Can a Boy See in a Wall Mirror?

Fig. 10 shows, in side view, a boy standing 1.50 m in front of a vertical plane mirror fixed on a wall. The boy is 1.72 m tall and his eyes E are 1.62 m above the floor. The upper edge T and the lower edge L of the mirror are 1.50 m and 0.70 m above the floor respectively. The mirror is wide enough for the boy.

Fig. 10 — Side view (drawn to scale: 1 m is represented by 100 px)

(a) State the distance between the boy’s eyes E and the image E′ of his eyes in the mirror. (1 mark)

(b) By considering similar triangles, show that the highest point on his body that he can see in the mirror is 1.38 m above the floor. You may draw on Fig. 10. (3 marks)

(c) Find the distance from the mirror of the nearest point on the floor that he can see in the mirror. Hence state, with a reason, whether he can see his feet. (3 marks)

(d) The upper edge of the mirror is raised so that the boy can just see the top of his head. Find the new height of the upper edge above the floor. (2 marks)

(e) With the original mirror, the boy walks backwards until he is 3.00 m from the mirror. Does the highest point he can see change? Support your answer with a calculation. (1 mark)

10 marks
04

Section B — Comprehensive Structured Question 3

Reflection of Light: Where Can the Image of a Lamp Be Seen?

Fig. 11 shows the top view of a room. A plane mirror of width 0.60 m is fixed on a wall; its ends are E1 and E2. A small lamp L, which may be treated as a point source, is 0.50 m in front of the mirror. The foot of the perpendicular from L to the wall is 0.20 m from E1. An observer walks along a straight path which is parallel to the wall and 2.50 m from the wall. N is the point on the path directly opposite L.

Fig. 11 — Top view (drawn to scale: 1 m is represented by 100 px)

(a) On Fig. 11, locate the image L′ of the lamp. Draw the two limiting rays from L which are reflected at E1 and E2 and reach the observer’s path, and mark the ends X and Y of the part of the path from which the image can be seen. (3 marks)

(b) By considering similar triangles, show that the observer can see the image of L along a length of 3.6 m of his path. (3 marks)

(c) Find the distance from N to each end of the 3.6 m length in (b). (2 marks)

(d) The half of the mirror nearer E2 (the right-hand half in Fig. 11) is now covered by a black card. Find the length of the path from which the image of L can now be seen, and state where this length lies relative to N. (2 marks)

10 marks
05

Section B — Comprehensive Structured Question 4

Refraction of Light: A Ray Passing Through a Glass Block

Fig. 12 shows a rectangular glass block of thickness 5.0 cm and refractive index 1.60. A narrow ray of monochromatic light of frequency 5.00 × 1014 Hz in air is incident on the upper face of the block at an angle of incidence i, where sin i = 0.800 (i = 53.1°).

air glass block (n = 1.60) i 5.0 cm
Fig. 12 — Side view (scale: 1 cm is represented by 20 px)

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

(b) Calculate the wavelength of the light (i) in air and (ii) in the glass. (2 marks)

(c) Show that the angle of refraction in the glass is 30.0°. (2 marks)

(d) Complete the path of the ray on Fig. 12 until it leaves the block, marking the angles at both faces. Hence calculate the lateral displacement of the emergent ray, i.e. the perpendicular distance between the emergent ray and the extension of the incident ray. (3 marks)

(e) Calculate the time taken for the light to travel through the block along its path. (2 marks)

10 marks
06

Section B — Comprehensive Structured Question 5

Refraction and Reflection: A Tank with Two Perpendicular Mirrors

Fig. 13 shows, in side view, a tank containing water to a depth of 12.0 cm. A plane mirror M1 lies on the bottom of the tank, and the right-hand wall of the tank is a plane mirror M2, perpendicular to M1. M2 is 12.0 cm horizontally from the point O where a narrow ray of light in air strikes the water surface. The angle of incidence at O is 53.1° and the angle of refraction in the water is 36.9°. The ray is reflected at P on M1, then at Q on M2, and finally leaves the water at R.

Use: sin 53.1° = 0.800; sin 36.9° = 0.600; cos 36.9° = 0.800; tan 36.9° = 0.750.

air water plane mirror M1 plane mirror M2 53.1° 36.9° O P Q R 12.0 cm 12.0 cm
Fig. 13 — Path of a ray in a tank of water (not drawn to scale)

(a) Show that the refractive index of water is 1.33. (1 mark)

(b) Show that Q is 4.00 cm above M1. (3 marks)

(c) Find the horizontal distance OR. (2 marks)

(d) Find the angle between the emergent ray in air and the normal. How is the direction of the emergent ray related to that of the incident ray? (2 marks)

(e) Calculate the time taken for light to travel from O to R in the water. (2 marks)

10 marks

Question 1 — Multiple Choice

1. A
2. C
3. D
4. B
5. C
6. D
7. B
8. C
9. D
10. B
11. C
12. A
13. A
14. A
15. C
16. D
17. B
18. A
19. D
20. C
  1. A — (1) and (2) only. (1) and (2) are correct. (3) is wrong: lateral inversion means that left and right are interchanged, not top and bottom.
  2. C — 8.0 m. Put M1 at 0, O at 1.0 m and M2 at 4.0 m. I1 is 1.0 m behind M1 (at −1.0 m). I2 is 3.0 m behind M2 (at 7.0 m). I1I2 = 8.0 m (= 2 × 4.0 m).
  3. D — length 0.88 m; lower edge at most 0.83 m high. The ray from a point at height h to the eye meets the mirror halfway between h and the eye level. Top of head: (1.76 + 1.66) ÷ 2 = 1.71 m. Feet: (0 + 1.66) ÷ 2 = 0.83 m. Length = 1.71 − 0.83 = 0.88 m (half his height). The lower edge can be at most 0.83 m high.
  4. B — 0.40 m s−1. The boy–mirror distance decreases at 0.50 − 0.30 = 0.20 m s−1. The boy–image distance is twice the boy–mirror distance, so it decreases at 2 × 0.20 = 0.40 m s−1. (Check: the image moves at 2(0.30) − 0.50 = 0.10 m s−1 in the direction of the boy, so relative to the boy it is 0.40 m s−1.)
  5. C — Virtual, upright and not laterally inverted. The final image is virtual, as it is formed by reflected rays that only appear to diverge. Each reflection laterally inverts the image, so two reflections cancel the inversion. The image is upright.
  6. D — 1.50 m. L′ is 0.60 m behind the mirror and 0.60 + 2.40 = 3.00 m from the path. By similar triangles, 0.30 ÷ ℓ = 0.60 ÷ 3.00, so ℓ = 1.50 m. (1.20 m comes from using 2.40 m instead of 3.00 m.)
  7. B — 5.7°. The reflected ray turns through twice the angle of the mirror. tan 2θ = 0.50 ÷ 2.50 = 0.200, so 2θ = 11.3° and θ = 5.7°.
  8. C — 24 m. The incident and reflected rays make equal angles with the ground, so the triangles formed with the girl and with the tree are similar: h ÷ 36 = 1.60 ÷ 2.4, so h = 24 m.
  9. D — A vertical arrow, pointing downwards. Take the mirror as the line y = x − 60 in page coordinates (y down). Reflecting in this line, the point (x, y) goes to (y + 60, x − 60). So B(108, 270) goes to B′(330, 48) and A(270, 270) goes to A′(330, 210). The image is vertical, with the head A′ below B′.
  10. B — 7.2 m. Take M1 as the line x = 0 and M2 as the line y = 0, with O at (3.0, 2.0). The image formed by two reflections is at (−3.0, −2.0). OI = √(6.0² + 4.0²) = √52 = 7.2 m. (5.0 m is 3.0 + 2.0; 10.0 m is 2 × 5.0.)
  11. C — (2) and (3) only. (1) is false: the frequency is fixed by the source and does not change. (2) is true: λ = v ÷ f and v decreases. (3) is true: light enters a medium of greater refractive index and bends towards the normal.
  12. A — 420 nm. λvacuum = 480 × 1.33 = 638 nm. λglass = 638 ÷ 1.52 = 420 nm. (549 nm comes from inverting the ratio.)
  13. A — 1.96 × 108 m s−1. The angles from the normal are i = 90° − 40° = 50° and r = 90° − 60° = 30°. n = sin 50° ÷ sin 30° = 1.53 and v = c ÷ n = 3.00 × 108 ÷ 1.53 = 1.96 × 108 m s−1.
  14. A — 2.6 cm. sin r = sin 60° ÷ 1.50 = 0.577, so r = 35.3°. Lateral displacement d = t sin(i − r) ÷ cos r = 5.0 × sin 24.7° ÷ cos 35.3° = 2.6 cm. (3.5 cm is the sideways shift inside the glass, 5.0 tan r. 5.1 cm is the separation along the exit face, 5.0(tan i − tan r). 6.1 cm is the path length in glass.)
  15. C — 37.0°. 1.60 sin 30° = 1.33 sin r, so sin r = 0.601 and r = 37.0°. (24.6° comes from inverting the ratio of indices.)
  16. D — 1.92 × 108 m s−1. The gradient of sin r against sin i is 1 ÷ n = 0.48 ÷ 0.75 = 0.64, so n = 1.56. v = c ÷ n = 3.00 × 108 ÷ 1.56 = 1.92 × 108 m s−1. (4.69 × 108 comes from using the gradient as n.)
  17. B — 5.0 cm. sin r = sin 45° ÷ 1.33 = 0.532, so r = 32.1°. x = 8.0 tan 32.1° = 5.0 cm. (4.3 cm uses sin instead of tan; 8.0 cm ignores refraction.)
  18. A — 28.8°. λair = c ÷ f = 3.00 × 108 ÷ 5.00 × 1014 = 600 nm. n = 600 ÷ 450 = 1.33. sin r = sin 40° ÷ 1.33 = 0.482, so r = 28.8°.
  19. D — 48.6°. At parallel boundaries n sin θ is constant, so the water layer does not affect the final angle: 1.50 sin 30° = 1.00 sin θ, so sin θ = 0.750 and θ = 48.6°. (34.3° is the angle in the water.)
  20. C — 1.46. The reflected ray also makes 30° with the normal. The angle between the reflected and refracted rays is 180° − 30° − r = 130°, so r = 20°. n = sin 30° ÷ sin 20° = 1.46.

Question 2 — Images Formed by Plane Mirrors: Ray-Diagram Drawing

A′ B′ P A′ B′ P A′ B′ P A′ B′ P
Model answer for Fig. 9 (gold dashed = image; grey dashed = virtual ray; green = real ray)
Marking: 1 mark for the image and 1 mark for the ray in each panel.
  • (a) A′B′ is vertical, 2.0 cm long, with A′ at the top. It is 3.0 cm behind the mirror, level with AB. The line from A′ to the eye cuts the mirror at P, about 1.7 cm below the level of A.
  • (b) A is 1.0 cm in front of the mirror, so A′ is 1.0 cm behind it at the same height; B′ is 3.0 cm behind the mirror. P is about 1.2 cm below the level of A.
  • (c) Each image point is the same distance below the mirror as the object point is above it: A (3.0 cm above the top edge line… i.e. 2.0 cm above the mirror) has A′ 2.0 cm below; B (1.0 cm above) has B′ 1.0 cm below. P is about 1.1 cm to the right of the point directly below A.
  • (d) For the 45° mirror, A′B′ is vertical (A′ is above B′), found by reflecting each point in the mirror. P is about 1.2 cm below A, almost directly beneath it.
Award the ray mark only if the real ray goes from A to a point P on the mirror and then to the eye, with the angles of incidence and reflection equal (the reflected ray lies on the line from A′ to the eye), and with arrows. Accept a dashed line from A′ to the eye through the mirror. Penalise a missing arrowhead on A′ only once.
(e)(i) B is 1.0 cm above the mirror, so B′ is 1.0 cm below it: BB′ = 2.0 cm (1)
(e)(ii) The mirror is at 45° to the horizontal object, so the image is vertical. The angle between AB and A′B′ is 90° (1)

Question 3 — Reflection of Light: What Can a Boy See in a Wall Mirror?

E′ H F
Model construction for Fig. 10 (H = highest visible point; F = nearest visible floor point)

(a) Distance EE′

The image is as far behind the mirror as the object is in front.
EE′ = 2 × 1.50 = 3.00 m (1)

(b) Highest visible point

The ray from the highest visible point H reflects at T and enters the eye, so it lies on the line from E′ through T. (1)
T is 1.62 − 1.50 = 0.12 m below eye level and 1.50 m from E′. H is 3.00 m from E′ (horizontally), so by similar triangles:
1.62 − h3.00 = 0.121.50 (1)
1.62 − h = 0.24, so h = 1.38 m (1)

(c) Nearest visible floor point

The limiting ray goes from E′ through the lower edge L and meets the floor at F, a distance x from the mirror. By similar triangles (heights 0.70 m and 1.62 m at distances x and x + 1.50 from F):
0.70x = 1.62x + 1.50 (1)
0.70(x + 1.50) = 1.62x, so 0.92x = 1.05 and x = 1.14 m (1)
The boy’s feet are 1.50 m from the mirror, which is farther than 1.14 m, so he can see his feet. (1)
Check: His feet are seen at a mirror height of 1.62 × 1.50 ÷ 3.00 = 0.81 m, which is between 0.70 m and 1.50 m, so the feet are indeed seen in the mirror.

(d) Raising the upper edge

The ray from the top of his head (1.72 m) meets the mirror halfway between 1.72 m and the eye level 1.62 m:
height = (1.72 + 1.62) ÷ 2 = 1.67 m (2)
(Award 1 mark for the method. The edge is raised by 0.17 m.)

(e) Boy at 3.00 m

1.62 − h6.00 = 0.123.00 , so h = 1.62 − 0.24 = 1.38 m.
The highest visible point does not change. It does not depend on how far he stands from the mirror. (1)

Question 4 — Reflection of Light: Where Can the Image of a Lamp Be Seen?

L′ X Y 3.6 m
Model construction for Fig. 11

(a) Construction

  • L′ is on the perpendicular from L to the wall, 0.50 m behind the mirror (1 mark).
  • Dashed lines from L′ through E1 and through E2 are extended (as solid reflected rays) to the observer’s path, meeting it at X and Y (1 mark).
  • Solid rays from L to E1 and E2 with the reflected rays, with arrows (1 mark).
X is 1.00 m to the left of the point opposite E1, and Y is 2.00 m to the right of the point opposite E2.

(b) Length of path (similar triangles)

L′ is 0.50 m + 2.50 m = 3.00 m from the path. (1)
The triangle formed by L′ and the mirror is similar to the triangle formed by L′ and XY:
0.60ℓ = 0.503.00 (1)
ℓ = 0.60 × 6.0 = 3.6 m (1)

(c) Distances from N

Left end (through E1, 0.20 m from the line LL′): 0.20a = 0.503.00 , so a = 1.20 m to the left of N (1)
Right end (through E2, 0.40 m from the line LL′): 0.40b = 0.503.00 , so b = 2.40 m to the right of N (1)
Check: 1.20 + 2.40 = 3.6 m, which agrees with (b).

(d) Right half covered

The uncovered part runs from E1 to the middle of the mirror (0.30 m wide):
ℓ = 0.30 × 6.0 = 1.8 m (1)
The middle of the mirror is 0.10 m from the line LL′, so the right-hand limit is 0.10 × 6.0 = 0.60 m to the right of N. The visible stretch runs from 1.20 m left of N to 0.60 m right of N. (1)

Question 5 — Refraction of Light: A Ray Passing Through a Glass Block

(a) Speed of light in glass

v = cn = 3.00 × 1081.60 = 1.88 × 108 m s−1 (1)

(b) Wavelengths

(i) λair = c ÷ f = 3.00 × 108 ÷ 5.00 × 1014 = 6.00 × 10−7 m (600 nm) (1)
(ii) λglass = v ÷ f = 1.875 × 108 ÷ 5.00 × 1014 = 3.75 × 10−7 m (375 nm) (1)
(The frequency is unchanged; λglass = 600 ÷ 1.60 = 375 nm.)

(c) Angle of refraction

sin r = sin i ÷ n = 0.800 ÷ 1.60 = 0.500 (1)
r = 30.0° (1)

(d) Path of the ray and lateral displacement

air glass block (n = 1.60) 53.1° 30.0° 30.0° d = 2.27 cm
Model answer for Fig. 12 (the emergent ray leaves at 53.1° to the normal, parallel to the incident ray)
Path: the ray bends towards the normal at the upper face (30.0°), then leaves the lower face at 53.1°, parallel to the incident ray but displaced sideways. (1)
d = t sin(i − r)cos r = 5.0 × sin 23.1°cos 30.0° (1)
d = 2.27 cm (1)
Alternative: The extended incident ray meets the lower face 5.0 tan 53.1° = 6.67 cm from A, while the ray in the glass reaches it 5.0 tan 30.0° = 2.89 cm from A. The separation along the face is 3.78 cm, and d = 3.78 cos 53.1° = 2.27 cm.

(e) Time in the glass

Path length in the glass = t ÷ cos r = 5.0 ÷ 0.866 = 5.77 cm = 0.0577 m (1)
time = distance ÷ speed = 0.0577 ÷ 1.875 × 108 = 3.08 × 10−10 s (1)

Question 6 — Refraction and Reflection: A Tank with Two Perpendicular Mirrors

(a) Refractive index of water

n = sin 53.1°sin 36.9° = 0.8000.600 = 1.33 (1)

(b) Height of Q

Horizontal distance from O to P = 12.0 tan 36.9° = 12.0 × 0.750 = 9.00 cm (1)
The ray is reflected at P with the same angle (36.9° to the vertical). Horizontal distance from P to M2 = 12.0 − 9.00 = 3.00 cm (1)
Height of Q = 3.00 ÷ tan 36.9° = 3.00 ÷ 0.750 = 4.00 cm (1)

(c) Distance OR

After reflection at M2 the horizontal component of the ray is reversed, and the ray still makes 36.9° with the vertical. Height remaining = 12.0 − 4.00 = 8.00 cm
Horizontal distance from Q to R = 8.00 × 0.750 = 6.00 cm (1)
OR = 12.0 − 6.00 = 6.00 cm (1)

(d) Emergent ray

sin θ = 1.33 × sin 36.9° = 1.33 × 0.600 = 0.800, so θ = 53.1° (1)
The emergent ray is parallel to the incident ray but travels in the opposite direction. Two perpendicular mirrors reverse both the horizontal and the vertical components of the ray’s direction. (1)

(e) Time from O to R

OP = 12.0 ÷ cos 36.9° = 15.0 cm; PQ = 4.00 ÷ 0.800 = 5.00 cm; QR = 8.00 ÷ 0.800 = 10.0 cm
Total path = 30.0 cm = 0.300 m (1)
v = c ÷ n = 3.00 × 108 ÷ 1.333 = 2.25 × 108 m s−1
t = 0.300 ÷ 2.25 × 108 = 1.33 × 10−9 s (1)
Common errors: (1) using the angle of incidence in air (53.1°) for the ray inside the water; (2) using sin instead of tan for horizontal distances; (3) forgetting that the vertical depth to be covered after Q is 12.0 − 4.00 = 8.00 cm, not 12.0 cm; (4) taking the time for only the straight vertical depth instead of the full path.
Mark weighting: Section A (20) + Question 2 (10) + Question 3 (10) + Question 4 (10) + Question 5 (10) + Question 6 (10) = 70 marks.
Scroll to Top