HKDSE Physics Recap: Plane Mirrors, Images and Ray Diagrams | Joe (2026)

HKDSE Physics — Plane Mirrors: Images and Ray Diagrams

Subject: Physics

Joe (2026)

Issue date: Thursday, 8 October 2026

Total: 50 marks | Suggested time: 60–70 minutes | Use a sharp pencil and a ruler. Draw virtual rays and virtual images with dashed lines. Figures are drawn on grid paper (print at 100% – actual size); the scale is stated in each question.

Comprehensive Structured Question 1

Image Formed by a Plane Mirror

Fig. 1 is drawn on a grid in which each large square represents 1.0 cm. A vertical plane mirror is placed with its reflecting surface facing left (the hatched side is the back). An arrow-shaped object AB stands in front of the mirror, with its tip at A. An eye E looks at the mirror. All the rays and the eye lie in the plane of the paper.

Fig. 1 — A plane mirror, an object AB and an eye E (1 large square = 1.0 cm)

(a) On Fig. 1, construct the image A′B′ of the object AB. Show your construction lines and draw the image as a dashed arrow. Label A′ and B′. (3 marks)

(b) On Fig. 1, draw a ray from A that is reflected by the mirror and enters the eye E. Mark the point of reflection P. Use arrows to show the direction of the light, and a dashed line for the part of the ray that is not real. (3 marks)

(c) State two characteristics of the image A′B′ (other than its position), and explain what is meant by saying that the image is virtual. (2 marks)

(d) The distance between A and the mirror is 1.5 cm. The mirror is now moved 2.0 cm directly away from the object, and the object stays fixed. Find

the new distance between A and its image A′; and
the distance moved by A′. (2 marks for (i) and (ii) together)

10 marks

Comprehensive Structured Question 2

Ray Diagrams for Horizontal and Inclined Mirrors

Fig. 2(a) and Fig. 2(b) are drawn on grids in which each large square represents 1.0 cm. In each figure, an arrow-shaped object AB (tip at A) is placed in front of a plane mirror and an eye E looks into the mirror. The hatched side of each mirror is its back.

Fig. 2(a) — Horizontal plane mirror (1 large square = 1.0 cm)

(a) Refer to Fig. 2(a).

Construct the image A′B′ of the object. Use a dashed arrow for the image and label A′ and B′. (2 marks)
Draw a ray from A which is reflected by the mirror and enters the eye E. Mark the point of reflection P. Show the direction of the light with arrows and use a dashed line for the virtual part of the ray. (2 marks)

Fig. 2(b) — Plane mirror inclined at 45° to the horizontal (1 large square = 1.0 cm)

(b) Refer to Fig. 2(b).

Construct the image A′B′ of the object. Use a dashed arrow for the image and label A′ and B′. (2 marks)
Draw a ray from B which is reflected by the mirror and enters the eye E. Mark the point of reflection Q. Show the direction of the light with arrows and use a dashed line for the virtual part of the ray. (2 marks)

(c) In Fig. 2(b), state the angle between the object AB and the image A′B′. Explain your answer. (2 marks)

10 marks

Comprehensive Structured Question 3

Full-length Mirror: How Long Must It Be?

A girl of height 1.60 m stands 0.80 m in front of a vertical wall mirror. Her eyes E are 1.50 m above the floor. T is the top of her head and F is her feet. Fig. 3 is a side view drawn to scale: 1 large square represents 0.20 m. The mirror in Fig. 3 is longer than necessary.

Fig. 3 — A girl in front of a wall mirror, side view (1 large square = 0.20 m)

(a) On Fig. 3, draw rays to show how the girl sees in the mirror (i) the top of her head T and (ii) her feet F. Mark the points of reflection P (for T) and Q (for F) on the mirror, and use arrows to show the direction of the light. Use dashed lines for the virtual parts of the rays. (4 marks)

(b) Calculate the minimum length of mirror that the girl needs to see her whole body, and the height of the lower edge of this mirror above the floor. (3 marks)

(c) The girl walks backwards until she is 2.0 m from the mirror. State whether the minimum length of mirror she needs changes. Explain your answer. (1 mark)

(d) In a shop, the girl stands 0.80 m in front of another mirror of length 0.80 m. Its lower edge is 0.90 m above the floor. Calculate the height above the floor of the lowest point of her body that she can see in this mirror. Can she see the top of her head? (2 marks)

10 marks

Comprehensive Structured Question 4

Two Plane Mirrors at Right Angles

Fig. 4 shows two plane mirrors M₁ and M₂ placed at right angles to each other, meeting at a corner. A small light source O is placed in front of the mirrors and an eye E looks into them. The figure is drawn on a grid in which each large square represents 1.0 cm. The hatched sides of the mirrors are their backs. Ignore the thickness of the mirrors.

Fig. 4 — Two plane mirrors M₁ and M₂ at 90° (1 large square = 1.0 cm)

(a) On Fig. 4, locate and mark the image I₁ of O formed by M₁ alone, and the image I₂ of O formed by M₂ alone. Show your construction lines (extend the mirror lines with dotted lines where necessary). (2 marks)

(b) Locate and mark a third image I₃, and explain how it is formed. (2 marks)

(c) The eye E sees the image I₃. Draw the ray from O that reaches E after being reflected once by each mirror. Mark the point of reflection P on M₁ and Q on M₂. Use arrows to show the direction of the light and dashed lines for the virtual parts. (3 marks)

(d) The angle of incidence of the ray at M₁ is θ. Show that the ray leaving M₂ is parallel to the ray that arrived at M₁, but travels in the opposite direction. (3 marks)

10 marks

Comprehensive Structured Question 5

A Mirror That Is Not Full-length: What Can Be Seen?

A boy of height 1.20 m stands 1.6 m in front of a small mirror fixed on a vertical wall. His eyes E are 1.10 m above the floor. The lower edge L of the mirror is 0.50 m above the floor and the upper edge U is 0.90 m above the floor. Fig. 5 is a side view drawn to scale: 1 large square represents 0.20 m.

Fig. 5 — A boy and a small wall mirror, side view (1 large square = 0.20 m)

(a) On Fig. 5, draw the ray from the highest point H of the boy’s body that he can see in the mirror to his eye E. Mark H on the boy and use arrows to show the direction of the light. Use a dashed line for the virtual part of the ray. (3 marks)

(b) Calculate the height of H above the floor. Hence, or otherwise, show that the boy can see his feet in the mirror. (3 marks)

(c) A coin lies on the floor between the boy and the mirror. By considering the image of the boy’s eyes, calculate the distance from the wall of the point on the floor nearest to the mirror that the boy can see in the mirror. (2 marks)

(d) The boy walks 0.80 m backwards (away from the mirror). State, with a reason, whether the height of H changes. Calculate the new distance from the wall of the nearest point on the floor that he can see in the mirror. (2 marks)

10 marks

Question 1 — Image Formed by a Plane Mirror

Model answer for Fig. 1 — green: real rays; gold dashed: virtual ray and image

(a) Constructing the image

The image is on the perpendicular from the object to the mirror, as far behind the mirror as the object is in front.
A is 1.5 squares from the mirror, so A′ is 1.5 squares behind it, at the same level as A (6.5 squares from the left edge, 1.5 squares from the top). (1)
B is 3 squares from the mirror, so B′ is 3 squares behind it, at the same level as B (8 squares from the left edge, 4 squares from the top). (1)
A′B′ is drawn as a dashed arrow from B′ to A′, with the tip at A′. (1)

(b) Ray from A to the eye

Join E to A′ with a straight line. It meets the mirror at P (on the mirror, 3 squares from the top of the grid). (1)
Draw the ray A → P → E as solid lines with arrows pointing from A to P and from P to E. (1)
Draw P → A′ as a dashed line behind the mirror (the light only appears to come from A′). (1)

Check: The angle of incidence at P equals the angle of reflection. The incident ray A → P and the reflected ray P → E make equal angles with the normal (the horizontal line through P).

(c) Characteristics of the image

Any two of: the image is virtual; the same size as the object; laterally inverted (left–right reversed); the line joining object and image is perpendicular to the mirror. (1)
Virtual means that the light rays do only appear to come from the image (they do not actually pass through it), so the image cannot be formed on a screen. (1)

(d) Moving the mirror

(i) New object distance = 1.5 + 2.0 = 3.5 cm. The image is 3.5 cm behind the mirror, so
AA′ = 2 × 3.5 = 7.0 cm (1)
(ii) Originally AA′ = 2 × 1.5 = 3.0 cm. The distance moved by A′ = 7.0 − 3.0 = 4.0 cm (twice the distance moved by the mirror) (1)

Marking note: When the paper is printed at 100%, one small square is 1 mm. Accept positions within 2 small squares (2 mm). For (b), do not award the second mark if the arrows point the wrong way or the ray does not end at E.

Question 2 — Ray Diagrams for Horizontal and Inclined Mirrors

Model answer for Fig. 2(a)

Model answer for Fig. 2(b)

(a) Horizontal mirror

(i) Reflect A and B in the mirror (mirror line is 4 squares from the top):
A′: 4.5 squares from the left, 6 squares from the top. (1)
B′: 2.5 squares from the left, 5 squares from the top; the image is drawn as a dashed arrow from B′ to A′. (1)
(ii) Join E to A′. It meets the mirror at P (6.5 squares from the left). (1)
Draw A → P → E with arrows, and the dashed line P → A′. (1)

(b) Mirror inclined at 45°

(i) Each image point lies on the perpendicular from the object point to the mirror, at an equal distance behind it:
A′: 6 squares from the left, 3 squares from the top. (1)
B′: 6 squares from the left, 6 squares from the top; the image is a dashed vertical arrow from B′ up to A′. (1)
(ii) Join E to B′. It meets the mirror at Q (3.5 squares from the left, 4.5 squares from the top). (1)
Draw B → Q → E with arrows, and the dashed line Q → B′. (1)

Check: The ray B → Q and the ray Q → E make equal angles with the normal at Q. The ray from A, drawn as a check, would reflect at 5 squares from the left, 3 squares from the top, and leave horizontally to the left (a vertical ray hitting a 45° mirror turns through 90°).

(c) Angle between object and image

90° (AB is horizontal and A′B′ is vertical). (1)
AB makes 45° with the mirror. The image is the mirror reflection of the object, so A′B′ also makes 45° with the mirror, on the other side. The angle between them is 45° + 45° = 90°. (1)

Common errors: (1) drawing the image on the same side of the mirror as the object; (2) joining the object (instead of the image) to the eye, or drawing the dashed line from the wrong point; (3) in Fig. 2(b), measuring the image distance vertically instead of along the perpendicular to the inclined mirror.

Question 3 — Full-length Mirror: How Long Must It Be?

Model answer for Fig. 3 — the gold section PQ of the mirror is the minimum length required

(a) Ray diagram

The image T′ is as far behind the mirror as T is in front (4 squares), at the same height. Join E to T′: it meets the mirror at P (1.55 m above the floor, i.e. 7.75 squares up). (1)
Draw T → P → E with arrows, and P → T′ dashed. (1)
Likewise, join E to F′: it meets the mirror at Q (0.75 m above the floor, i.e. 3.75 squares up). (1)
Draw F → Q → E with arrows, and Q → F′ dashed. (1)

(b) Minimum length and position of the mirror

The ray reflects at the height halfway between the object point and the eye (similar triangles).
For the head: height of P = 1.60 + 1.502 = 1.55 m | For the feet: height of Q = 0 + 1.502 = 0.75 m (1)
Minimum length = 1.55 − 0.75 = 0.80 m (1)
Lower edge of the mirror: 0.75 m above the floor (1)

(c) Walking backwards

No change. The minimum length is half of her height, whatever her distance from the mirror, because the heights of P and Q depend only on the heights of T, F and E (by similar triangles), not on the distance. (1)

(d) Shop mirror

Lowest visible point (height h) is seen at the lower edge:
h + 1.502 = 0.90, so h = 0.30 m (1)
Top of head: it is seen at 1.55 m, which is below the upper edge of the mirror (0.90 + 0.80 = 1.70 m), so she can see the top of her head. (1)

Common errors: (1) taking the mirror length equal to the girl’s full height; (2) measuring the reflection point from the top of the head instead of from the eye (the mirror must reach only to the midpoint between the eye and the top of the head); (3) thinking that a farther distance needs a shorter (or longer) mirror.

Question 4 — Two Plane Mirrors at Right Angles

Model answer for Fig. 4 — gold circles: images; gold dashed lines: virtual rays

(a) Images formed by single reflections

Each image is on the perpendicular from O to the mirror (extended if necessary), at equal distance behind it.
I₁: 8 squares from the left, 6 squares from the top (directly below O, 1 square below M₁… i.e. 1 square behind it, since O is 1 square above M₁). (1)
I₂: 2 squares from the left, 4 squares from the top (level with O, 3 squares behind M₂). (1)

(b) The third image

I₃: 2 squares from the left, 6 squares from the top. (1)
I₃ is the image of I₁ formed by M₂ (extended). Equivalently, it is the image of I₂ formed by M₁ (extended). It is formed by light that has been reflected twice, once by each mirror. (1)

(c) Ray diagram

Join E to I₃. It meets M₂ at Q (5 squares from the left, 4.5 squares from the top). (1)
Join Q to I₁. It meets M₁ at P (6 squares from the left). (1)
Draw O → P → Q → E with arrows in this direction, and the dashed lines P → I₁ and Q → I₃. (1)

(d) The emergent ray is antiparallel to the incident ray

At M₁: angle of incidence = angle of reflection = θ, measured from the normal to M₁. The reflected ray therefore makes an angle θ with the normal to M₁. (1)
The normal to M₁ is parallel to M₂ (since M₁ ⊥ M₂), so this ray makes angle θ with the surface of M₂. The angle of incidence at M₂ is 90° − θ. (1)
Total deviation = (180° − 2θ) + [180° − 2(90° − θ)] = (180° − 2θ) + 2θ = 180°.
So the emergent ray is parallel to the incident ray but in the opposite direction. (1)

Numerical check (from the diagram): At M₁ the ray O → P travels 2 squares horizontally for every 1 square vertically, so tan θ = 2 and θ = 63.4°. At M₂ the ray P → Q makes tan θ′ = 0.5, so θ′ = 26.6° = 90° − 63.4°. The emergent ray Q → E has the same slope as O → P (1 square up for every 2 squares across), in the opposite direction.

Common errors: (1) forgetting the third image I₃, or placing it by reflecting O twice in the same mirror; (2) not extending the mirror lines when constructing images; (3) drawing the ray as a solid line behind the mirror.

Question 5 — A Mirror That Is Not Full-length: What Can Be Seen?

Model answer for Fig. 5 — ray from H (via U) and ray from the nearest visible floor point P (via L)

(a) The highest point he can see

The highest visible point is seen by the ray that reflects at the upper edge U of the mirror. (1)
Draw E → U, and the reflected ray makes equal angles with the normal at U (or use the image of E, 1.6 m = 8 squares behind the mirror). Going back along the incident ray from U, it meets the boy at H (3.5 squares above the floor line). (1)
Draw H → U → E with arrows pointing from H to U and from U to E, and a dashed line behind the mirror. (1)

(b) Height of H; the feet

The ray reflects at the height halfway between the object point and the eye:
h + 1.102 = 0.90 (1)
h = 0.70 m (1)
Feet: the ray from his feet reflects at 0 + 1.102 = 0.55 m, which lies between the lower edge (0.50 m) and the upper edge (0.90 m) of the mirror. So he can see his feet. (1)

Note: He can see his body from his feet up to 0.70 m above the floor. The top of his head would be seen at (1.20 + 1.10) ÷ 2 = 1.15 m, which is above the upper edge of the mirror (0.90 m), so it cannot be seen.

(c) Nearest floor point he can see

The image E′ of his eyes is 1.6 m behind the mirror and 1.10 m above the floor. The line from E′ through L (0.50 m above the floor) falls 0.60 m over 1.6 m, so its gradient is 0.60 ÷ 1.6 = 0.375. (1)
In front of the mirror it falls a further 0.50 m to reach the floor:
distance from the wall = 0.500.375 = 1.33 m (1)

Check on the diagram: P is at 126 px, so its distance from the wall is (486 − 126) ÷ 270 = 1.33 m. The floor between P and the wall (hidden below L) cannot be seen.

(d) Walking backwards

No change. The reflection height 0.90 m = (h + 1.10) ÷ 2 does not depend on the distance from the mirror. (1)
New distance from the mirror: 1.6 + 0.80 = 2.4 m. Gradient = 0.60 ÷ 2.4 = 0.25, so
distance from the wall = 0.500.25 = 2.0 m (1)

Common errors: (1) using the mirror height (0.90 m) as the height of the highest visible point H; (2) forgetting to halve (h + eye height); (3) measuring the distance in (c) from the boy instead of from the wall.

Mark weighting: Question 1 (10) + Question 2 (10) + Question 3 (10) + Question 4 (10) + Question 5 (10) = 50 marks.