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Physics Recap – Reflection and Plane Mirrors | Vanilla (2026)
Vanilla (2026)

Student Questions

Reflection and Plane Mirrors — Recap

To be used on Sunday, 23 August 2026

Coverage: pages 2–21 of the physics notes, including the nature of light, the ray model, laws and types of reflection, plane-mirror images, full-length mirrors, and movement of objects, mirrors and images. Show all working and give appropriate units.

Understanding checks Advanced thinking
01

Nature of Light

Light and the ray model

Evaluate the following statements about light.

  1. Light is a form of energy.
  2. Light requires a material medium in order to travel.
  3. Light travels in straight lines in a uniform transparent medium.
  4. The speed of light in a vacuum is approximately 3.0 × 108 m s−1.
  5. A light ray represents the path and direction in which light travels.
  6. A ray diagram shows the actual width of a beam of light.
  1. Identify every correct statement.
  2. Correct each incorrect statement.
  3. Distinguish between a light ray and a light beam.
  4. State the three common types of light beam and describe how the rays are arranged in each.
Advanced thinking

A student says, “Since a ray has no physical width, the ray model is useless for studying real beams of light.” Evaluate this statement.

8 marks
02

Laws of Reflection

Angles measured from the mirror and normal

Plane mirror Normal 28° Incident ray Reflected ray

The incident ray makes an angle of 28° with the mirror surface.

  1. State the two laws of reflection.
  2. Calculate the angle of incidence.
  3. Calculate the angle of reflection.
  4. Determine the angle between the reflected ray and the mirror surface.
  5. Imagine that the incident ray continued straight through the mirror. Calculate the smaller angle through which the light is turned by reflection.
Advanced thinking

The incident ray is changed so that it makes a smaller angle with the normal. Explain what happens to the angle between the incident and reflected rays.

7 marks
03

Mirror Rotation

Rotation of a mirror and its reflected ray

A narrow incident ray remains fixed while a plane mirror is rotated clockwise through 6° about the point of incidence.

  1. Through what angle does the normal rotate?
  2. State the direction in which the normal rotates.
  3. Through what angle does the reflected ray rotate?
  4. State the direction in which the reflected ray rotates.
  5. The original angle between the incident and reflected rays is 70°. Determine the new angle between the rays.
  6. Explain why the reflected ray rotates through twice the angle of rotation of the mirror.
Advanced thinking

A reflected light spot is observed on a screen 8.0 m from the mirror. Explain why a very small mirror rotation can produce an easily observed movement of the spot.

8 marks
04

Types of Reflection

Regular and diffuse reflection

Consider the following statements.

  1. Diffuse reflection does not obey the laws of reflection.
  2. Only regular reflection normally forms a clear image.
  3. An object undergoing diffuse reflection cannot be seen.
  4. Parallel rays incident on a smooth surface remain parallel after reflection.
  5. Parallel rays incident on a rough surface are reflected in different directions.
  1. Identify every correct statement.
  2. Correct each incorrect statement.
  3. Explain why every individual ray in diffuse reflection still obeys the laws of reflection.
  4. Classify each surface as producing mainly regular or diffuse reflection:
    • calm water;
    • a blackboard;
    • polished metal;
    • a sheet of paper;
    • rough wood;
    • flat glass.
Advanced thinking

Explain why a person can read words printed on paper from many viewing directions, but cannot usually see a clear image of their face in the paper.

8 marks
05

Plane-Mirror Images

Image formed by calm water

A bird with a height of 0.40 m is hovering 2.5 m vertically above a calm water surface. Treat the water surface as a horizontal plane mirror.

  1. State the distance of the image below the water surface.
  2. Calculate the bird–image separation.
  3. State whether the image is real or virtual.
  4. State whether the image is upright or inverted relative to the bird.
  5. Determine the height of the image.
  6. Explain why the image cannot be formed on a screen placed below the water surface.
  7. The bird descends to 0.8 m above the water. Calculate the new bird–image separation.
Advanced thinking

A student claims that the image becomes larger as the bird moves closer to the water. Evaluate the claim and distinguish between actual image size and apparent angular size.

9 marks
06

Image Construction

Locating a virtual image

Plane mirror Object 2.4 m

A 1.2 m tall object is placed 2.4 m in front of a vertical plane mirror.

  1. Determine the image distance behind the mirror.
  2. Determine the image height.
  3. Calculate the object–image separation.
  4. State what is meant by lateral inversion.
  5. Describe how two reflected rays may be used to locate the image of the top of the object.
  6. State four characteristics of a plane-mirror image.
Advanced thinking

A student draws the reflected rays themselves meeting behind the mirror. Explain the error and state what should be drawn behind the mirror instead.

9 marks
07

Full-Length Mirror

Minimum mirror length and position

A student is 1.68 m tall. Her eyes are 1.58 m above the floor. She wishes to see her complete image, from the top of her head to her feet, in a vertical plane mirror.

  1. Calculate the minimum length of the mirror.
  2. Calculate the required height of the lower edge of the mirror above the floor.
  3. Calculate the required height of the upper edge of the mirror above the floor.
  4. Check that your upper-edge and lower-edge answers give the required mirror length.
  5. State what happens to the required minimum mirror length if the student moves farther from the mirror.
  6. State what happens to the required minimum mirror length if the student moves closer to the mirror.
  7. Explain why moving towards or away from the mirror does not change the minimum mirror length.
Advanced thinking

The available mirror is 0.80 m long and extends from 0.79 m to 1.59 m above the floor. Determine whether the student can see both the top of her head and her feet. Explain using the required reflection points.

9 marks
08

Object Movement

Object moving relative to a fixed mirror

A person is initially 9.0 m in front of a stationary plane mirror. The person walks directly towards the mirror at a constant speed of 0.75 m s−1.

  1. State the initial distance of the image behind the mirror.
  2. Calculate the initial person–image separation.
  3. State the speed and direction of the image relative to the mirror.
  4. Determine the rate at which the person–image separation decreases.
  5. Calculate the person–mirror distance after 4.0 s.
  6. Calculate the person–image separation after 4.0 s.
  7. Determine the time taken for the person–image separation to decrease to 6.0 m.
Advanced thinking

The person now walks parallel to the mirror at 0.75 m s−1. Describe the velocity of the image and state whether the perpendicular person–image separation changes.

8 marks
09

Mirror Movement

A moving mirror and a stationary object

A small object remains stationary. A plane mirror is initially 5.0 m from the object and moves directly towards it at 0.20 m s−1.

  1. Calculate the initial object–image separation.
  2. Determine the speed and direction of the image relative to the ground.
  3. Determine the rate at which the object–image separation decreases.
  4. Calculate the object–mirror distance after 6.0 s.
  5. Calculate the object–image separation after 6.0 s.
  6. Explain why the image speed is twice the mirror speed in this situation.
Advanced thinking

Suppose the mirror instead moves parallel to its surface while remaining in the same plane. Explain why this idealised motion has no effect on the position of the image of the stationary object.

8 marks
10

Integrated Movement

Object and mirror moving towards each other

Object Mirror 0.40 m s⁻¹ 0.15 m s⁻¹ 5.0 m

A small object is initially 5.0 m in front of a plane mirror. The object moves towards the mirror at 0.40 m s−1, while the mirror moves towards the object at 0.15 m s−1.

  1. Calculate the rate at which the object–mirror distance decreases.
  2. State the rate at which the image–mirror distance decreases.
  3. Determine the speed and direction of the image relative to the ground.
  4. Determine the rate at which the object–image separation decreases.
  5. Calculate the object–mirror distance after 4.0 s.
  6. Calculate the object–image separation after 4.0 s.
  7. Explain why the object–image separation is always twice the object–mirror distance.
Advanced thinking

Using the relation image position = 2 × mirror position − object position, explain why the image speed depends on both the mirror speed and the object speed.

10 marks

End of Student Questions

Total: 84 marks

Check all angles carefully.
Include directions and units where appropriate.
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Question 1 — Light and the ray model

  1. Statements 1, 3, 4 and 5 are correct.
    • Statement 2 is incorrect. Light can travel through a vacuum and does not require a material medium.
    • Statement 6 is incorrect. A ray is an ideal line showing the path and direction of light; it does not represent the actual width of a beam.
  3. A light ray is an ideal straight line showing the direction of travel. A light beam is a collection of light rays.
    • Parallel beam: rays remain parallel and the beam width stays constant.
    • Converging beam: rays move towards one another and meet at a point.
    • Diverging beam: rays spread apart from one another.
Advanced answer

The statement is incorrect. A model does not need to reproduce every physical detail to be useful. Rays allow the paths, directions, intersections and reflections of light to be represented simply. A real beam can be modelled using several rays, such as its boundary rays and central ray.

Question 2 — Angles measured from the mirror and normal

    • The angle of incidence equals the angle of reflection.
    • The incident ray, reflected ray and normal at the point of incidence lie in the same plane.
  2. Angles of incidence are measured from the normal.
    i = 90° − 28° = 62°
    Angle of incidence = 62°.
  3. By the law of reflection:
    r = i = 62°
    Angle of reflection = 62°.
  4. Angle with mirror = 90° − 62° = 28°
    The reflected ray makes 28° with the mirror.
  5. The straight continuation would make 28° with the mirror below it. The reflected ray makes 28° with the mirror above it.
    Turning angle = 28° + 28° = 56°
    The light is turned through 56°.
Advanced answer

The angle between the incident and reflected rays is 2i. If the angle of incidence becomes smaller, the angle between the two rays also becomes smaller. The reflected ray moves closer to the incident ray.

Question 3 — Rotation of a mirror and reflected ray

  1. The normal rotates through 6°.
  2. The normal rotates clockwise.
  3. Reflected-ray rotation = 2 × 6° = 12°
    The reflected ray rotates through 12°.
  4. The reflected ray rotates clockwise.
  5. The reflected ray rotates 12° away from its original direction.
    New angle = 70° + 12° = 82°
    The new angle between the rays is 82°.
  6. When the mirror and normal rotate through θ, the angle of incidence changes by θ. The reflected ray must remain at the same angle on the opposite side of the new normal. It therefore changes direction by another θ, giving a total change of 2θ.
Advanced answer

The reflected ray rotates through twice the mirror angle. The displacement of the spot also increases with the distance to the screen. Therefore, even a small angular change can produce a measurable linear displacement on a distant screen.

Question 4 — Regular and diffuse reflection

  1. Statements 2, 4 and 5 are correct.
    • Statement 1 is incorrect. Every individual ray in diffuse reflection obeys the laws of reflection.
    • Statement 3 is incorrect. Diffuse reflection allows an object to be seen from many directions.
  3. A rough surface contains many small surface sections with different orientations. Each ray has its own local normal and reflects with i = r. Since the normals point in different directions, the reflected rays spread in different directions.
  4. Surface Main type of reflection
    Calm water Regular
    Blackboard Mainly diffuse
    Polished metal Regular
    Sheet of paper Diffuse
    Rough wood Diffuse
    Flat glass Regular
Advanced answer

Paper sends reflected light in many directions, so light from the printed words can reach observers at different positions. However, the reflected rays do not preserve the ordered geometrical relationship needed to form a clear image.

Question 5 — Image formed by calm water

  1. Image distance equals object distance.
    Image distance = 2.5 m
    The image is 2.5 m below the water surface.
  2. Separation = 2.5 + 2.5 = 5.0 m
    Bird–image separation = 5.0 m.
  3. The image is virtual.
  4. The image is upright relative to the bird.
  5. A plane mirror produces an image of the same size.
    Image height = 0.40 m
  6. The reflected rays do not actually pass through the image position. They only appear to come from that position when extended backwards. Therefore, the image cannot be projected onto a screen.
  7. New separation = 0.8 + 0.8 = 1.6 m
    New bird–image separation = 1.6 m.
Advanced answer

The actual image height remains 0.40 m because the linear magnification of a plane mirror is 1. However, as the bird and its image become closer to an observer, the image may subtend a larger angle at the eye and therefore appear larger. Apparent angular size is not the same as actual image size.

Question 6 — Locating a virtual image

  1. Image distance = object distance = 2.4 m
    The image is 2.4 m behind the mirror.
  2. Image height = object height = 1.2 m
    Image height = 1.2 m.
  3. Object–image separation = 2.4 + 2.4 = 4.8 m
    Separation = 4.8 m.
  4. Lateral inversion is the apparent interchange of left and right in the mirror image.
  5. Draw two rays from the top of the object to two different points on the mirror. Reflect each ray using i = r. Extend the reflected rays backwards behind the mirror using dashed lines. Their backward extensions meet at the image of the top of the object.
  6. Any four:
    • virtual;
    • upright;
    • same size as the object;
    • laterally inverted;
    • same distance behind the mirror as the object is in front.
Advanced answer

The reflected rays remain in front of the mirror and do not physically travel behind it. Only their backward extensions should be drawn behind the mirror. These extensions are conventionally shown as dashed lines.

Question 7 — Minimum mirror length and position

  1. Minimum mirror length = 1.68 ÷ 2 = 0.84 m
    Minimum length = 0.84 m.
  2. The lower reflection point is halfway between the eyes and feet.
    Lower edge height = 1.58 ÷ 2 = 0.79 m
    Lower edge = 0.79 m above the floor.
  3. The upper reflection point is halfway between the eyes and top of the head.
    Upper edge height = (1.58 + 1.68) ÷ 2 = 1.63 m
    Upper edge = 1.63 m above the floor.
  4. Length = 1.63 − 0.79 = 0.84 m
    This agrees with part (a).
  5. The minimum length remains unchanged.
  6. The minimum length remains unchanged.
  7. The required reflection points are the midpoints between the eyes and the parts of the body being viewed. These midpoint heights depend on the person’s height and eye position, not on the person’s distance from the mirror.
Advanced answer

The lower edge is correctly positioned at 0.79 m, so the student can see her feet.

However, the required upper edge is 1.63 m, while the actual upper edge is:

0.79 + 0.80 = 1.59 m

Therefore, the mirror is 0.04 m too short at the top. The student cannot see the top of her head while also seeing her feet.

Question 8 — Object moving relative to a fixed mirror

  1. The image is initially 9.0 m behind the mirror.
  2. Initial separation = 9.0 + 9.0 = 18.0 m
    Initial person–image separation = 18.0 m.
  3. The image moves towards the mirror at 0.75 m s−1.
  4. The person and image approach the mirror from opposite sides.
    Rate of decrease = 0.75 + 0.75 = 1.50 m s−1
    Separation decreases at 1.50 m s−1.
  5. Distance travelled = 0.75 × 4.0 = 3.0 m
    New person–mirror distance = 9.0 − 3.0 = 6.0 m
    Person–mirror distance = 6.0 m.
  6. Person–image separation = 2 × 6.0 = 12.0 m
    Separation after 4.0 s = 12.0 m.
  7. 18.0 − 1.50t = 6.0
    1.50t = 12.0
    t = 8.0 s
    Time taken = 8.0 s.
Advanced answer

The image moves parallel to the mirror at the same speed and in the same direction as the person: 0.75 m s−1. The perpendicular distance from the person to the mirror is unchanged, so the perpendicular person–image separation is also unchanged.

Question 9 — A moving mirror and a stationary object

  1. Initial object–image separation = 2 × 5.0 = 10.0 m
    Initial separation = 10.0 m.
  2. Image speed = 2 × mirror speed = 2 × 0.20 = 0.40 m s−1
    The image moves towards the object at 0.40 m s−1.
  3. Since the object is stationary:
    Rate of decrease = 0.40 m s−1
    The object–image separation decreases at 0.40 m s−1.
  4. Mirror movement = 0.20 × 6.0 = 1.2 m
    New object–mirror distance = 5.0 − 1.2 = 3.8 m
    Object–mirror distance = 3.8 m.
  5. Object–image separation = 2 × 3.8 = 7.6 m
    Object–image separation = 7.6 m.
  6. If the mirror moves a distance d towards the object, the image must remain the same distance behind the new mirror position as the object is in front. Relative to the ground, the image therefore moves through 2d. Its speed is twice the mirror speed.
Advanced answer

An ideal infinite plane mirror is defined by the plane containing its reflecting surface. Sliding it parallel to itself does not change that plane. The perpendicular object distance and the corresponding image position therefore remain unchanged.

Question 10 — Object and mirror moving towards each other

  1. The object and mirror move towards one another.
    Rate of decrease = 0.40 + 0.15 = 0.55 m s−1
    Object–mirror distance decreases at 0.55 m s−1.
  2. Image distance always equals object distance from the mirror.

    The image–mirror distance also decreases at 0.55 m s−1.
  3. Take the direction from the object towards the mirror as positive. Then:
    • object velocity = +0.40 m s−1;
    • mirror velocity = −0.15 m s−1.
    Using vimage = 2vmirror − vobject:
    vimage = 2(−0.15) − 0.40 = −0.70 m s−1
    The image moves towards the object at 0.70 m s−1.
  4. The object and image approach each other.
    Closing speed = 0.40 + 0.70 = 1.10 m s−1
    Object–image separation decreases at 1.10 m s−1.
  5. Decrease in object–mirror distance = 0.55 × 4.0 = 2.2 m
    New distance = 5.0 − 2.2 = 2.8 m
    Object–mirror distance after 4.0 s = 2.8 m.
  6. Object–image separation = 2 × 2.8 = 5.6 m
    Object–image separation after 4.0 s = 5.6 m.
  7. The image is the same perpendicular distance behind the mirror as the object is in front. Therefore:
    Object–image distance = object–mirror distance + mirror–image distance
    = d + d = 2d
Advanced answer

For positions measured along the normal to the mirror:

ximage = 2xmirror − xobject

Differentiating with respect to time gives:

vimage = 2vmirror − vobject

The image position must continuously remain symmetric to the object position about the moving mirror. Consequently, movement of either the object or mirror changes the image velocity.

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