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

Student Questions

Lesson Recap Questions to be used on 18/8/2026

Show all working clearly. Diagrams are not necessarily drawn to scale. Give appropriate units with numerical answers.

01

Reflection

Reflection from a plane surface

Plane mirror Normal 32°

The incident ray makes an angle of 32° with the mirror surface.

  1. Determine the angle of incidence.
  2. Determine the angle of reflection.
  3. Imagine that the incident ray continued straight through the mirror. Calculate the smaller angle between this straight continuation and the actual reflected ray.
4 marks
02

Reflection Experiment

Testing the law of reflection

A student investigates the reflection of light by a plane mirror. The student measures the angle of incidence i and the angle of reflection r. The results are shown below.

Angle of incidence, i / ° 20 30 40 50
Angle of reflection, r / ° 19 31 54 49
  1. State the law of reflection.
  2. Identify the anomalous result in the table.
  3. State the expected angle of reflection when the angle of incidence is 40°.
  4. Suggest one experimental reason for the anomalous result.
  5. Suggest one way to improve the reliability of the results.
5 marks
03

Mirror Rotation

Rotation of a plane mirror

A narrow light ray is incident on a plane mirror. The direction of the incident ray remains fixed. The mirror is then rotated clockwise through 7° about the point at which the ray strikes it.

  1. Through what angle does the normal rotate?
  2. Through what angle does the reflected ray rotate?
  3. State the direction of rotation of the reflected ray.
  4. The reflected ray produces a light spot on a distant screen. Explain why a small rotation of the mirror can produce a large movement of the light spot.
5 marks
04

Types of Reflection

Regular and diffuse reflection

Consider the following statements.

  1. Diffuse reflection does not obey the law of reflection.
  2. Only regular reflection can produce a clear image.
  3. An object cannot be seen if its surface produces diffuse reflection.
  4. Each ray reflected from a rough surface obeys the law of reflection.
  1. Identify every correct statement.
  2. Correct each incorrect statement.
  3. Explain why writing on a sheet of paper can be seen from many directions, but a clear image is not normally formed by the paper.
  4. Classify each surface as producing mainly regular or diffuse reflection:
    • calm water;
    • rough wood;
    • polished metal;
    • ordinary paper.
7 marks
05

Image Formation

A bird above calm water

A bird is hovering 3.2 m vertically above a calm water surface. Treat the water surface as a horizontal plane mirror.

  1. How far below the water surface is the image?
  2. Calculate the distance between the bird and its image.
  3. State whether the image is real or virtual.
  4. Compare the image size with the size of the bird.
  5. The bird descends until it is 1.0 m above the surface. Determine the new bird–image separation.
  6. A student claims that the image becomes larger as the bird approaches the water. Evaluate this claim.
7 marks
06

Image Position

Object in front of a plane mirror

Plane mirror Object 3.0 m

An object 3.0 m tall is placed 3.0 m in front of a vertical plane mirror.

  1. Determine the distance of the image behind the mirror.
  2. Determine the height of the image.
  3. Calculate the distance between the object and its image.
  4. State three other characteristics of the image formed.
6 marks
07

Mirror Size

Minimum length of a mirror

A student is 1.72 m tall. Her eyes are 1.56 m above the floor. A vertical plane mirror is used so that she can see her complete image, from the top of her head to her feet.

  1. Determine the minimum length of the mirror.
  2. Determine the height of the lower edge of the mirror above the floor.
  3. Determine the height of the upper edge of the mirror above the floor.
  4. She moves farther away from the mirror. State whether the required minimum mirror length increases, decreases or remains unchanged. Explain your answer.
  5. Would jumping vertically allow her to see more of herself in a mirror that is too short? Explain briefly.
7 marks
08

Object Motion

Person moving towards a fixed mirror

A runner is initially 12.0 m in front of a stationary plane mirror. She runs directly towards it at a constant speed of 1.20 m s−1.

  1. State the speed and direction of her image relative to the mirror.
  2. Determine the speed at which the distance between the runner and her image decreases.
  3. Calculate the runner–image separation after 3.0 s.
  4. Determine the time at which the runner–image separation becomes 4.8 m.
6 marks
09

Mirror Motion

Moving mirror and stationary object

A small stationary object is placed in front of a plane mirror. The mirror moves directly towards the object at 0.35 m s−1.

  1. Determine the speed and direction of the image relative to the ground.
  2. At what rate does the object–image separation change?
  3. The initial object–mirror distance is 4.20 m. Calculate the object–image separation after 5.0 s.
  4. Explain why the image speed is twice the mirror speed in this situation.
6 marks
10

Combined Motion

An object and a mirror moving towards each other

Object Mirror 0.10 m s⁻¹ 0.20 m s⁻¹ 4.50 m

A small object is initially 4.50 m in front of a vertical plane mirror. The object and the mirror move directly towards each other, as shown.

  1. Calculate the rate at which the distance between the object and the mirror decreases.
  2. State the rate at which the distance between the image and the mirror decreases.
  3. Determine the speed and direction of the image relative to the ground.
  4. Calculate the object–image separation after 5.0 s.
  5. Explain why the object–image separation decreases at twice the rate of the object–mirror distance.
7 marks
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Question 1 — Reflection from a plane surface

  1. i = 90° − 32° = 58°
    Angle of incidence = 58°.
  2. By the law of reflection, r = i.
    r = 58°
    Angle of reflection = 58°.
  3. Smaller angle = 2(32°) = 64°
    The smaller angle is 64°.

Question 2 — Testing the law of reflection

  1. The angle of incidence is equal to the angle of reflection. Both angles are measured from the normal.
  2. The anomalous result is i = 40° and r = 54°.
  3. r = i = 40°
    The expected angle of reflection is 40°.
  4. One acceptable reason is:
    • the ray or normal was drawn inaccurately;
    • the protractor was positioned incorrectly;
    • the angle was measured from the mirror instead of the normal;
    • the centre of the protractor was not placed at the point of incidence.
  5. Repeat each measurement several times and calculate the average result.

Question 3 — Rotation of a plane mirror

  1. The normal rotates 7° clockwise.
  2. Reflected-ray rotation = 2(7°) = 14°
    The reflected ray rotates through 14°.
  3. The reflected ray rotates clockwise.
  4. The reflected ray rotates through twice the angle of the mirror. A small angular change also produces a large displacement when the screen is far from the mirror.

Question 4 — Regular and diffuse reflection

  1. Statements 2 and 4 are correct.
    • Statement 1 is incorrect. Each individual ray undergoing diffuse reflection obeys the law of reflection.
    • Statement 3 is incorrect. Diffuse reflection allows an object to be seen from many directions.
  3. The paper reflects light in many directions, so it can be seen from different positions. The reflected rays are not kept in an orderly arrangement, so a clear image is not formed.
  4. Surface Main type of reflection
    Calm water Regular
    Rough wood Diffuse
    Polished metal Regular
    Ordinary paper Diffuse

Question 5 — A bird above calm water

  1. The image is 3.2 m below the water surface.
  2. Bird–image distance = 3.2 + 3.2 = 6.4 m
    The separation is 6.4 m.
  3. The image is virtual.
  4. The image is the same size as the bird.
  5. New separation = 1.0 + 1.0 = 2.0 m
    The new separation is 2.0 m.
  6. The claim is incorrect. A plane mirror has a linear magnification of 1. The image remains the same size as the bird.

Question 6 — Object in front of a plane mirror

  1. Image distance equals object distance.
    Image distance = 3.0 m
    The image is 3.0 m behind the mirror.
  2. A plane mirror produces an image of the same size as the object.
    Image height = 3.0 m
    The image height is 3.0 m.
  3. Object–image distance = 3.0 + 3.0 = 6.0 m
    The distance is 6.0 m.
  4. Any three:
    • virtual;
    • upright;
    • laterally inverted;
    • the same distance behind the mirror as the object is in front.

Question 7 — Minimum length of a mirror

  1. Minimum length = 1.72 ÷ 2 = 0.86 m
    Minimum length = 0.86 m.
  2. Lower edge = 1.56 ÷ 2 = 0.78 m
    The lower edge is 0.78 m above the floor.
  3. Upper edge = (1.72 + 1.56) ÷ 2 = 1.64 m
    The upper edge is 1.64 m above the floor.
  4. It remains unchanged. The minimum mirror length depends on the person’s height and eye position, not on the distance from the mirror.
  5. No. The person and the person’s eyes move vertically together, so jumping does not increase the fraction of the body visible in the mirror.

Question 8 — Person moving towards a fixed mirror

  1. The image moves towards the mirror at 1.20 m s−1.
  2. Closing speed = 1.20 + 1.20 = 2.40 m s−1
    The separation decreases at 2.40 m s−1.
  3. Initial separation = 2(12.0) = 24.0 m

    Decrease = 2.40 × 3.0 = 7.2 m

    New separation = 24.0 − 7.2 = 16.8 m
    The separation is 16.8 m.
  4. 24.0 − 2.40t = 4.8
    2.40t = 19.2
    t = 8.0 s
    The time is 8.0 s.

Question 9 — Moving mirror and stationary object

  1. Image speed = 2(0.35) = 0.70 m s−1
    The image moves at 0.70 m s−1 towards the object.
  2. The object–image separation decreases at 0.70 m s−1.
  3. Initial separation = 2(4.20) = 8.40 m

    Decrease = 0.70 × 5.0 = 3.50 m

    New separation = 8.40 − 3.50 = 4.90 m
    The new separation is 4.90 m.
  4. When the mirror moves a distance d towards the object, the object–mirror distance decreases by d. The image must also move d towards the mirror from behind it. The image therefore moves a total distance of 2d relative to the ground.

Question 10 — An object and a mirror moving towards each other

  1. The object and mirror move towards each other.
    Rate of decrease = 0.10 + 0.20 = 0.30 m s−1
    The object–mirror distance decreases at 0.30 m s−1.
  2. The image is always the same distance behind the mirror as the object is in front.

    The image–mirror distance decreases at 0.30 m s−1.
  3. The mirror moves towards the object at 0.20 m s−1. The image also moves towards the mirror at 0.30 m s−1 relative to the mirror.
    Image speed = 0.20 + 0.30 = 0.50 m s−1
    The image moves towards the object at 0.50 m s−1 relative to the ground.
  4. Initial object–image separation = 2(4.50) = 9.00 m
    In 5.0 s:
    Decrease in object–mirror distance = 0.30 × 5.0 = 1.50 m
    New object–mirror distance = 4.50 − 1.50 = 3.00 m
    Object–image separation = 2(3.00) = 6.00 m
    The object–image separation after 5.0 s is 6.00 m.
  5. The image is the same distance behind the mirror as the object is in front. Therefore, the object–image separation is always twice the object–mirror distance.

    The object–image separation consequently decreases at twice the rate of the object–mirror distance.
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