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DSE Physics Recap – Reflection of Light | Vanilla (2026)

Reflection of Light

DSE Physics

Vanilla (2026)

For use on Saturday, 5 September 2026

01

Fundamental Principles

Laws of Reflection

A narrow ray of light travels towards a plane mirror at point O. The angle between the incident ray and the mirror surface is 32°. The incident ray, the normal and the reflected ray are all in the same vertical plane.

Incident ray Reflected ray Normal O 32° Plane mirror

The diagram is not drawn to scale.

  1. State the two laws of reflection.
  2. Define:
    1. the angle of incidence;
    2. the angle of reflection.
  3. Calculate the angle of incidence of the ray.
  4. Hence determine the angle between the incident ray and the reflected ray.
  5. The incident ray is kept fixed. The mirror is rotated clockwise through 7° about point O, while the incident ray continues to strike O.
    1. State the angle through which the normal rotates.
    2. Determine the angle through which the reflected ray rotates.
  6. A student says, “The angle between the incident ray and the mirror is the angle of incidence.” Explain why this statement is incorrect.
12 marks
02

Reflection from Surfaces

Regular and Diffuse Reflection

Two parallel light rays are directed onto surface P and surface Q. Surface P is a polished metal surface, while surface Q is a sheet of rough white paper.

Surface P Polished metal Surface Q Rough white paper Regular reflection Diffuse reflection

The short dashed lines represent local normals at the points of incidence.

  1. Name the type of reflection occurring at P and at Q.
  2. Explain why the reflected rays from P remain parallel, while those from Q travel in different directions.
  3. A student claims that the laws of reflection are not obeyed at Q because the reflected rays are not parallel. Comment on the claim.
  4. Explain why a clear image may be seen in polished metal but not in a sheet of ordinary white paper.
  5. A white page can be seen by students sitting at many different positions in a classroom. Explain this observation using diffuse reflection.
  6. Under certain conditions, a polished surface produces strong glare. Explain why roughening the surface can reduce the glare.
  7. State one similarity and one difference between regular reflection and diffuse reflection.
13 marks
03

Ray Construction

Formation of an Image by a Plane Mirror

An illuminated object AB is placed in front of a vertical plane mirror. Point A is at the top of the object and point B is at its base. The object is 1.6 m tall and is 2.4 m in front of the mirror.

A B Object Plane mirror 2.4 m 1.6 m Behind mirror

Complete the required construction on a printed copy of the diagram where appropriate.

  1. Describe how to construct the position of the image A′B′ using perpendicular distances from the mirror.
  2. State the distance between the image and the mirror.
  3. Calculate the distance between the object and its image.
  4. State the height of the image.
  5. Describe how two rays from point A can be used to locate A′ by a ray diagram. Your answer should refer to reflected rays and backward extensions.
  6. The object moves 0.70 m directly towards the stationary mirror.
    1. Determine the distance moved by the image.
    2. Determine the decrease in the separation between the object and the image.
  7. Vanilla raises her right hand while facing the mirror. The image appears to raise its left hand. Explain why this effect is described as lateral inversion and not an actual rotation of the image.
14 marks
04

Image Classification

Image Properties: Virtual and Real Images

A student investigates images formed in several optical situations. The observations are recorded below.

Situation Observation
P An image is seen behind a plane mirror, but no clear image is obtained when a screen is placed at that apparent position.
Q Reflected light rays actually meet at a point, and a sharp image can be obtained on a screen placed there.
R Reflected rays spread out, but their backward extensions meet at a point.
  1. Classify the image in each of P, Q and R as real or virtual.
  2. Define a real image in terms of the paths of actual light rays.
  3. Define a virtual image in terms of the paths of actual light rays and their backward extensions.
  4. Explain why the image formed by a plane mirror cannot be projected onto a screen placed behind the mirror.
  5. State four properties of the image formed by a plane mirror, excluding the fact that it is virtual.
  6. Light does not actually travel behind an ordinary plane mirror. Explain why an observer nevertheless judges the image to be behind the mirror.
  7. A student says, “A virtual image is imaginary and therefore cannot be photographed.” Explain why this statement is incorrect.
  8. Give one experimental observation that can distinguish a real image from a virtual image.
15 marks
05

Practical Applications

Plane Mirrors in a Periscope

A simple periscope contains two parallel plane mirrors, M₁ and M₂. Each mirror is inclined at 45° to the vertical axis of the tube. A horizontal ray from a distant object enters the top opening.

M₁ M₂ Incoming ray Emergent ray Simple periscope

The periscope diagram is schematic and is not drawn to scale.

  1. Explain why the first mirror changes the direction of the horizontal incoming ray through 90°.
  2. Describe the complete path of the ray through the periscope until it enters the observer’s eye.
  3. Compare the direction of the incoming ray with that of the emergent ray.
  4. Explain why the two mirrors should be parallel if the emergent ray is to be parallel to the incoming ray.
  5. A technician accidentally rotates M₁ clockwise through 3° while M₂ remains fixed. State the angle through which the ray reflected from M₁ changes direction.
  6. The reflective surfaces of the mirrors are placed on the front rather than behind a thick glass layer. Suggest why this can improve image quality.
  7. State one practical use of a periscope and explain why mirrors make the application useful.
  8. Apart from a periscope, give one application of a plane mirror and explain how one property of its image is useful in that application.
14 marks

Question 1

  1. The two laws of reflection are:
    • The angle of incidence is equal to the angle of reflection.
    • The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
    1. The angle of incidence is the angle between the incident ray and the normal at the point of incidence.
    2. The angle of reflection is the angle between the reflected ray and the normal at the point of incidence.
  3. The mirror surface is perpendicular to the normal.
    Angle of incidence
    = 90° − 32°
    = 58°
  4. By the law of reflection, the angle of reflection is also 58°.
    Angle between the incident and reflected rays
    = 58° + 58°
    = 116°
    1. The normal remains perpendicular to the mirror. Therefore, when the mirror rotates through 7°, the normal also rotates through 7° clockwise.
    2. For a fixed incident ray, when a plane mirror rotates through an angle θ, the reflected ray rotates through 2θ.
      Change in direction of reflected ray
      = 2 × 7°
      = 14° clockwise
  6. The angle of incidence is measured from the normal, not from the mirror surface. Since the normal is perpendicular to the mirror, the angle of incidence is complementary to the angle between the incident ray and the mirror.
Important: All angles of incidence and reflection are measured from the normal. If the angle between a ray and a mirror surface is θ, the angle between that ray and the normal is 90° − θ.

Question 2

  1. P produces regular reflection.
    Q produces diffuse reflection.
  2. Surface P is smooth, so the normals at the different points of incidence are parallel. The parallel incident rays have equal angles of incidence and are reflected in the same direction. Therefore, the reflected rays remain parallel.

    Surface Q is rough at a microscopic level. Its local normals point in different directions. Although the incident rays are parallel, they have different angles of incidence relative to the different local normals. They are therefore reflected in different directions.
  3. The claim is incorrect. Each ray at Q still obeys the law of reflection: its angle of incidence is equal to its angle of reflection when both angles are measured from the local normal. The rays are not parallel because the local normals are not parallel.
  4. Regular reflection from polished metal preserves the orderly arrangement of rays coming from different points of an object. The eye can trace the reflected rays back to definite image points, so a clear image is formed.

    Ordinary paper causes diffuse reflection. Light from each object point is scattered in many directions, so the orderly ray pattern needed to form a clear image is not preserved.
  5. The rough surface of the page reflects light in many directions. Therefore, some reflected light can enter the eyes of students at many different viewing positions. This allows the page to be seen over a wide range of directions.
  6. A polished surface may direct a large amount of reflected light into a particular direction, producing intense glare. Roughening the surface causes diffuse reflection, so the reflected light is spread over many directions and less light reaches the eye from any one direction.
  7. Similarity: Every individual light ray obeys the laws of reflection in both cases.

    Difference: regular reflection from a smooth surface sends parallel incident rays into one common direction, while diffuse reflection from a rough surface sends them in different directions.
Diffuse reflection does not mean random violation of the laws of reflection. The surface consists of many small regions with different orientations, and each region has its own local normal.

Question 3

  1. Draw a perpendicular line from A to the mirror and extend it behind the mirror. Mark A′ so that its perpendicular distance behind the mirror equals the perpendicular distance of A in front of the mirror.

    Repeat the process for B to locate B′. Join A′ and B′ to obtain the complete image.
  2. A plane mirror forms an image as far behind the mirror as the object is in front of it.
    Image distance from mirror = 2.4 m
  3. Object–image separation
    = object distance + image distance
    = 2.4 m + 2.4 m
    = 4.8 m
  4. A plane mirror forms an image of the same size as the object.
    Image height = 1.6 m
  5. Draw two rays from A to two different points on the mirror. At each point:
    • draw the normal;
    • draw the reflected ray such that the angle of reflection equals the angle of incidence;
    • extend the reflected ray backward behind the mirror using a dashed line.
    The backward extensions meet at A′. Their intersection gives the apparent position of the image of A.
    1. The image remains the same distance behind the mirror as the object is in front. When the object moves 0.70 m towards the mirror, the image also moves 0.70 m towards the mirror on the other side.
    2. Both the object and its image move 0.70 m towards the mirror.
      Decrease in separation
      = 0.70 m + 0.70 m
      = 1.40 m
  7. A plane mirror reverses the apparent front–back direction. This makes the observer interpret the image as having left and right interchanged. However, the image remains upright and is not physically turned around. The effect is an apparent sideways reversal, called lateral inversion, rather than an actual rotation.
For a stationary plane mirror:
image distance = object distance.
Therefore, if an object moves a distance d directly towards the mirror, the object–image separation decreases by 2d.

Question 4

    • P: virtual image.
    • Q: real image.
    • R: virtual image.
  2. A real image is formed at a position where actual light rays converge or meet after reflection or refraction. Since light really passes through the image position, the image can normally be projected onto a screen.
  3. A virtual image is formed at a position from which light rays appear to diverge. The actual rays do not meet at the virtual image position; instead, the backward extensions of the rays meet there.
  4. The reflected rays remain in front of the plane mirror and enter the observer’s eyes. They only appear to come from behind the mirror. No actual light rays pass through or converge at the image position behind the mirror. A screen placed there therefore cannot receive the required rays to form the image.
  5. Any four:
    • upright;
    • same size as the object;
    • the same perpendicular distance behind the mirror as the object is in front;
    • laterally inverted;
    • the image and object are symmetrical about the mirror plane.
  6. The human eye assumes that light travels in straight lines. The reflected rays entering the eye are therefore traced backward in straight lines. Their backward extensions meet behind the mirror, so the brain interprets the light as coming from an object at that position.
  7. The statement is incorrect. A virtual image has a definite apparent position and can send reflected or refracted light into a camera. A camera lens can focus the rays entering it onto the camera sensor, so a virtual image can be photographed. It is called virtual because the rays do not actually meet at the apparent image position, not because the image cannot be observed or recorded.
  8. Place a screen at the image position:
    • if a sharp image can be obtained on the screen, the image is real;
    • if no image can be obtained there even though the image can be seen by looking into the optical system, the image is virtual.
Real image: actual rays meet; it can be projected onto a screen.
Virtual image: actual rays do not meet at the image position; backward extensions meet, and the image cannot be projected directly onto a screen there.

Question 5

  1. Since the mirror is at 45° to the incoming horizontal ray, its normal is also at 45° to the ray. Therefore:
    Angle of incidence = angle of reflection = 45°
    Total change in direction
    = 45° + 45°
    = 90°
  2. The horizontal incoming ray strikes M₁ and is reflected vertically down the periscope. It then strikes M₂ and is reflected horizontally towards the observer. The ray enters the observer’s eye after undergoing two reflections.
  3. The emergent ray travels in the same direction as and parallel to the incoming ray, although it has been displaced vertically.
  4. Parallel mirrors have parallel normals. The second reflection reverses the directional change produced by the first reflection. Consequently, the final ray is parallel to the original incoming ray. If the mirrors are not parallel, their normals have different relative orientations, so the second reflection does not produce the required final direction.
  5. For a fixed incident ray, rotating a plane mirror through an angle θ changes the direction of the reflected ray through 2θ.
    Change in reflected-ray direction
    = 2 × 3°
    = 6° clockwise
  6. In an ordinary back-silvered mirror, a small amount of light may reflect from the front glass surface while most reflects from the metal coating behind the glass. These separate reflections may produce faint multiple or ghost images.

    A front-surface mirror reflects light at one main surface and therefore gives a sharper image.
  7. One example is a submarine periscope. The mirrors redirect light from an object above the water to an observer below the water, allowing the observer to see above the surface while remaining in a protected or concealed position.

    Other acceptable examples include observation from a trench or viewing over a tall obstacle.
  8. Any suitable application and explanation, for example:
    • Dressing mirror: the upright image and unchanged image size allow a person to inspect their appearance.
    • Bathroom mirror: the upright, same-sized virtual image provides a natural view of the user.
    • Interior rear-view mirror: the mirror redirects light from vehicles behind towards the driver. The upright image allows the driver to identify traffic behind.
    • Optical alignment: the predictable equality of the angles of incidence and reflection allows a beam to be redirected accurately.
Two parallel plane mirrors in a simple periscope produce two reflections. The emergent ray is parallel to the incoming ray, allowing the observer to see from a displaced position.
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