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GRADE 8 · LIGHT PHENOMENA

Reflection of light and mirrors

The law of reflection and the construction of an image in a plane mirror.

22 minutes4 theory cards
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Understanding step by step

A mirror makes space appear behind its surface, yet the reflected light does not travel there. How does the same reflection rule for each ray produce a whole image?

Object and virtual imageLampimageeyeLight reflects in front of the mirrorDashes: ray extensions
The lamp and its virtual image are equally far from the mirror. Two reflected paths from one lamp point are shown; one reaches the eye. Dashed extensions meet at the corresponding image point. The grey icon on the right represents the image, not a second lamp.

Start with the normal

To compare light falling on different surfaces, we need a common reference line. At the point where the ray meets the mirror, draw the normal: a line perpendicular to the surface. The angle of incidence α is measured between the incident ray and this normal, in degrees. If the ray follows the normal, α = 0°, while its angle to the surface is 90°. This simple check prevents confusing the two angles.

αβNormalα = β
The normal passes through the shared incidence/reflection point, perpendicular to the mirror. Angle α between the incident ray and normal equals angle β between the reflected ray and normal. All three directions lie in the picture plane; both arcs are centred at the reflection point.

Construct the reflected ray

Experiments with a smooth mirror give the law: the reflection angle β equals the incidence angle α, so β = α. Both are measured from the normal; the incident ray, reflected ray and normal lie in one plane. Draw the outgoing ray symmetrically to the incident ray about the normal. Increasing α by 10° also increases β by 10°; at α = 0° light returns along the same line. Arrows distinguish travel towards and away from the mirror.

Symmetry about the normalnormalαβα = βRays and normal lie in one plane
The red incident ray and blue reflected ray meet the mirror at the same point. The normal passes through it, and angles α and β are equal. Rays and normal lie in the picture plane; arrowheads show travel towards and away from the mirror.

Roughness changes the normals

Why does a mirror reflect light in an ordered way while paper sends it in many directions? Neighbouring regions of a smooth plane mirror have parallel normals, so incident rays with the same direction also reflect in the same direction. Small regions of a rough surface tilt differently: their normals and reflected directions differ. Each region still obeys β = α. Diffuse reflection does not mean that the law is randomly broken.

SmoothRoughDashed lines: local normals
Incident rays with the same direction reflect according to the equal-angle law. On the left the normals are parallel; on the right magnified surface regions have different tilts. Dashed normals are perpendicular to their own regions, and reflected directions are calculated from these normals. Surface irregularities are shown schematically enlarged.

Why a matte object is visible

White paper can be seen from many places because its reflected light spreads in different directions. A mirror can also send light into the eye, but the directions remain linked to each object point in a way that forms a sharp image. Imagine roughening the mirror: light does not disappear, but this ordered relation between directions is disrupted and the image blurs. Preserving that relation distinguishes image-forming reflection from simply illuminating a surface.

Matte paperMirrormany directionsone directionFor the incident beam shownColours distinguish incident and reflected light
Reflection of one narrow incident beam is compared. Paper scatters some light in many directions, including paths to the eyes shown. A smooth mirror reflects this beam in a particular direction. Different incident light gives different directions; observers at different positions can see a mirror image when suitable light paths exist.

Where the light seems to come from

Take an object point and two rays that enter the eye after reflection. Their directions make the light appear to have travelled straight from a point behind the mirror. Extend the reflected rays backwards: their extensions meet there. This is the original point's virtual image. Light really enters the eye and the image is visible, but those rays do not actually meet behind the mirror.

ObjectImagedd
Two incident and two reflected rays are shown for the object's tip. Their dashed backward extensions meet at the symmetric point behind the mirror. Object and upright image have the same height and equal perpendicular distances d to the mirror. Dashes on the right show apparent position, not a light path.

Distance and size

Equal angles imply symmetry: each image point lies behind a plane mirror at the same perpendicular distance as the object point in front. The image is therefore upright and the same size. Move the object one metre farther away and the image moves one metre farther behind the mirror; their separation grows by two metres. Do not confuse the image's size with how large it looks from a chosen viewpoint.

Symmetric about the mirrorobjectimageddmeasure perpendicularly, not diagonally
The image is symmetric about the mirror plane: the same distance behind it and the same height. Distances are measured perpendicular to the mirror, not diagonally.

Why call it ‘virtual’?

Putting a screen where the image appears to be is not enough: the real reflected rays do not arrive there. The virtual image therefore cannot be projected onto a screen at that position. Draw actual paths as solid arrowed lines and backward extensions as dashed lines. This distinction checks the construction's physical meaning: an image seen by the eye need not be formed by light gathering at its apparent location.

No image forms on the screenobjectИscreenSolid lines: light pathsDashes: extensions only
Light from the object point reflects in front of the mirror; only dashed extensions meet at the symmetric point И. A screen at that point receives no image, because the real rays do not converge there. Other sources may illuminate the screen, so absence of an image does not necessarily mean complete darkness.

Angle from the surface

An incident ray makes an angle of 30° with a mirror surface. Find the angles of incidence and reflection.

Known quantitiesnormal30°α — ?30° measured from the surfaceFind α and β from the normal
The incident ray makes 30° with the mirror surface. It reaches the mirror–normal intersection exactly. The green arc marks the given surface angle; α and the future reflection angle β are measured from the normal. The setup stays unchanged through all steps.

Worked example · 1

The given 30° is measured from the surface. The normal is perpendicular to it, so the angle of incidence is α = 90° − 30° = 60°.

Surface ⟂ normalnormal30°α = 60°α = 90° − 30° = 60°
With the ray and mirror unchanged, α complements 30° to make a right angle. The mark shows that normal and surface are perpendicular. Both arcs start from the same incidence point; α = 90° − 30° = 60°.

Worked example · 2

By the reflection law, β = α = 60°. Measure this second angle from the normal too, keeping the same reference line.

Law of reflectionnormal30°α = 60°β = 60°β = α = 60°
A reflected ray is added at the same incidence point. It is symmetric to the incident ray about the same normal: β = α = 60°. The original ray and given 30° from the surface remain unchanged.

Worked example · 3

Symmetry check: the reflected ray also makes 30° with the surface. Thus the answer 60° agrees with both the given angle and the law.

One diagram, two reference linesnormal30°α = 60°β = 60°30°From the normal: 60°From the surface: 30°
Mirror, normal and both rays stay fixed. Each ray makes 60° with the normal and 30° with the surface. Red and blue arcs are measured from the normal; green arcs from the surface. Their centres coincide with the reflection point.

Check the result

Incidence angle: 60°; reflection angle: 60°. Both are measured from the normal to the mirror.

Key idea · 1

The law β = α concerns angles from the normal and also applies to individual regions of a rough surface.

Key idea · 2

A plane mirror preserves the relation between ray directions and gives an image symmetric to the object about its plane.

Key idea · 3

Arriving rays make a virtual image visible, but only their extensions meet at its position, so a screen there cannot receive it.

What comes next

Light can change direction on entering another transparent medium. This enables lenses to form images.

Basic Concepts

3 concepts

Law of reflection

The incident ray, reflected ray and normal lie in one plane, and the reflection angle equals the incidence angle.

A plane mirror forms a virtual, upright image as far behind the mirror as the object is in front.

Specular and diffuse reflection

A smooth surface reflects parallel rays in an orderly manner, while a rough surface scatters them in different directions.

Thanks to diffuse reflection, a matte object can be seen from different sides, while a mirror produces a directional beam.

Image in a plane mirror

A plane mirror gives a virtual, upright image equal in size to the object and symmetric about the mirror plane.

The extensions of the reflected rays intersect behind the mirror at a distance equal to the distance from the object to the mirror.

Physical quantities

1 magnitude

Angle of incidence

\alpha

The angle between the incident ray and the perpendicular to the reflecting surface.

Unit: degree · {}^\circ

Test the relationship

Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.

Reflection and refraction

θотр = θ₁; n₁ sin θ₁ = n₂ sin θ₂

Plane boundary between transparent uniform media; ray optics. Line thickness does not encode power. Under total internal reflection there is no propagating transmitted ray.

At the same incidence angle, n₂ increases from 1.5 to 2. How will the refraction angle from the normal change?

Your prediction

Initial values

n₁ = 1n₂ = 1.5⊥

For reflection, follow the blue ray and the normal. The green ray belongs to the next topic: refraction.

Calculated result

θотр
30 °
θ₂
19.47 °
Explaining the result

Angles are measured from the normal. Entering a larger n bends the ray towards the normal. Total internal reflection requires n₁ > n₂ and incidence above the critical angle.

Problems on this topic

Statements, answer options and the walkthrough.