GRADE 8 · LIGHT PHENOMENA
Reflection of light and mirrors
The law of reflection and the construction of an image in a plane mirror.
Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.
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?
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.
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.
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.
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.
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.
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.
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.
Angle from the surface
An incident ray makes an angle of 30° with a mirror surface. Find the angles of incidence and reflection.
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°.
Worked example · 2
By the reflection law, β = α = 60°. Measure this second angle from the normal too, keeping the same reference line.
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.
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
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
Angle of incidence
\alphaThe 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?
Initial values
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.