The key idea
The most diagram-heavy topic on the paper. The physics lives in three rules of reflection, one law of refraction, and two special rays through a lens. Draw with a ruler; freehand rays lose marks that cost nothing to keep.
Understand it · 01
Construct a plane-mirror image
At a plane mirror, the angle of reflection equals the angle of incidence. Both angles are measured from the normal, a line drawn perpendicular to the mirror at the point where the ray strikes. If a question gives an angle measured from the mirror surface, subtract it from 90° before using the reflection law. Put arrows on rays so their direction of travel is unambiguous.
A plane-mirror image is virtual because the reflected rays only appear to come from behind the mirror. Draw their backward extensions as dashed lines. The image is the same distance behind the mirror as the object is in front, has the same size and is upright. A screen behind the mirror cannot collect this image because the reflected light does not actually travel there.
Understand it · 02
Explain bending through a change of speed
When light enters glass from air, it usually slows and bends towards the normal if it meets the boundary at an angle. Its frequency remains fixed by the source, so its wavelength decreases. When it returns to air it speeds up and bends away from the normal. At normal incidence the speed changes but the direction does not, so a speed change does not always produce a visible bend.
For light entering a material from air, refractive index is approximately n = sin i/sin r, where i is the angle in air and r is the angle in the material, both measured from the normal. Keep the calculator in degree mode. A rectangular block produces an emergent ray parallel to the incident ray when its opposite faces are parallel and the medium is the same on both sides. The ray is displaced sideways, rather than permanently rotated.
Understand it · 03
Check both conditions for total internal reflection
Total internal reflection requires light to travel towards a boundary with a lower refractive index, such as glass to air, and to have an incidence angle greater than the critical angle. At the critical angle the refracted ray travels along the boundary. Below it, some light is transmitted. Reversing the direction to air-to-glass removes the required condition, however large the incidence angle becomes.
An optical fibre guides light through repeated total internal reflections at the core boundary. In an ideal ray model the light stays inside; real fibres still have losses from absorption and imperfect construction. For a material-to-air boundary, sin c = 1/n. That expression should not be used unchanged for every pair of materials: it assumes the second refractive index is approximately one.
Understand it · 04
Use two reliable rays to locate an image
For a thin converging lens, a ray parallel to the principal axis refracts through the far principal focus. A ray through the optical centre continues approximately straight in the thin-lens model. Start both rays at the same point on the object. Where the actual refracted rays cross is the corresponding point on a real image; extend the construction to determine its size and orientation.
With an object outside the focal length, a converging lens can form a real inverted image that a screen can receive. With the object inside the focal length, the outgoing rays diverge and their backward extensions meet at a virtual upright magnified image. Do not draw solid rays backwards through the lens to force an intersection: the dashed extensions represent apparent origins, not additional physical light paths.
Another view of the physics
A converging lens: object beyond 2F
Key definitions
- Law of reflection
- The angle of incidence equals the angle of reflection, both measured from the normal.
- Refractive index (n)
- The ratio of the speed of light in vacuum to its speed in a medium. For light entering from air, n ≈ sin i ÷ sin r, with both angles measured from the normal.
- Critical angle
- The angle of incidence in the higher-refractive-index medium for which the refracted ray lies along the boundary. Total internal reflection occurs above this angle when light is travelling towards a lower-refractive-index medium.
- Principal focus (focal point)
- The point at which rays parallel to the principal axis converge after passing through a converging lens.
Equations, units & conditions
Refractive index
n ≈ sin i ÷ sin r
Entering the material from air; angles from the normal.
Critical angle
sin c ≈ 1 ÷ n
Material-to-air boundary.
Worked examples, step by step
Example 1
The refractive index of glass is 1.5. Calculate the critical angle for the glass-air boundary.
Follow the method
- sin c = 1 ÷ n = 1 ÷ 1.5 = 0.667.
- c = sin⁻¹(0.667).
c ≈ 42°
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 3 suggested marks
1. Find refractive index
A ray enters a transparent block from air at an incidence angle of 42° and a refraction angle of 27°. Calculate its refractive index.
Show solution and self-check: question 1
- Use n = sin i/sin r.
- n = sin 42°/sin 27° = 1.4739… .
n ≈ 1.47, with no unit.
Self-check · one mark per point
- Use the sine ratio.
- Substitute the angles in the correct order using degree mode.
- Calculate approximately 1.47 without a unit.
Build · 3 suggested marks
2. Apply both TIR conditions
The critical angle of a glass–air boundary is 41°. Light inside the glass reaches the boundary at 46° to the normal. State what happens and explain why; then state what changes if the angle is reduced to 35°.
Show solution and self-check: question 2
- Light travels from glass towards air, a lower-index medium.
- 46° exceeds the critical angle, so total internal reflection occurs.
- At 35°, below the critical angle, some light refracts into the air.
Total internal reflection at 46°; transmission by refraction is possible at 35°.
Self-check · one mark per point
- Identify the glass-to-air condition.
- Compare 46° with 41° and conclude total internal reflection.
- State that a refracted ray is present at 35°.
Common mistakes & how to avoid them
- Measuring angles from the surface instead of the normal. Every angle in optics is measured from the normal — the perpendicular line at the point of incidence.
- Describing the plane-mirror image incompletely. Mark scheme: same size, same distance behind the mirror as the object is in front, virtual, and laterally inverted.
- Total internal reflection answers missing a condition. Two are required: the light must travel from the denser medium towards the less dense, and the angle of incidence must exceed the critical angle.
- Calling the image 'virtual' or 'real' by guesswork. If the rays actually meet, it is real and can be projected; if they only appear to come from a point, it is virtual — a magnifying glass makes a virtual image, a projector makes a real one.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
How do optical fibres work?
An optical fibre guides light by repeated total internal reflection at a boundary from a higher refractive index to a lower one. The incidence angle must exceed the critical angle. The reflections retain light within the guiding region, although real fibres still have losses.
A useful ray sketch shows each angle measured from the normal to the boundary, not from the fibre’s length or its surface. At the critical angle the transmitted ray would run along the boundary; above it, the ideal model has no transmitted ray. A fibre can guide light around gentle curves while maintaining the relevant angles. Bending too tightly can allow light to escape. Do not describe the fibre as a mirror-coated tunnel unless the question actually specifies reflective coatings: ordinary optical guidance uses refractive-index differences.
How do you draw a converging lens ray diagram?
Draw two rays from the same point on the object: one parallel to the principal axis, refracting through the far focus, and one through the optical centre, continuing approximately straight. Their intersection locates the corresponding image point in the thin-lens model.
Use a ruler, place both focal points at the stated focal distance, and mark the lens position before tracing rays. Starting the two rays from different object points creates an invalid image construction. For an object outside the focal length, actual refracted rays can meet on the far side and form a real image. If they diverge, extend them backwards with dashed lines to locate a virtual image. Keep arrowheads on the physical rays so a reader can distinguish real travel from a construction line.
Set up a scale construction
Draw a converging lens with focal length 5 cm and an object 15 cm to its left using 1 cm on paper for 5 cm in reality.
- Place the focal points 1 cm from the lens on the drawing.
- Place the object 3 cm to the left.
- Trace the parallel-to-focus ray and the optical-centre ray from its tip.
Use the intersection of those rays to read the image position and height from the scale drawing.
When does a converging lens act as a magnifying glass?
A converging lens acts as a magnifying glass when the object lies inside its focal length. It then produces an upright, enlarged virtual image on the object’s side. The rays reaching the eye diverge as if they originated from that larger image.
An enlarged image is not always virtual: an object just beyond the focal point can form an enlarged real image on the opposite side. The distinguishing feature of the magnifying-glass arrangement is the object being closer to the lens than the focal distance. A screen cannot collect its virtual image because light does not actually converge at that apparent location. In a diagram, extend the outgoing rays backwards rather than drawing them physically travelling backwards. Moving the object changes the image, so the focal position matters more than the lens’s name.
Why must refraction angles be measured from the normal?
Angles of incidence and refraction are measured from the normal, perpendicular to the boundary. If an angle is given from the surface, subtract it from 90°. For light entering a material from air, its refractive index is approximately sin i divided by sin r.
A common numerical error uses a surface angle as though it were the angle to the normal. Draw a short normal at the point of entry and label the actual incident and refracted angles before touching the calculator. Use degree mode and take each sine separately. For the reverse journey, the angle in the material and the angle in air exchange incident/refraction roles, so the same ratio cannot be applied unchanged. Refractive index is a speed ratio and therefore has no unit.
Convert a surface angle
A ray in air makes 30° with a glass surface and refracts at 35° to the normal. Calculate the glass refractive index.
- The incidence angle is 90° − 30° = 60°.
- n ≈ sin 60°/sin 35° ≈ 1.51.
The refractive index is approximately 1.51.
What is the difference between a real image and a virtual image?
A real image forms where light rays actually converge and can be received on a screen. A virtual image forms where diverging rays appear to originate when extended backwards. “Virtual” describes ray geometry, not whether the image can be seen by an eye.
A plane-mirror image is visible because reflected rays enter the eye, and the brain interprets their directions as coming from behind the mirror. No reflected light needs to travel through the mirror to that apparent position. A projected image on a cinema screen, by contrast, is formed by real convergence of light onto the screen. Do not use upright versus inverted as the definition: these are common features of particular arrangements, while the fundamental test is whether actual rays meet or only their extensions meet.
Why is a ray emerging from a rectangular glass block parallel to the incident ray?
For a rectangular block with parallel faces and the same surrounding medium on both sides, the emergent ray is parallel to the incident ray but displaced sideways. Refraction at the second boundary reverses the first angular change; it does not necessarily undo the sideways displacement.
Draw normals at both entry and exit points. Because the faces are parallel, those normals are parallel too, and the angle inside the block at the exit matches the corresponding refracted angle at entry. Returning to the original medium restores the original direction. A prism behaves differently because its faces are not parallel, so this conclusion cannot be transferred to every glass shape. At normal incidence there is no sideways displacement, even though light still changes speed and wavelength while inside the material.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Use the law of reflection (angle of incidence = angle of reflection) and describe the image in a plane mirror
- Describe refraction at a boundary and define refractive index n = sin i ÷ sin r
- Explain total internal reflection and the critical angle; apply to optical fibres
- Draw ray diagrams for a thin converging lens and describe the images formed
- Describe dispersion of white light by a prism
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Light. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/light
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