What are the basic properties of light in the context of Cambridge IGCSE Coordinated Science?

In the Cambridge IGCSE Coordinated Science curriculum, light is described as a form of electromagnetic radiation that travels in waves. Key properties include its speed in a vacuum (approximately 300,000 km/s), its ability to travel in straight lines, and its dual nature, exhibiting both wave-like and particle-like properties. Light can be reflected, refracted, and dispersed, which are essential concepts in understanding how it interacts with different materials.

How does refraction affect the path of light?

Refraction is the bending of light as it passes from one medium to another with a different density, such as from air to water. This occurs because light changes speed when it enters a new medium. In the Cambridge IGCSE Coordinated Science syllabus, students learn that the angle of incidence and the angle of refraction are related by Snell's Law. This principle explains phenomena like the apparent bending of a straw in a glass of water.

What is the significance of the electromagnetic spectrum in understanding light?

The electromagnetic spectrum encompasses all types of electromagnetic radiation, including visible light, which is only a small part of the spectrum. In the Cambridge IGCSE Coordinated Science curriculum, students explore how different wavelengths correspond to different types of radiation, such as ultraviolet, infrared, and X-rays. Understanding the electromagnetic spectrum is crucial for comprehending how light interacts with matter and the various applications of different types of electromagnetic waves.

How does the concept of reflection explain the behavior of light on surfaces?

Reflection occurs when light bounces off a surface. In the Cambridge IGCSE Coordinated Science course, students learn about the laws of reflection, which state that the angle of incidence equals the angle of reflection. This principle helps explain how mirrors work and why we see images. Reflection is also key in understanding phenomena like the formation of echoes in sound waves, highlighting the interconnectedness of wave properties.

Why is the study of light important in the Cambridge IGCSE Coordinated Science curriculum?

Studying light is crucial because it helps students understand fundamental scientific principles and their applications in real life. Light is involved in various technologies, from optical instruments like microscopes and telescopes to everyday devices like cameras and glasses. The Cambridge IGCSE Coordinated Science curriculum emphasizes these applications, preparing students for further studies in physics and related fields, and fostering an appreciation for the role of light in the natural world.

Why is "Measuring angles of incidence/refraction from the surface instead of the normal" a common mistake in Light?

Why it happens: The surface is the visible line; students measure from what they can see. How to avoid it: Always draw the normal first (dashed line perpendicular to the surface), then measure angles from it.

Why is "Forgetting both conditions for total internal reflection" a common mistake in Light?

Why it happens: Students remember 'angle must be large' but not 'must be in the denser medium'. How to avoid it: TIR requires: (1) light travelling in the denser medium, AND (2) angle of incidence greater than the critical angle. Both must be stated.

Why is "Inverting Snell's law, writing $n = \sin r / \sin i$" a common mistake in Light?

Why it happens: Confusion about which angle is in the incident medium. How to avoid it: n = i / r where i is in air (less dense). Remember: entering glass, bends toward normal, so r r, so n > 1.

Properties of Waves, including Light and SoundCambridge IGCSE Coordinated Science (0654)

Light

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Light — Cambridge IGCSE Coordinated Science 0654

Light is a transverse electromagnetic wave. Cambridge tests reflection in plane mirrors, refraction (Snell's law), total internal reflection, and how lenses form images — including real/virtual image distinctions.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

P4.2a — Describe reflection of light at plane mirrors.
P4.2b — Define and use refractive index; apply Snell's law.
P4.2c — Describe total internal reflection and its applications.
P4.2d — Describe how converging lenses form images; distinguish real from virtual.

Reflection and Refraction of Light

Know the law of reflection and Snell's law, and how to draw ray diagrams.

Reflection in plane mirrors:

Angle of incidence = angle of reflection (both measured from normal to surface)
Image in plane mirror: same size as object; same distance behind mirror as object in front; laterally inverted; VIRTUAL (cannot be projected)

Refraction:

Light bends when it enters a medium of different optical density
Going from less dense to more dense medium (e.g., air → glass): slows down → bends toward normal
Going from more dense to less dense (e.g., glass → air): speeds up → bends away from normal

Refractive index (n):

n = sin i / sin r (Snell's law, at a boundary) n = c / v (where c = speed in vacuum, v = speed in medium)

n has no unit (it's a ratio)
n > 1 for all real media (light slows in any medium compared to vacuum)

Example: Light travels from air into glass. The angle of incidence is 40° and the angle of refraction is 25°.

n = sin 40° / sin 25° = 0.643 / 0.423 = 1.52

Reflection: angle i = angle r (from normal).
Refraction: slows + bends toward normal entering denser medium.
n = sin i / sin r = c/v. Greater n → greater bending.

Total Internal Reflection

TIR occurs when light hits a boundary at an angle greater than the critical angle — used in optical fibres.

Total internal reflection (TIR):

Occurs when light travels from a denser medium to a less dense medium (e.g., glass → air)
When angle of incidence > critical angle (C): ALL light is reflected back; no refraction occurs

Critical angle:

sin C = 1/n (or n = 1/sin C)

For glass (n ≈ 1.5): C ≈ 42°
For diamond (n ≈ 2.4): C ≈ 25° → lots of TIR → brilliant sparkle

Applications of TIR:

Optical fibres: light enters one end → undergoes repeated TIR along the fibre → exits at other end
- Used in: medical endoscopes, telecommunications (internet cables)
- Advantages over copper wire: no electrical interference, very fast data transmission, signal carries further
Periscopes and binoculars: glass prisms (45°-45°-90°) reflect light by TIR (better than mirrors — 100% reflection)
Diamonds: high refractive index → small critical angle → TIR occurs for most angles of incidence → bright internal reflections → sparkling appearance

TIR: light in denser medium; angle i > critical angle C.
sin C = 1/n. Greater n → smaller C → TIR occurs more easily.
Optical fibres use TIR to carry light signals; used in telecoms and medicine.

Lenses and Image Formation

Converging lenses form real or virtual images depending on object distance relative to focal length.

Converging (convex) lens:

Parallel rays converge (meet) at the focal point F
Focal length f = distance from lens to focal point

Ray diagrams — three rules:

Ray parallel to axis → refracted through F (focal point)
Ray through centre of lens → passes straight through (undeviated)
Ray through F on object side → refracted parallel to axis

Image types:

Object position	Image type	Characteristics	Use
Beyond 2F	Real, inverted, diminished	Same side as image from 2F to F	Camera
At 2F	Real, inverted, same size	At 2F on other side	—
Between F and 2F	Real, inverted, magnified	Beyond 2F on other side	Projector
At F	No image	Parallel rays emerge	Searchlight
Inside F	Virtual, upright, magnified	Same side as object	Magnifying glass

Real vs virtual images:

Real: formed where light rays actually cross; can be projected onto a screen; on opposite side of lens to object
Virtual: light rays appear to diverge from this point but don't actually cross there; cannot be projected; on same side of lens as object

The eye as a lens system:

Converging lens (cornea + crystalline lens) focuses image on retina
Near objects: ciliary muscles contract → lens becomes thicker → shorter focal length (more converging)
Far objects: ciliary muscles relax → lens thinner → longer focal length (less converging)

Converging lens: parallel rays → focal point F.
Object beyond F → real, inverted image. Object inside F → virtual, upright, magnified.
Real image: rays actually cross; can project onto screen.

How it’s examined

Paper 4: 'Light strikes a glass surface at 40°. The glass has refractive index 1.5. Calculate the angle of refraction' (2 marks — sin r = sin 40°/1.5 = 0.429; r = 25.4°). 'Calculate the critical angle for glass (n = 1.5)' (2 marks — sin C = 1/1.5 = 0.667; C = 41.8°). 'Describe ONE use of optical fibres and explain how TIR enables this use' (3 marks). 'State the characteristics of the image formed by a converging lens of a very distant object' (2 marks — real, inverted, diminished, at focal point).

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (P4); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

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Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

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Step-by-step worked examples — Light

Step-by-step solutions to past-paper-style questions on light, written exactly the way a tutor would explain them at the board.

1Snell's law — finding angle of refraction
Core• Snell's law, refraction, refractive index
Question
A ray of light hits the boundary between air and glass at an angle of incidence of $40°$ . The refractive index of the glass is $1.5$ . Calculate the angle of refraction.
Step-by-step solution
1. Step 1
  Apply Snell's law: $n = \dfrac{\sin i}{\sin r}$ .
  $\sin r = \dfrac{\sin 40°}{1.5} = \dfrac{0.643}{1.5} = 0.429$
2. Step 2
  Find $r$ .
  $r = \arcsin(0.429) = 25.4° \approx 25°$
Answer
Angle of refraction $\approx 25°$
Examiner tip
The angle is always measured from the normal, not from the surface. Remind yourself of this before drawing any ray diagram.
2Finding the critical angle from refractive index
Extended• critical angle, total internal reflection, refractive index
Question
The refractive index of a glass fibre is $1.5$ . Calculate the critical angle for light travelling from glass to air.
Step-by-step solution
1. Step 1
  Use $n = \dfrac{1}{\sin c}$ , where $c$ is the critical angle.
  $\sin c = \dfrac{1}{n} = \dfrac{1}{1.5} = 0.667$
2. Step 2
  Find $c$ .
  $c = \arcsin(0.667) = 41.8° \approx 42°$
Answer
Critical angle $\approx 42°$
Examiner tip
Total internal reflection occurs when the angle of incidence inside the denser medium exceeds this critical angle.
3Converging lens — ray diagram and image characteristics
Challenge• converging lens, ray diagram, real image, virtual image
Question
An object is placed $30\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . Describe the image formed using a ray diagram approach and state the image characteristics.
Step-by-step solution
1. Step 1
  Draw three principal rays from the top of the object: (1) Parallel to the axis → refracts through the far focal point. (2) Through the optical centre → passes straight through undeviated. (3) Through the near focal point → refracts parallel to the axis.
2. Step 2
  The object is at $30\,\text{cm}$ ; focal length $= 10\,\text{cm}$ ; so object is beyond $2f$ ( $2 \times 10 = 20\,\text{cm}$ ).
3. Step 3
  For an object beyond $2f$ : image is real (rays converge on the far side), inverted, and diminished (smaller than object), formed between $f$ and $2f$ on the far side.
Answer
Image is real, inverted, and diminished, formed between $f$ and $2f$ on the opposite side of the lens from the object.

Key Formulae — Light

The formulae you need to memorise for light on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Snell's law / refractive index
$n = \dfrac{\sin i}{\sin r}$
$n$
refractive index (no units)
$i$
angle of incidence (from normal, in air)
$r$
angle of refraction (from normal, in denser medium)
When to use
Refraction at a boundary between air and a denser medium.
Critical angle
$n = \dfrac{1}{\sin c}$
$n$
refractive index of the denser medium
$c$
critical angle
When to use
Finding the critical angle for total internal reflection from a medium with known refractive index.

Key Definitions and Keywords — Light

Definitions to memorise and the exact keywords mark schemes credit for light answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Normal
Examiner keyword
An imaginary line drawn perpendicular to a surface at the point of incidence. All angles of incidence and refraction are measured from the normal.
Refraction
Examiner keyword
The change in direction of a wave when it crosses a boundary between two media of different speeds. The wave bends toward the normal when entering a denser medium.
Total internal reflection (TIR)
Examiner keyword
When light travelling in a denser medium hits the boundary at an angle of incidence greater than the critical angle, all light is reflected back into the denser medium. No refraction occurs.
Related: critical angle, optical fibre
Critical angle
Examiner keyword
The angle of incidence (inside the denser medium) at which the refracted ray travels exactly along the boundary (angle of refraction $= 90°$ ). Above this angle, total internal reflection occurs.
Real image
Examiner keyword
An image formed where light rays actually converge. It can be projected onto a screen. Produced by a converging lens when the object is beyond the focal point.
Related: virtual image

Common Mistakes and Misconceptions — Light

The traps other students keep falling into on light questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Measuring angles of incidence/refraction from the surface instead of the normal
Why it happens
The surface is the visible line; students measure from what they can see.
How to avoid it
Always draw the normal first (dashed line perpendicular to the surface), then measure angles from it.
✕Forgetting both conditions for total internal reflection
Why it happens
Students remember 'angle must be large' but not 'must be in the denser medium'.
How to avoid it
TIR requires: (1) light travelling in the denser medium, AND (2) angle of incidence greater than the critical angle. Both must be stated.
✕Inverting Snell's law, writing $n = \sin r / \sin i$
Why it happens
Confusion about which angle is in the incident medium.
How to avoid it
$n = \sin i / \sin r$ where $i$ is in air (less dense). Remember: entering glass, bends toward normal, so $r < i$ , so $\sin i > \sin r$ , so $n > 1$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Light — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Light

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Snell's law — finding angle of refraction

2Finding the critical angle from refractive index

3Converging lens — ray diagram and image characteristics

Snell's law / refractive index

Critical angle

Normal

Refraction

Total internal reflection (TIR)

Critical angle

Real image

✕Measuring angles of incidence/refraction from the surface instead of the normal

✕Forgetting both conditions for total internal reflection

✕Inverting Snell's law, writing n=sin⁡r/sin⁡in = \sin r / \sin in=sinr/sini

Practice questions

Past paper style quiz

Assess your understanding

Light — frequently asked questions

✕Inverting Snell's law, writing $n = \sin r / \sin i$