What is the nature of light in the context of Cambridge IGCSE Physics?

In Cambridge IGCSE Physics, light is described as a form of electromagnetic wave. It exhibits properties of both waves and particles, a concept known as wave-particle duality. Light travels in straight lines and can be reflected, refracted, and dispersed, demonstrating its wave-like behavior.

How does refraction of light occur and what are its effects?

Refraction of light occurs when light waves pass from one medium to another, causing a change in speed and direction. This bending of light is due to the difference in optical density between the two media. Common effects of refraction include the bending of a straw in a glass of water and the focusing of light by lenses.

What is the electromagnetic spectrum and where does visible light fit in?

The electromagnetic spectrum encompasses all types of electromagnetic radiation, from radio waves to gamma rays. Visible light is a small portion of this spectrum, ranging from approximately 400 nm (violet) to 700 nm (red) in wavelength. It is the only part of the spectrum visible to the human eye.

How does the speed of light vary in different media?

The speed of light is fastest in a vacuum, approximately 299,792,458 meters per second. When light enters a medium like air, water, or glass, its speed decreases due to the medium's optical density. This change in speed is responsible for phenomena such as refraction.

What is the principle of reflection of light and its applications?

The principle of reflection states that when light waves strike a surface, they bounce back at the same angle at which they hit the surface, known as the angle of incidence equaling the angle of reflection. This principle is used in various applications, such as periscopes, mirrors, and optical instruments.

Why is "Measuring angles from the surface, not the normal" a common mistake in Light?

Why it happens: Ruler reads from the surface. How to avoid it: ALWAYS draw the normal first; measure from it. Source: 0625/42 — every series

Why is "Mis-orienting which is $i$ and which is $r$" a common mistake in Light?

Why it happens: Light direction changes at the boundary. How to avoid it: i is in the FIRST medium, r is in the SECOND. Always check direction of travel.

Why is "Saying TIR happens when going from less dense to denser medium" a common mistake in Light?

Why it happens: Forgetting the prerequisite. How to avoid it: TIR requires going FROM the denser medium INTO the less dense one (e.g. glass to air).

Why is "Drawing real image rays as dashed (and vice versa)" a common mistake in Light?

Why it happens: Mixing the conventions. How to avoid it: Real image: SOLID converging rays meet. Virtual image: DASHED back-projected rays.

WavesCambridge IGCSE Physics (0625)

Light

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Light — Cambridge IGCSE 0625 Physics Extended (2026)

Reflection in plane mirrors, refraction with $n = \sin i / \sin r$ , total internal reflection and the critical angle, and image formation by lenses.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

3.3 — State and apply the law of reflection.
3.4 — Recall and use $n = \sin i / \sin r$ and $\sin c = 1/n$ .
3.4 — Describe total internal reflection and its application in optical fibres and prisms.
3.5 — Describe converging and diverging lenses and image formation.

Reflection at a plane mirror

Angle in = angle out. Image: virtual, upright, same size, laterally inverted, equidistant behind.

Law of reflection. Angle of incidence $=$ angle of reflection (both from the normal).

Image properties (plane mirror).

Virtual — light only APPEARS to come from behind the mirror; it doesn't actually arrive there.
Upright.
Same size as object.
Laterally inverted — left↔right swap.
Equidistant — image distance behind mirror = object distance in front.

Tip. When you stand in front of a mirror, your right hand appears to be on your reflection's "left". That's lateral inversion.

Worked qualitative. Why don't ambulances write "AMBULANCE" mirror-imaged on their bonnet? They DO — so that when a car ahead sees it in their rear-view mirror, it reads correctly forwards.

Angle of incidence = angle of reflection.
Plane mirror image: virtual, same size, upright, laterally inverted.
Image distance behind = object distance in front.

Refraction and refractive index

$n = \dfrac{\sin i}{\sin r}$ . Slow medium → bend toward normal.

Snell's law. When light passes from one medium to another: $n = \dfrac{\sin i}{\sin r},$ where $n$ is the refractive index of the second medium (relative to the first), $i$ is angle of incidence, $r$ is angle of refraction.

Both angles measured from the normal.

Also: $n = c_{\text{air}} / c_{\text{medium}}$ . Light is slower in denser media; the bigger $n$ , the slower light goes.

Worked. Light enters glass at $i = 40\degree$ . Glass has $n = 1.5$ . Find $r$ .

$\sin r = \sin 40\degree / 1.5 \approx 0.643/1.5 = 0.428$ .
$r = \sin^{-1}(0.428) \approx 25.4\degree$ .

Common refractive indices.

Air: $\approx 1$ .
Water: $1.33$ .
Glass: $1.5$ .
Diamond: $2.4$ .

Direction. Going INTO a denser medium (higher $n$ ) → light slows, bends TOWARD normal. Going OUT to a less dense medium → speeds up, bends AWAY from normal.

$n = \sin i / \sin r$ .
$n = c_{\text{air}}/c_{\text{medium}}$ .
Higher $n$ = denser optically = slower.
Always measure angles from normal.

Total internal reflection

Above critical angle (dense → less dense), all light reflects back.

Total internal reflection (TIR). When light travels from a DENSER medium toward a LESS dense one (e.g. glass → air), at angles larger than the critical angle, ALL light reflects back. None refracts out.

Critical angle. $\sin c = \dfrac{1}{n}.$

For glass ( $n = 1.5$ ): $c = \sin^{-1}(1/1.5) \approx 41.8\degree$ .

Conditions for TIR.

Light going from DENSER to less dense medium.
Angle of incidence GREATER than critical angle.

Applications.

Optical fibres: light enters at one end, hits the wall at $> c$ → TIR → bounces along the fibre. Used in telecommunications, endoscopes.
Prisms in periscopes / binoculars: 45-45-90 prisms reflect light internally at $45\degree$ (which is greater than glass's critical angle).

Worked. Find critical angle for water ( $n = 1.33$ ).

$c = \sin^{-1}(1/1.33) \approx 48.6\degree$ .

$\sin c = 1/n$ .
TIR: dense → less dense AND $\theta_{i} > c$ .
Optical fibres, prisms.
Glass critical angle: $\sim 42\degree$ .

Lenses

Converging: brings parallel rays together. Diverging: spreads them.

Converging (convex) lens. Thicker in the middle. Parallel rays meet at a single point — the focal point ( $F$ ). Distance from lens to focal point is the focal length ( $f$ ).

Diverging (concave) lens. Thinner in the middle. Parallel rays spread out as if coming from a focal point on the SAME side as the incoming rays (so it's "virtual").

Image formation by a converging lens. Depends on object distance vs $f$ :

Object beyond $2f$ : image real, inverted, smaller (camera).
Object at $2f$ : real, inverted, same size.
Between $f$ and $2f$ : real, inverted, magnified (projector).
At $f$ : image at infinity (parallel beam, used in spotlights).
Inside $f$ : virtual, upright, magnified (magnifying glass).

Diverging lenses always produce: virtual, upright, smaller image (used to correct short-sightedness).

Cambridge tip. Drawing a ray diagram: use the three principal rays.

Ray through optical centre — goes straight.
Ray parallel to axis — refracts through $F$ on the far side.
Ray through $F$ on the near side — emerges parallel to axis.

Converging: parallel rays meet at $F$ .
Diverging: parallel rays appear to spread from virtual $F$ .
Magnifier: object inside $f$ on a converging lens.
Camera: object beyond $2f$ .

How it’s examined

Light is heavily examined: every Paper 4 has a refraction-Snell's law calculation (3-4 marks), every other Paper 4 has a TIR / optical-fibre question, plus lens ray-diagram questions. Examiner reports flag measuring from the surface and missing the dense→less-dense direction requirement for TIR.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (3.3-3.5); 0625/42 Oct/Nov 2024 — Q14 (refractive index, TIR); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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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.

1Apply Snell's law
Extended• Adapted from 0625/42 May/Jun 2024 Q13• refraction
Question
Light enters glass from air with angle of incidence $40\degree$ . The refractive index of glass is $1.5$ . Find the angle of refraction.
Step-by-step solution
1. Step 1
  $n = \dfrac{\sin i}{\sin r}$ .
  $\sin r = \dfrac{\sin 40\degree}{1.5}$
2. Step 2
  Compute.
  $\sin r = \dfrac{0.643}{1.5} = 0.428 \Rightarrow r \approx 25.4\degree$
Answer
$r \approx 25.4\degree$
2Critical angle
Extended• TIR
Question
A material has refractive index $1.5$ . Find the critical angle for the material-air boundary.
Step-by-step solution
1. Step 1
  $\sin c = \dfrac{1}{n}$ .
  $\sin c = \dfrac{1}{1.5} \approx 0.667$
2. Step 2
  Apply $\sin^{-1}$ .
  $c \approx 41.8\degree$
Answer
$c \approx 41.8\degree$
3Law of reflection
Core• reflection
Question
A ray hits a plane mirror at $30\degree$ from the surface. Find the angle of reflection (measured from the normal).
Step-by-step solution
1. Step 1
  Angles in reflection are measured from the NORMAL.
2. Step 2
  Angle of incidence = $90\degree - 30\degree = 60\degree$ .
3. Step 3
  Angle of reflection = angle of incidence.
  $= 60\degree$
Answer
$60\degree$ (from the normal)
Examiner tip
Always measure ray angles from the NORMAL (perpendicular to the surface), not from the surface itself.
4Image from a converging lens
Extended• lens
Question
An object is placed $30\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . Describe the image (real/virtual, magnification, orientation).
Step-by-step solution
1. Step 1
  Object beyond $2f$ ( $30 > 20$ ): image is real, inverted, diminished.
Answer
Real, inverted, diminished, between $f$ and $2f$ on the other side.
5Image properties in a plane mirror
Core• reflection, image
Question
An object stands $20\,\text{cm}$ in front of a plane mirror. Describe the image: its position, size, orientation, and whether it is real or virtual.
Step-by-step solution
1. Step 1
  A plane mirror produces an image the SAME distance behind the mirror as the object is in front.
  $d_{\text{image}} = 20\,\text{cm behind}$
2. Step 2
  Same size as the object; upright; laterally inverted (left-right swapped).
3. Step 3
  The rays only APPEAR to come from behind the mirror; no rays actually meet there → virtual image.
Answer
Virtual, upright, same size, $20\,\text{cm}$ behind the mirror, laterally inverted.
6Find refractive index from angles
Extended• Adapted from 0625/42 Oct/Nov 2023 Q14• refraction, n
Question
Light passes from air into a transparent block with angle of incidence $50\degree$ and angle of refraction $30\degree$ . Find the refractive index of the block.
Step-by-step solution
1. Step 1
  $n = \dfrac{\sin i}{\sin r}$ .
  $n = \dfrac{\sin 50\degree}{\sin 30\degree} = \dfrac{0.766}{0.500}$
2. Step 2
  Compute.
  $n \approx 1.53$
Answer
$n \approx 1.53$
Examiner tip
The examiner report flags candidates often invert the ratio (using $\sin r / \sin i$ ). For light entering a denser medium $i > r$ , so $n > 1$ — use this as a sanity check.
7Critical angle from refractive index
Extended• TIR, critical angle
Question
A material has refractive index $n = 1.40$ . Find the critical angle for light passing from this material into air.
Step-by-step solution
1. Step 1
  $\sin c = 1/n$ .
  $\sin c = \dfrac{1}{1.40} = 0.714$
2. Step 2
  Apply $\sin^{-1}$ .
  $c \approx 45.6\degree$
Answer
$c \approx 45.6\degree$
8Total internal reflection in an optical fibre
Extended• Adapted from 0625/42 May/Jun 2023 Q14• TIR, fibre
Question
An optical fibre's core has refractive index $1.50$ . (a) Find the critical angle at the core-air interface. (b) Light enters the fibre and strikes the inner wall at $50\degree$ from the normal. Does it pass through or reflect? Explain.
Step-by-step solution
1. Step 1
  Critical angle from $\sin c = 1/n$ .
  $\sin c = \dfrac{1}{1.50} \approx 0.667 \Rightarrow c \approx 41.8\degree$
2. Step 2
  The angle of incidence at the wall is $50\degree > c = 41.8\degree$ .
3. Step 3
  Therefore total internal reflection occurs — all the light is reflected back into the core.
Answer
(a) $c \approx 41.8\degree$ . (b) Since $50\degree > c$ , the light undergoes total internal reflection and stays in the fibre.
9Converging lens ray diagram: object beyond 2F
Extended• lens, ray diagram
Question
An object of height $4.0\,\text{cm}$ is placed at $30\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . By scale drawing or by using lens properties, find the image distance and image height (image is at $15\,\text{cm}$ on the far side).
Step-by-step solution
1. Step 1
  Draw two key rays: one parallel to the axis (refracts through $F$ on far side) and one through the lens centre (undeviated).
2. Step 2
  The rays cross at the image position, between $F$ and $2F$ on the far side.
  $v = 15\,\text{cm}$
3. Step 3
  Linear magnification.
  $m = \dfrac{v}{u} = \dfrac{15}{30} = 0.5$
4. Step 4
  Image height.
  $h_{i} = m\,h_{o} = 0.5 \times 4.0 = 2.0\,\text{cm}$
Answer
Real, inverted, diminished image, $v = 15\,\text{cm}$ on the far side, height $2.0\,\text{cm}$ .
Examiner tip
The examiner report flags candidates often draw the parallel ray bending at the lens centre instead of at the principal axis. Refract the ray AT the lens line.
10Object between F and 2F (projector setup)
Extended• lens, ray diagram
Question
A slide of height $2.0\,\text{cm}$ is placed at $15\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . Describe the image and explain why this arrangement is used in a projector.
Step-by-step solution
1. Step 1
  Object distance $u = 15\,\text{cm}$ is between $F$ ( $10\,\text{cm}$ ) and $2F$ ( $20\,\text{cm}$ ).
2. Step 2
  Ray construction puts the image beyond $2F$ on the far side: real, inverted, MAGNIFIED.
3. Step 3
  In a projector the small slide produces a large image on a screen; the slide is therefore placed just outside $F$ . Slides are loaded UPSIDE DOWN so the screen image appears upright.
Answer
Real, inverted, magnified image beyond $2F$ on the far side — used in projectors.
11Dispersion in a prism
Extended• dispersion, prism
Question
White light enters a glass prism and emerges as a spectrum. (a) Explain why the light splits up. (b) State which colour is refracted most and why.
Step-by-step solution
1. Step 1
  White light is a mixture of all visible colours, each with a different frequency / wavelength.
2. Step 2
  The refractive index of glass is slightly different for each frequency (greater for higher frequency / shorter wavelength).
3. Step 3
  Each colour is therefore refracted by a different amount on entering and leaving the prism, fanning out into the spectrum red-orange-yellow-green-blue-indigo-violet.
4. Step 4
  (b) Violet has the shortest wavelength / highest frequency → largest $n$ in glass → refracted MOST. Red is refracted least.
Answer
(a) Different colours have different refractive indices in glass, so they refract by different amounts. (b) Violet is refracted most; red least.
12A* — Right-angled prism reflectors
Challenge• Adapted from 0625/42 Oct/Nov 2024 Q14• TIR, synoptic
Question
A $45\degree$ - $45\degree$ - $90\degree$ glass prism ( $n = 1.50$ ) is used as a periscope reflector. Light enters one short face along the normal. Explain, with reference to the critical angle, why total internal reflection occurs at the hypotenuse, and state one advantage of using such a prism instead of a silvered mirror.
Step-by-step solution
1. Step 1
  Find the critical angle.
  $\sin c = \dfrac{1}{1.50} \approx 0.667 \Rightarrow c \approx 41.8\degree$
2. Step 2
  Light entering normally goes straight through, then meets the hypotenuse face at $45\degree$ from the normal.
3. Step 3
  $45\degree > c$ ( $41.8\degree$ ) → TOTAL internal reflection: all light reflects, none refracts out.
4. Step 4
  Advantage: a silvered mirror absorbs a fraction of the light and the silver layer can tarnish or get scratched. TIR in a prism is 100% efficient and the glass surface is durable.
Answer
At the hypotenuse the $45\degree$ angle exceeds $c \approx 41.8\degree$ so all light undergoes TIR. Advantage: 100% reflection, no tarnishing.
13A* — Magnifying glass (virtual image)
Challenge• lens, synoptic
Question
An object of height $5.0\,\text{mm}$ is placed $6.0\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . (a) State whether the image is real or virtual and how you know. (b) From a scale ray diagram the image distance is $15\,\text{cm}$ on the same side as the object. Find the magnification and the image height. (c) Why is this used as a magnifying glass?
Step-by-step solution
1. Step 1
  Object distance $u = 6.0\,\text{cm}$ is LESS than $f = 10\,\text{cm}$ . The refracted rays diverge after the lens.
2. Step 2
  Tracing the diverging rays back, they meet on the SAME side as the object → virtual, upright, magnified image.
3. Step 3
  Magnification.
  $m = \dfrac{v}{u} = \dfrac{15}{6.0} = 2.5$
4. Step 4
  Image height.
  $h_{i} = m\,h_{o} = 2.5 \times 5.0 = 12.5\,\text{mm}$
5. Step 5
  (c) Placing the object inside $F$ gives a magnified upright virtual image — the eye sees an enlarged version of the object.
Answer
(a) Virtual (object inside $F$ ). (b) $m = 2.5$ ; image height $12.5\,\text{mm}$ . (c) Object inside $F$ → enlarged virtual image.
Examiner tip
The examiner report flags candidates often shade real images as if they were on a screen when the object is inside $F$ . Use DASHED rays back-projected to mark a virtual image.

Key Formulae — Light

The formulae you need to memorise for light on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Snell's law
$n = \dfrac{\sin i}{\sin r}$
$n$
refractive index of the second medium relative to the first
$i$
angle of incidence (from normal)
$r$
angle of refraction (from normal)
When to use
Light crossing a boundary between two media.
Critical angle
$\sin c = \dfrac{1}{n}$
When to use
Boundary from a denser medium back to air. Above $c$ , total internal reflection.
Law of reflection
$\theta_{i} = \theta_{r}$
When to use
Plane mirror; angles measured from the normal.

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 0625 sitting.

Normal
Examiner keyword
A line perpendicular to a surface at the point where the ray meets it. ALL ray angles are measured from this line.
Refraction
Examiner keyword
Change in the direction of light when it crosses a boundary, caused by a change in speed.
Refractive index ( $n$ )
Examiner keyword
$n = \dfrac{c_{\text{vacuum}}}{c_{\text{medium}}}$ — how much slower light is in the medium.
Total internal reflection
Examiner keyword
When light strikes a denser-medium boundary at an angle larger than the critical angle, it reflects back entirely.
Focal length
Examiner keyword
Distance from the lens centre to the focal point ( $F$ ); a measure of how strongly the lens converges/diverges light.

Common Mistakes and Misconceptions — Light

The traps other students keep falling into on light questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Measuring angles from the surface, not the normal
0625/42 — every series
Why it happens
Ruler reads from the surface.
How to avoid it
ALWAYS draw the normal first; measure from it.
✕Mis-orienting which is $i$ and which is $r$
Why it happens
Light direction changes at the boundary.
How to avoid it
$i$ is in the FIRST medium, $r$ is in the SECOND. Always check direction of travel.
✕Saying TIR happens when going from less dense to denser medium
Why it happens
Forgetting the prerequisite.
How to avoid it
TIR requires going FROM the denser medium INTO the less dense one (e.g. glass to air).
✕Drawing real image rays as dashed (and vice versa)
Why it happens
Mixing the conventions.
How to avoid it
Real image: SOLID converging rays meet. Virtual image: DASHED back-projected rays.

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Light — frequently asked questions

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WavesCambridge IGCSE Physics (0625)

Light

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Light — Cambridge IGCSE 0625 Physics Extended (2026)

Reflection in plane mirrors, refraction with $n = \sin i / \sin r$ , total internal reflection and the critical angle, and image formation by lenses.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

3.3 — State and apply the law of reflection.
3.4 — Recall and use $n = \sin i / \sin r$ and $\sin c = 1/n$ .
3.4 — Describe total internal reflection and its application in optical fibres and prisms.
3.5 — Describe converging and diverging lenses and image formation.

Reflection at a plane mirror

Angle in = angle out. Image: virtual, upright, same size, laterally inverted, equidistant behind.

Law of reflection. Angle of incidence $=$ angle of reflection (both from the normal).

Image properties (plane mirror).

Virtual — light only APPEARS to come from behind the mirror; it doesn't actually arrive there.
Upright.
Same size as object.
Laterally inverted — left↔right swap.
Equidistant — image distance behind mirror = object distance in front.

Tip. When you stand in front of a mirror, your right hand appears to be on your reflection's "left". That's lateral inversion.

Worked qualitative. Why don't ambulances write "AMBULANCE" mirror-imaged on their bonnet? They DO — so that when a car ahead sees it in their rear-view mirror, it reads correctly forwards.

Angle of incidence = angle of reflection.
Plane mirror image: virtual, same size, upright, laterally inverted.
Image distance behind = object distance in front.

Refraction and refractive index

$n = \dfrac{\sin i}{\sin r}$ . Slow medium → bend toward normal.

Both angles measured from the normal.

Also: $n = c_{\text{air}} / c_{\text{medium}}$ . Light is slower in denser media; the bigger $n$ , the slower light goes.

Worked. Light enters glass at $i = 40\degree$ . Glass has $n = 1.5$ . Find $r$ .

$\sin r = \sin 40\degree / 1.5 \approx 0.643/1.5 = 0.428$ .
$r = \sin^{-1}(0.428) \approx 25.4\degree$ .

Common refractive indices.

Air: $\approx 1$ .
Water: $1.33$ .
Glass: $1.5$ .
Diamond: $2.4$ .

Direction. Going INTO a denser medium (higher $n$ ) → light slows, bends TOWARD normal. Going OUT to a less dense medium → speeds up, bends AWAY from normal.

$n = \sin i / \sin r$ .
$n = c_{\text{air}}/c_{\text{medium}}$ .
Higher $n$ = denser optically = slower.
Always measure angles from normal.

Total internal reflection

Above critical angle (dense → less dense), all light reflects back.

Critical angle. $\sin c = \dfrac{1}{n}.$

For glass ( $n = 1.5$ ): $c = \sin^{-1}(1/1.5) \approx 41.8\degree$ .

Conditions for TIR.

Light going from DENSER to less dense medium.
Angle of incidence GREATER than critical angle.

Applications.

Optical fibres: light enters at one end, hits the wall at $> c$ → TIR → bounces along the fibre. Used in telecommunications, endoscopes.
Prisms in periscopes / binoculars: 45-45-90 prisms reflect light internally at $45\degree$ (which is greater than glass's critical angle).

Worked. Find critical angle for water ( $n = 1.33$ ).

$c = \sin^{-1}(1/1.33) \approx 48.6\degree$ .

$\sin c = 1/n$ .
TIR: dense → less dense AND $\theta_{i} > c$ .
Optical fibres, prisms.
Glass critical angle: $\sim 42\degree$ .

Lenses

Converging: brings parallel rays together. Diverging: spreads them.

Converging (convex) lens. Thicker in the middle. Parallel rays meet at a single point — the focal point ( $F$ ). Distance from lens to focal point is the focal length ( $f$ ).

Diverging (concave) lens. Thinner in the middle. Parallel rays spread out as if coming from a focal point on the SAME side as the incoming rays (so it's "virtual").

Image formation by a converging lens. Depends on object distance vs $f$ :

Object beyond $2f$ : image real, inverted, smaller (camera).
Object at $2f$ : real, inverted, same size.
Between $f$ and $2f$ : real, inverted, magnified (projector).
At $f$ : image at infinity (parallel beam, used in spotlights).
Inside $f$ : virtual, upright, magnified (magnifying glass).

Diverging lenses always produce: virtual, upright, smaller image (used to correct short-sightedness).

Cambridge tip. Drawing a ray diagram: use the three principal rays.

Ray through optical centre — goes straight.
Ray parallel to axis — refracts through $F$ on the far side.
Ray through $F$ on the near side — emerges parallel to axis.

Converging: parallel rays meet at $F$ .
Diverging: parallel rays appear to spread from virtual $F$ .
Magnifier: object inside $f$ on a converging lens.
Camera: object beyond $2f$ .

How it’s examined

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (3.3-3.5); 0625/42 Oct/Nov 2024 — Q14 (refractive index, TIR); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Master this Topic

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

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Light, ready to print or save as PDF.

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.

1Apply Snell's law
Extended• Adapted from 0625/42 May/Jun 2024 Q13• refraction
Question
Light enters glass from air with angle of incidence $40\degree$ . The refractive index of glass is $1.5$ . Find the angle of refraction.
Step-by-step solution
1. Step 1
  $n = \dfrac{\sin i}{\sin r}$ .
  $\sin r = \dfrac{\sin 40\degree}{1.5}$
2. Step 2
  Compute.
  $\sin r = \dfrac{0.643}{1.5} = 0.428 \Rightarrow r \approx 25.4\degree$
Answer
$r \approx 25.4\degree$
2Critical angle
Extended• TIR
Question
A material has refractive index $1.5$ . Find the critical angle for the material-air boundary.
Step-by-step solution
1. Step 1
  $\sin c = \dfrac{1}{n}$ .
  $\sin c = \dfrac{1}{1.5} \approx 0.667$
2. Step 2
  Apply $\sin^{-1}$ .
  $c \approx 41.8\degree$
Answer
$c \approx 41.8\degree$
3Law of reflection
Core• reflection
Question
A ray hits a plane mirror at $30\degree$ from the surface. Find the angle of reflection (measured from the normal).
Step-by-step solution
1. Step 1
  Angles in reflection are measured from the NORMAL.
2. Step 2
  Angle of incidence = $90\degree - 30\degree = 60\degree$ .
3. Step 3
  Angle of reflection = angle of incidence.
  $= 60\degree$
Answer
$60\degree$ (from the normal)
Examiner tip
Always measure ray angles from the NORMAL (perpendicular to the surface), not from the surface itself.
4Image from a converging lens
Extended• lens
Question
An object is placed $30\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . Describe the image (real/virtual, magnification, orientation).
Step-by-step solution
1. Step 1
  Object beyond $2f$ ( $30 > 20$ ): image is real, inverted, diminished.
Answer
Real, inverted, diminished, between $f$ and $2f$ on the other side.
5Image properties in a plane mirror
Core• reflection, image
Question
An object stands $20\,\text{cm}$ in front of a plane mirror. Describe the image: its position, size, orientation, and whether it is real or virtual.
Step-by-step solution
1. Step 1
  A plane mirror produces an image the SAME distance behind the mirror as the object is in front.
  $d_{\text{image}} = 20\,\text{cm behind}$
2. Step 2
  Same size as the object; upright; laterally inverted (left-right swapped).
3. Step 3
  The rays only APPEAR to come from behind the mirror; no rays actually meet there → virtual image.
Answer
Virtual, upright, same size, $20\,\text{cm}$ behind the mirror, laterally inverted.
6Find refractive index from angles
Extended• Adapted from 0625/42 Oct/Nov 2023 Q14• refraction, n
Question
Light passes from air into a transparent block with angle of incidence $50\degree$ and angle of refraction $30\degree$ . Find the refractive index of the block.
Step-by-step solution
1. Step 1
  $n = \dfrac{\sin i}{\sin r}$ .
  $n = \dfrac{\sin 50\degree}{\sin 30\degree} = \dfrac{0.766}{0.500}$
2. Step 2
  Compute.
  $n \approx 1.53$
Answer
$n \approx 1.53$
Examiner tip
The examiner report flags candidates often invert the ratio (using $\sin r / \sin i$ ). For light entering a denser medium $i > r$ , so $n > 1$ — use this as a sanity check.
7Critical angle from refractive index
Extended• TIR, critical angle
Question
A material has refractive index $n = 1.40$ . Find the critical angle for light passing from this material into air.
Step-by-step solution
1. Step 1
  $\sin c = 1/n$ .
  $\sin c = \dfrac{1}{1.40} = 0.714$
2. Step 2
  Apply $\sin^{-1}$ .
  $c \approx 45.6\degree$
Answer
$c \approx 45.6\degree$
8Total internal reflection in an optical fibre
Extended• Adapted from 0625/42 May/Jun 2023 Q14• TIR, fibre
Question
An optical fibre's core has refractive index $1.50$ . (a) Find the critical angle at the core-air interface. (b) Light enters the fibre and strikes the inner wall at $50\degree$ from the normal. Does it pass through or reflect? Explain.
Step-by-step solution
1. Step 1
  Critical angle from $\sin c = 1/n$ .
  $\sin c = \dfrac{1}{1.50} \approx 0.667 \Rightarrow c \approx 41.8\degree$
2. Step 2
  The angle of incidence at the wall is $50\degree > c = 41.8\degree$ .
3. Step 3
  Therefore total internal reflection occurs — all the light is reflected back into the core.
Answer
(a) $c \approx 41.8\degree$ . (b) Since $50\degree > c$ , the light undergoes total internal reflection and stays in the fibre.
9Converging lens ray diagram: object beyond 2F
Extended• lens, ray diagram
Question
An object of height $4.0\,\text{cm}$ is placed at $30\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . By scale drawing or by using lens properties, find the image distance and image height (image is at $15\,\text{cm}$ on the far side).
Step-by-step solution
1. Step 1
  Draw two key rays: one parallel to the axis (refracts through $F$ on far side) and one through the lens centre (undeviated).
2. Step 2
  The rays cross at the image position, between $F$ and $2F$ on the far side.
  $v = 15\,\text{cm}$
3. Step 3
  Linear magnification.
  $m = \dfrac{v}{u} = \dfrac{15}{30} = 0.5$
4. Step 4
  Image height.
  $h_{i} = m\,h_{o} = 0.5 \times 4.0 = 2.0\,\text{cm}$
Answer
Real, inverted, diminished image, $v = 15\,\text{cm}$ on the far side, height $2.0\,\text{cm}$ .
Examiner tip
The examiner report flags candidates often draw the parallel ray bending at the lens centre instead of at the principal axis. Refract the ray AT the lens line.
10Object between F and 2F (projector setup)
Extended• lens, ray diagram
Question
A slide of height $2.0\,\text{cm}$ is placed at $15\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . Describe the image and explain why this arrangement is used in a projector.
Step-by-step solution
1. Step 1
  Object distance $u = 15\,\text{cm}$ is between $F$ ( $10\,\text{cm}$ ) and $2F$ ( $20\,\text{cm}$ ).
2. Step 2
  Ray construction puts the image beyond $2F$ on the far side: real, inverted, MAGNIFIED.
3. Step 3
  In a projector the small slide produces a large image on a screen; the slide is therefore placed just outside $F$ . Slides are loaded UPSIDE DOWN so the screen image appears upright.
Answer
Real, inverted, magnified image beyond $2F$ on the far side — used in projectors.
11Dispersion in a prism
Extended• dispersion, prism
Question
White light enters a glass prism and emerges as a spectrum. (a) Explain why the light splits up. (b) State which colour is refracted most and why.
Step-by-step solution
1. Step 1
  White light is a mixture of all visible colours, each with a different frequency / wavelength.
2. Step 2
  The refractive index of glass is slightly different for each frequency (greater for higher frequency / shorter wavelength).
3. Step 3
  Each colour is therefore refracted by a different amount on entering and leaving the prism, fanning out into the spectrum red-orange-yellow-green-blue-indigo-violet.
4. Step 4
  (b) Violet has the shortest wavelength / highest frequency → largest $n$ in glass → refracted MOST. Red is refracted least.
Answer
(a) Different colours have different refractive indices in glass, so they refract by different amounts. (b) Violet is refracted most; red least.
12A* — Right-angled prism reflectors
Challenge• Adapted from 0625/42 Oct/Nov 2024 Q14• TIR, synoptic
Question
A $45\degree$ - $45\degree$ - $90\degree$ glass prism ( $n = 1.50$ ) is used as a periscope reflector. Light enters one short face along the normal. Explain, with reference to the critical angle, why total internal reflection occurs at the hypotenuse, and state one advantage of using such a prism instead of a silvered mirror.
Step-by-step solution
1. Step 1
  Find the critical angle.
  $\sin c = \dfrac{1}{1.50} \approx 0.667 \Rightarrow c \approx 41.8\degree$
2. Step 2
  Light entering normally goes straight through, then meets the hypotenuse face at $45\degree$ from the normal.
3. Step 3
  $45\degree > c$ ( $41.8\degree$ ) → TOTAL internal reflection: all light reflects, none refracts out.
4. Step 4
  Advantage: a silvered mirror absorbs a fraction of the light and the silver layer can tarnish or get scratched. TIR in a prism is 100% efficient and the glass surface is durable.
Answer
At the hypotenuse the $45\degree$ angle exceeds $c \approx 41.8\degree$ so all light undergoes TIR. Advantage: 100% reflection, no tarnishing.
13A* — Magnifying glass (virtual image)
Challenge• lens, synoptic
Question
An object of height $5.0\,\text{mm}$ is placed $6.0\,\text{cm}$ from a converging lens of focal length $10\,\text{cm}$ . (a) State whether the image is real or virtual and how you know. (b) From a scale ray diagram the image distance is $15\,\text{cm}$ on the same side as the object. Find the magnification and the image height. (c) Why is this used as a magnifying glass?
Step-by-step solution
1. Step 1
  Object distance $u = 6.0\,\text{cm}$ is LESS than $f = 10\,\text{cm}$ . The refracted rays diverge after the lens.
2. Step 2
  Tracing the diverging rays back, they meet on the SAME side as the object → virtual, upright, magnified image.
3. Step 3
  Magnification.
  $m = \dfrac{v}{u} = \dfrac{15}{6.0} = 2.5$
4. Step 4
  Image height.
  $h_{i} = m\,h_{o} = 2.5 \times 5.0 = 12.5\,\text{mm}$
5. Step 5
  (c) Placing the object inside $F$ gives a magnified upright virtual image — the eye sees an enlarged version of the object.
Answer
(a) Virtual (object inside $F$ ). (b) $m = 2.5$ ; image height $12.5\,\text{mm}$ . (c) Object inside $F$ → enlarged virtual image.
Examiner tip
The examiner report flags candidates often shade real images as if they were on a screen when the object is inside $F$ . Use DASHED rays back-projected to mark a virtual image.

Key Formulae — Light

The formulae you need to memorise for light on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Snell's law
$n = \dfrac{\sin i}{\sin r}$
$n$
refractive index of the second medium relative to the first
$i$
angle of incidence (from normal)
$r$
angle of refraction (from normal)
When to use
Light crossing a boundary between two media.
Critical angle
$\sin c = \dfrac{1}{n}$
When to use
Boundary from a denser medium back to air. Above $c$ , total internal reflection.
Law of reflection
$\theta_{i} = \theta_{r}$
When to use
Plane mirror; angles measured from the normal.

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 0625 sitting.

Normal
Examiner keyword
A line perpendicular to a surface at the point where the ray meets it. ALL ray angles are measured from this line.
Refraction
Examiner keyword
Change in the direction of light when it crosses a boundary, caused by a change in speed.
Refractive index ( $n$ )
Examiner keyword
$n = \dfrac{c_{\text{vacuum}}}{c_{\text{medium}}}$ — how much slower light is in the medium.
Total internal reflection
Examiner keyword
When light strikes a denser-medium boundary at an angle larger than the critical angle, it reflects back entirely.
Focal length
Examiner keyword
Distance from the lens centre to the focal point ( $F$ ); a measure of how strongly the lens converges/diverges light.

Common Mistakes and Misconceptions — Light

The traps other students keep falling into on light questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Measuring angles from the surface, not the normal
0625/42 — every series
Why it happens
Ruler reads from the surface.
How to avoid it
ALWAYS draw the normal first; measure from it.
✕Mis-orienting which is $i$ and which is $r$
Why it happens
Light direction changes at the boundary.
How to avoid it
$i$ is in the FIRST medium, $r$ is in the SECOND. Always check direction of travel.
✕Saying TIR happens when going from less dense to denser medium
Why it happens
Forgetting the prerequisite.
How to avoid it
TIR requires going FROM the denser medium INTO the less dense one (e.g. glass to air).
✕Drawing real image rays as dashed (and vice versa)
Why it happens
Mixing the conventions.
How to avoid it
Real image: SOLID converging rays meet. Virtual image: DASHED back-projected rays.

Practice questions

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

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

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 Study Notes

What you’ll learn

How it’s examined

1Apply Snell's law

2Critical angle

3Law of reflection

4Image from a converging lens

5Image properties in a plane mirror

6Find refractive index from angles

7Critical angle from refractive index

8Total internal reflection in an optical fibre

9Converging lens ray diagram: object beyond 2F

10Object between F and 2F (projector setup)

11Dispersion in a prism

12A* — Right-angled prism reflectors

13A* — Magnifying glass (virtual image)

Snell's law

Critical angle

Law of reflection

Normal

Refraction

Refractive index (nnn)

Total internal reflection

Focal length

✕Measuring angles from the surface, not the normal

✕Mis-orienting which is iii and which is rrr

✕Saying TIR happens when going from less dense to denser medium

✕Drawing real image rays as dashed (and vice versa)

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Light — frequently asked questions

Light

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Apply Snell's law

2Critical angle

3Law of reflection

4Image from a converging lens

5Image properties in a plane mirror

6Find refractive index from angles

7Critical angle from refractive index

8Total internal reflection in an optical fibre

9Converging lens ray diagram: object beyond 2F

10Object between F and 2F (projector setup)

11Dispersion in a prism

12A* — Right-angled prism reflectors

13A* — Magnifying glass (virtual image)

Snell's law

Critical angle

Law of reflection

Normal

Refraction

Refractive index (nnn)

Total internal reflection

Focal length

✕Measuring angles from the surface, not the normal

✕Mis-orienting which is iii and which is rrr

✕Saying TIR happens when going from less dense to denser medium

✕Drawing real image rays as dashed (and vice versa)

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Light — frequently asked questions

Refractive index ( $n$ )

✕Mis-orienting which is $i$ and which is $r$

Refractive index ( $n$ )

✕Mis-orienting which is $i$ and which is $r$