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.

Cambridge IGCSE Physics (0625)

Light

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Short Notes - Light

Detailed Study Notes

Detailed notes on Waves for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

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.

At a glance

Reflection in a plane mirror: image is virtual, upright, same size, laterally inverted, equidistant behind the mirror.
Snell's law / refractive index: $n = \dfrac{\sin i}{\sin r}$ .
$n$ also equals $c_{\text{air}}/c_{\text{medium}}$ — light slows in denser media.
Critical angle $c$ : $\sin c = 1/n$ .
Total internal reflection (TIR) when $\theta_{i} > c$ AND going from dense to less-dense medium.
Converging (convex) lens: parallel rays meet at focal point. Used in cameras, eyes.
Diverging (concave) lens: parallel rays appear to spread from a virtual focal point.
Correcting sight: short sight → diverging lens; long sight → converging lens.

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.
3.2.3 — Describe the use of converging and diverging lenses to correct long-sightedness and short-sightedness.

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.

The image is upright and the same size, lying as far behind the mirror as the object is in front (d = d).

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.

Entering the denser glass the light slows and bends toward the normal — the angle of refraction is smaller than the angle of incidence.

$n = \sin i / \sin r$ .
$n = c_{\text{air}}/c_{\text{medium}}$ .
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.

Below the critical angle the ray refracts out, bending away from the normal; at the critical angle it grazes the surface; above it, the light is totally internally reflected.

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.

A convex lens converges parallel rays to a real focal point; a concave lens makes them diverge as if from a virtual focal point on the near side.

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.

Object beyond 2F: the two principal rays cross to form a real, inverted, smaller image (as in a camera).

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 $2f$ .

Correcting short-sightedness and long-sightedness

Short sight → diverging lens. Long sight → converging lens.

A healthy eye focuses light from objects, both near and far, exactly onto the retina. Two common defects move that focus point off the retina, and each is corrected with a different spectacle lens. The 2026 syllabus (§3.2.3) requires both.

Short-sightedness (myopia).

The person cannot focus clearly on distant objects.
Light from a distant object is brought to a focus in front of the retina — the eye converges the rays too strongly (or the eyeball is too long). The retinal image is blurred.
Correction: a diverging (concave) lens. It spreads the rays apart slightly before they enter the eye, so the eye then brings them to a focus a little further back — exactly on the retina.

Long-sightedness (hypermetropia).

The person cannot focus clearly on near objects.
Light from a near object is brought to a focus behind the retina — the eye cannot converge the rays strongly enough (or the eyeball is too short). The retinal image is blurred.
Correction: a converging (convex) lens. It bends the rays towards each other before they enter the eye, so the eye has less converging to do — the rays then focus exactly on the retina.

Memory hook. Short sight = image short of (in front of) the retina → diverging lens. Long sight = image beyond (behind) the retina → converging lens.

Short sight: image in front of retina → diverging (concave) lens.
Long sight: image behind retina → converging (convex) lens.
Both lenses move the focus point back onto the retina.

Quick recap

Plane mirror: image virtual, equidistant behind.
Refractive index: $n = \sin i / \sin r = c_{\text{air}}/c_{\text{medium}}$ .
Critical angle: $\sin c = 1/n$ .
TIR: dense→less dense, $\theta > c$ .
Converging lens: many image possibilities; depends on $u$ vs $f$ .
Diverging lens: always virtual upright smaller.
Short sight → diverging lens; long sight → converging lens.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Refractive index — ratio of speed of light in vacuum (or air) to speed in the medium; equals $\sin i / \sin r$ .
Critical angle — angle of incidence above which total internal reflection occurs.
Total internal reflection — all light reflected back into the denser medium when $\theta > c$ .
Focal length — distance from a lens to its focal point.

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.

Take this whole topic with you

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.

Question type:

Question patterns to master — Light

Almost every light exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single find or calculate instruction using one optics formula — Snell's law $n = \sin i/\sin r$ , the critical angle $\sin c = 1/n$ , or $\theta_i = \theta_r$ .

How to approach it

Draw the normal, measure every angle from it, write the formula and substitute; for a denser medium check $i > r$ and $n > 1$ as a sanity test.

Common trap

Examiner reports flag candidates measuring angles from the surface instead of the normal, or inverting Snell's law to $\sin r/\sin i$ .

Identify & classify

Recognise it by

Describe the image instructions — state whether it is real or virtual, upright or inverted, magnified or diminished.

How to approach it

Compare the object distance to $f$ and $2f$ for a lens, or use the equal-distance rule for a plane mirror, then read off the standard set of image properties.

Common trap

Examiner reports flag candidates calling a plane-mirror image 'real' — no rays actually meet behind the mirror, so it is always virtual.

Multi-step problem

Recognise it by

Two stages chained — calculate a critical angle, then compare it to an angle of incidence to decide whether light reflects or refracts.

How to approach it

Compute the critical angle first, then compare: if the angle of incidence exceeds $c$ , total internal reflection occurs.

Common trap

Examiner reports flag candidates forgetting that total internal reflection only happens going from the denser medium into the less dense one.

Graph or diagram

Recognise it by

Draw the ray diagram or by scale drawing — construct the image of an object through a converging lens.

How to approach it

Draw the two standard rays (one parallel to the axis refracting through $F$ , one straight through the lens centre), refracting them AT the lens line, and read the image where they cross.

Common trap

Examiner reports flag candidates bending the parallel ray at the lens centre instead of the lens line, and shading a virtual image with solid rays — virtual images use DASHED back-projected rays.

Show that / prove

Recognise it by

Explain why — why white light disperses in a prism, or why a right-angled prism gives total internal reflection.

How to approach it

Name the cause (refractive index varying with frequency, or an angle of incidence exceeding the critical angle) and link it step by step to the observed effect.

Common trap

Examiner reports flag candidates explaining dispersion by 'different speeds' alone without linking it to a different refractive index and hence a different angle of refraction for each colour.

1Apply Snell's law
ExtendedDirect calculation• 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
ExtendedDirect calculation• 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
CoreDirect calculation• 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
ExtendedIdentify & classify• 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
CoreIdentify & classify• 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
ExtendedDirect calculation• 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
ExtendedDirect calculation• 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
ExtendedMulti-step problem• 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.
9Beyond 0625 — Converging lens ray diagram with magnification (object beyond 2F)
ExtendedGraph or diagram• 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
Enrichment beyond Cambridge IGCSE 0625 — the numeric linear magnification $m = v/u$ used here (steps 3-4) is NOT in the 0625 syllabus. 0625 §3.2.3 describes images only qualitatively, using the terms enlarged/diminished, upright/inverted, real/virtual — there is no magnification formula. The magnification calculation belongs to AS/A-level (and other-board IGCSE) physics. The ray-diagram construction and the qualitative image description (real, inverted, diminished) ARE on-syllabus; the $m = v/u$ calculation will not be examined on 0625.
10Object between F and 2F (projector setup)
ExtendedGraph or diagram• 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
ExtendedShow that / prove• 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
ChallengeShow that / prove• 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.
13Correcting long-sightedness with a converging lens
Extended• lens, long-sightedness, vision
Question
A long-sighted person cannot focus clearly on nearby objects. (a) State, with a reason, the cause of long-sightedness in terms of where the image forms. (b) Name the type of lens used to correct it and explain how the lens works.
Step-by-step solution
1. Step 1
  (a) In a long-sighted eye, light from a near object is brought to a focus BEHIND the retina (the eye lens cannot converge the rays strongly enough, or the eyeball is too short). The image on the retina is therefore blurred.
2. Step 2
  (b) A converging (convex) lens is used. It is a converging lens because near-sight correction needs the rays to be bent inwards before they reach the eye.
3. Step 3
  The converging spectacle lens bends the rays from a near object towards each other before they enter the eye. The eye then has less converging to do, so the rays come to a focus exactly on the retina, forming a sharp image.
Answer
(a) Long-sightedness: the image of a near object forms behind the retina, so it is blurred. (b) A converging (convex) lens corrects it — it converges the rays before they enter the eye so the final image falls on the retina.
Examiner tip
Pair this with the short-sight case: short-sightedness (image in front of the retina) is corrected with a diverging lens; long-sightedness (image behind the retina) with a converging lens. The 2026 syllabus §3.2.3 requires BOTH.
14Correcting short-sightedness with a diverging lens
Extended• lens, short-sightedness, vision
Question
A short-sighted person cannot focus clearly on distant objects. (a) State where the image of a distant object forms in the uncorrected eye. (b) Name the lens used to correct it and explain how it works.
Step-by-step solution
1. Step 1
  (a) In a short-sighted eye, light from a distant object is brought to a focus IN FRONT OF the retina (the eye converges the rays too strongly, or the eyeball is too long). The image on the retina is blurred.
2. Step 2
  (b) A diverging (concave) lens is used.
3. Step 3
  The diverging spectacle lens spreads the rays apart slightly before they enter the eye. The eye then converges them a little later, so the image forms exactly on the retina.
Answer
(a) Short-sightedness: the image of a distant object forms in front of the retina. (b) A diverging (concave) lens corrects it — it diverges the rays before they enter the eye so the image falls on the retina.
15Beyond 0625 — Magnifying glass with magnification calculation (virtual image)
ChallengeGraph or diagram• 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
Enrichment beyond Cambridge IGCSE 0625 — the numeric linear magnification $m = v/u$ used here (steps 3-4) is NOT in the 0625 syllabus. 0625 §3.2.3 describes images only qualitatively (enlarged/diminished, upright/inverted, real/virtual) — there is no magnification formula. The magnification calculation belongs to AS/A-level (and other-board IGCSE) physics. Parts (a) and (c) — recognising a virtual, upright, enlarged image when the object is inside $F$ , and the magnifying-glass application — ARE on-syllabus; the $m = v/u$ calculation in part (b) will not be examined on 0625.

Model Answers — Light

High-scoring sample answers for light on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State the law of reflection for a plane mirror.
Model answer
The angle of incidence equals the angle of reflection (both measured between the ray and the normal at the point of incidence).
Why this scores
One mark for $i = r$ with reference to the normal. Stating the angles are measured from the normal (not the mirror surface) is the detail examiners check.
Question 2
Paper 2/4 style2 marks
A ray of light passes from air into glass of refractive index $1.5$ with an angle of incidence of $30\,\text{°}$ . Calculate the angle of refraction.
Model answer
Using $n = \dfrac{\sin i}{\sin r}$ , rearranged: $\sin r = \dfrac{\sin i}{n} = \dfrac{\sin 30\,\text{°}}{1.5} = \dfrac{0.500}{1.5} = 0.333$ .

$r = \sin^{-1}(0.333) \approx \mathbf{19.5\,\text{°}}.$
Why this scores
One mark for correct use of Snell's law, one for $r \approx 19.5\,\text{°}$ . Entering a denser medium bends the ray towards the normal, so $r < i$ — a useful check.
Question 3
Paper 4 structured style3 marks
A glass block has a refractive index of $1.5$ . (a) Calculate the critical angle for the glass–air boundary. (b) State the condition for total internal reflection to occur.
Model answer
(a) $\sin c = \dfrac{1}{n} = \dfrac{1}{1.5} = 0.667$ , so $c = \sin^{-1}(0.667) \approx \mathbf{41.8\,\text{°}}$ .

(b) Total internal reflection occurs when light travelling inside the denser medium strikes the boundary at an angle of incidence greater than the critical angle.
Why this scores
Three marks: use of $\sin c = 1/n$ (1); $c \approx 41.8\,\text{°}$ (1); the condition — angle of incidence greater than $c$ , going from denser to less dense medium (1).
Question 4
Paper 4 structured style4 marks
An object is placed $20\,\text{cm}$ in front of a plane mirror. Describe the image formed, including its position, size, orientation and whether it is real or virtual.
Model answer
The image is:

the same distance behind the mirror as the object is in front — i.e. $20\,\text{cm}$ behind the mirror;

the same size as the object;

upright (and laterally inverted — left and right appear swapped);

virtual, because the reflected rays only appear to come from behind the mirror — no light actually passes through the image position.
Why this scores
Four marks: image $20\,\text{cm}$ behind / same distance (1); same size (1); upright/laterally inverted (1); virtual with a reason (1). Calling a plane-mirror image 'real' is the standard error — no rays actually meet behind the mirror.
Question 5
Paper 4 application style5 marks
An optical fibre has a core of refractive index $1.5$ . (a) Explain, using the critical angle, how light is kept inside the fibre. (b) Show by calculation that light striking the inside wall at $50\,\text{°}$ to the normal stays in the fibre. (c) State one use of optical fibres.
Model answer
(a) Light travels along the fibre and repeatedly hits the inside wall. If it strikes at an angle of incidence greater than the critical angle, it undergoes total internal reflection — all the light reflects back into the core with none escaping, so it is guided along the fibre.

(b) Critical angle: $\sin c = \dfrac{1}{1.5} = 0.667$ , so $c \approx 41.8\,\text{°}$ . The ray strikes at $50\,\text{°}$ , and since $50\,\text{°} > 41.8\,\text{°}$ , the angle exceeds the critical angle, so total internal reflection occurs and the light stays in the fibre.

(c) Telecommunications / transmitting data (telephone and internet signals), or medical endoscopes for looking inside the body.
Why this scores
Five marks: TIR keeps light in when the angle exceeds the critical angle (1); calculate $c \approx 41.8\,\text{°}$ (1); compare $50\,\text{°} > c$ (1); conclude TIR (1); a valid use (1).
Question 6
Paper 4 extended-response style6 marks
(a) Explain the cause of short-sightedness and how a lens corrects it. (b) Explain the cause of long-sightedness and how a lens corrects it. Refer in each case to where the image forms relative to the retina and the type of lens used.
Model answer
(a) Short-sightedness: the person cannot focus on distant objects because the eye converges the light too strongly (or the eyeball is too long), so the image of a distant object forms in front of the retina and is blurred. It is corrected with a diverging (concave) lens, which spreads the rays apart slightly before they enter the eye, so the image is pushed back to form exactly on the retina.

(b) Long-sightedness: the person cannot focus on near objects because the eye does not converge the light strongly enough (or the eyeball is too short), so the image of a near object forms behind the retina and is blurred. It is corrected with a converging (convex) lens, which converges the rays before they enter the eye, so the image is brought forward to form exactly on the retina.
Why this scores
Six marks: short sight — image in front of retina (1), diverging lens (1), diverges rays so image falls on retina (1); long sight — image behind retina (1), converging lens (1), converges rays so image falls on retina (1). Matching the correct lens to each defect is essential — the 2026 syllabus §3.2.3 requires both.

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.

Light — frequently asked questions

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

Cambridge IGCSE Physics (0625)

Light

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Short Notes - Light

Detailed Study Notes

Detailed notes on Waves for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

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.

At a glance

Reflection in a plane mirror: image is virtual, upright, same size, laterally inverted, equidistant behind the mirror.
Snell's law / refractive index: $n = \dfrac{\sin i}{\sin r}$ .
$n$ also equals $c_{\text{air}}/c_{\text{medium}}$ — light slows in denser media.
Critical angle $c$ : $\sin c = 1/n$ .
Total internal reflection (TIR) when $\theta_{i} > c$ AND going from dense to less-dense medium.
Converging (convex) lens: parallel rays meet at focal point. Used in cameras, eyes.
Diverging (concave) lens: parallel rays appear to spread from a virtual focal point.
Correcting sight: short sight → diverging lens; long sight → converging lens.

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.
3.2.3 — Describe the use of converging and diverging lenses to correct long-sightedness and short-sightedness.

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.

The image is upright and the same size, lying as far behind the mirror as the object is in front (d = d).

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.

Entering the denser glass the light slows and bends toward the normal — the angle of refraction is smaller than the angle of incidence.

$n = \sin i / \sin r$ .
$n = c_{\text{air}}/c_{\text{medium}}$ .
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.

Below the critical angle the ray refracts out, bending away from the normal; at the critical angle it grazes the surface; above it, the light is totally internally reflected.

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.

A convex lens converges parallel rays to a real focal point; a concave lens makes them diverge as if from a virtual focal point on the near side.

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.

Object beyond 2F: the two principal rays cross to form a real, inverted, smaller image (as in a camera).

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 $2f$ .

Correcting short-sightedness and long-sightedness

Short sight → diverging lens. Long sight → converging lens.

Short-sightedness (myopia).

The person cannot focus clearly on distant objects.
Light from a distant object is brought to a focus in front of the retina — the eye converges the rays too strongly (or the eyeball is too long). The retinal image is blurred.
Correction: a diverging (concave) lens. It spreads the rays apart slightly before they enter the eye, so the eye then brings them to a focus a little further back — exactly on the retina.

Long-sightedness (hypermetropia).

The person cannot focus clearly on near objects.
Light from a near object is brought to a focus behind the retina — the eye cannot converge the rays strongly enough (or the eyeball is too short). The retinal image is blurred.
Correction: a converging (convex) lens. It bends the rays towards each other before they enter the eye, so the eye has less converging to do — the rays then focus exactly on the retina.

Memory hook. Short sight = image short of (in front of) the retina → diverging lens. Long sight = image beyond (behind) the retina → converging lens.

Short sight: image in front of retina → diverging (concave) lens.
Long sight: image behind retina → converging (convex) lens.
Both lenses move the focus point back onto the retina.

Quick recap

Plane mirror: image virtual, equidistant behind.
Refractive index: $n = \sin i / \sin r = c_{\text{air}}/c_{\text{medium}}$ .
Critical angle: $\sin c = 1/n$ .
TIR: dense→less dense, $\theta > c$ .
Converging lens: many image possibilities; depends on $u$ vs $f$ .
Diverging lens: always virtual upright smaller.
Short sight → diverging lens; long sight → converging lens.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Refractive index — ratio of speed of light in vacuum (or air) to speed in the medium; equals $\sin i / \sin r$ .
Critical angle — angle of incidence above which total internal reflection occurs.
Total internal reflection — all light reflected back into the denser medium when $\theta > c$ .
Focal length — distance from a lens to its focal point.

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.

Take this whole topic with you

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.

Question type:

Question patterns to master — Light

Almost every light exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single find or calculate instruction using one optics formula — Snell's law $n = \sin i/\sin r$ , the critical angle $\sin c = 1/n$ , or $\theta_i = \theta_r$ .

How to approach it

Draw the normal, measure every angle from it, write the formula and substitute; for a denser medium check $i > r$ and $n > 1$ as a sanity test.

Common trap

Examiner reports flag candidates measuring angles from the surface instead of the normal, or inverting Snell's law to $\sin r/\sin i$ .

Identify & classify

Recognise it by

Describe the image instructions — state whether it is real or virtual, upright or inverted, magnified or diminished.

How to approach it

Compare the object distance to $f$ and $2f$ for a lens, or use the equal-distance rule for a plane mirror, then read off the standard set of image properties.

Common trap

Examiner reports flag candidates calling a plane-mirror image 'real' — no rays actually meet behind the mirror, so it is always virtual.

Multi-step problem

Recognise it by

Two stages chained — calculate a critical angle, then compare it to an angle of incidence to decide whether light reflects or refracts.

How to approach it

Compute the critical angle first, then compare: if the angle of incidence exceeds $c$ , total internal reflection occurs.

Common trap

Examiner reports flag candidates forgetting that total internal reflection only happens going from the denser medium into the less dense one.

Graph or diagram

Recognise it by

Draw the ray diagram or by scale drawing — construct the image of an object through a converging lens.

How to approach it

Draw the two standard rays (one parallel to the axis refracting through $F$ , one straight through the lens centre), refracting them AT the lens line, and read the image where they cross.

Common trap

Examiner reports flag candidates bending the parallel ray at the lens centre instead of the lens line, and shading a virtual image with solid rays — virtual images use DASHED back-projected rays.

Show that / prove

Recognise it by

Explain why — why white light disperses in a prism, or why a right-angled prism gives total internal reflection.

How to approach it

Name the cause (refractive index varying with frequency, or an angle of incidence exceeding the critical angle) and link it step by step to the observed effect.

Common trap

Examiner reports flag candidates explaining dispersion by 'different speeds' alone without linking it to a different refractive index and hence a different angle of refraction for each colour.

1Apply Snell's law
ExtendedDirect calculation• 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
ExtendedDirect calculation• 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
CoreDirect calculation• 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
ExtendedIdentify & classify• 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
CoreIdentify & classify• 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
ExtendedDirect calculation• 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
ExtendedDirect calculation• 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
ExtendedMulti-step problem• 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.
9Beyond 0625 — Converging lens ray diagram with magnification (object beyond 2F)
ExtendedGraph or diagram• 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
Enrichment beyond Cambridge IGCSE 0625 — the numeric linear magnification $m = v/u$ used here (steps 3-4) is NOT in the 0625 syllabus. 0625 §3.2.3 describes images only qualitatively, using the terms enlarged/diminished, upright/inverted, real/virtual — there is no magnification formula. The magnification calculation belongs to AS/A-level (and other-board IGCSE) physics. The ray-diagram construction and the qualitative image description (real, inverted, diminished) ARE on-syllabus; the $m = v/u$ calculation will not be examined on 0625.
10Object between F and 2F (projector setup)
ExtendedGraph or diagram• 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
ExtendedShow that / prove• 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
ChallengeShow that / prove• 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.
13Correcting long-sightedness with a converging lens
Extended• lens, long-sightedness, vision
Question
A long-sighted person cannot focus clearly on nearby objects. (a) State, with a reason, the cause of long-sightedness in terms of where the image forms. (b) Name the type of lens used to correct it and explain how the lens works.
Step-by-step solution
1. Step 1
  (a) In a long-sighted eye, light from a near object is brought to a focus BEHIND the retina (the eye lens cannot converge the rays strongly enough, or the eyeball is too short). The image on the retina is therefore blurred.
2. Step 2
  (b) A converging (convex) lens is used. It is a converging lens because near-sight correction needs the rays to be bent inwards before they reach the eye.
3. Step 3
  The converging spectacle lens bends the rays from a near object towards each other before they enter the eye. The eye then has less converging to do, so the rays come to a focus exactly on the retina, forming a sharp image.
Answer
(a) Long-sightedness: the image of a near object forms behind the retina, so it is blurred. (b) A converging (convex) lens corrects it — it converges the rays before they enter the eye so the final image falls on the retina.
Examiner tip
Pair this with the short-sight case: short-sightedness (image in front of the retina) is corrected with a diverging lens; long-sightedness (image behind the retina) with a converging lens. The 2026 syllabus §3.2.3 requires BOTH.
14Correcting short-sightedness with a diverging lens
Extended• lens, short-sightedness, vision
Question
A short-sighted person cannot focus clearly on distant objects. (a) State where the image of a distant object forms in the uncorrected eye. (b) Name the lens used to correct it and explain how it works.
Step-by-step solution
1. Step 1
  (a) In a short-sighted eye, light from a distant object is brought to a focus IN FRONT OF the retina (the eye converges the rays too strongly, or the eyeball is too long). The image on the retina is blurred.
2. Step 2
  (b) A diverging (concave) lens is used.
3. Step 3
  The diverging spectacle lens spreads the rays apart slightly before they enter the eye. The eye then converges them a little later, so the image forms exactly on the retina.
Answer
(a) Short-sightedness: the image of a distant object forms in front of the retina. (b) A diverging (concave) lens corrects it — it diverges the rays before they enter the eye so the image falls on the retina.
15Beyond 0625 — Magnifying glass with magnification calculation (virtual image)
ChallengeGraph or diagram• 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
Enrichment beyond Cambridge IGCSE 0625 — the numeric linear magnification $m = v/u$ used here (steps 3-4) is NOT in the 0625 syllabus. 0625 §3.2.3 describes images only qualitatively (enlarged/diminished, upright/inverted, real/virtual) — there is no magnification formula. The magnification calculation belongs to AS/A-level (and other-board IGCSE) physics. Parts (a) and (c) — recognising a virtual, upright, enlarged image when the object is inside $F$ , and the magnifying-glass application — ARE on-syllabus; the $m = v/u$ calculation in part (b) will not be examined on 0625.

Model Answers — Light

High-scoring sample answers for light on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State the law of reflection for a plane mirror.
Model answer
The angle of incidence equals the angle of reflection (both measured between the ray and the normal at the point of incidence).
Why this scores
One mark for $i = r$ with reference to the normal. Stating the angles are measured from the normal (not the mirror surface) is the detail examiners check.
Question 2
Paper 2/4 style2 marks
A ray of light passes from air into glass of refractive index $1.5$ with an angle of incidence of $30\,\text{°}$ . Calculate the angle of refraction.
Model answer
Using $n = \dfrac{\sin i}{\sin r}$ , rearranged: $\sin r = \dfrac{\sin i}{n} = \dfrac{\sin 30\,\text{°}}{1.5} = \dfrac{0.500}{1.5} = 0.333$ .

$r = \sin^{-1}(0.333) \approx \mathbf{19.5\,\text{°}}.$
Why this scores
One mark for correct use of Snell's law, one for $r \approx 19.5\,\text{°}$ . Entering a denser medium bends the ray towards the normal, so $r < i$ — a useful check.
Question 3
Paper 4 structured style3 marks
A glass block has a refractive index of $1.5$ . (a) Calculate the critical angle for the glass–air boundary. (b) State the condition for total internal reflection to occur.
Model answer
(a) $\sin c = \dfrac{1}{n} = \dfrac{1}{1.5} = 0.667$ , so $c = \sin^{-1}(0.667) \approx \mathbf{41.8\,\text{°}}$ .

(b) Total internal reflection occurs when light travelling inside the denser medium strikes the boundary at an angle of incidence greater than the critical angle.
Why this scores
Three marks: use of $\sin c = 1/n$ (1); $c \approx 41.8\,\text{°}$ (1); the condition — angle of incidence greater than $c$ , going from denser to less dense medium (1).
Question 4
Paper 4 structured style4 marks
An object is placed $20\,\text{cm}$ in front of a plane mirror. Describe the image formed, including its position, size, orientation and whether it is real or virtual.
Model answer
The image is:

the same distance behind the mirror as the object is in front — i.e. $20\,\text{cm}$ behind the mirror;

the same size as the object;

upright (and laterally inverted — left and right appear swapped);

virtual, because the reflected rays only appear to come from behind the mirror — no light actually passes through the image position.
Why this scores
Four marks: image $20\,\text{cm}$ behind / same distance (1); same size (1); upright/laterally inverted (1); virtual with a reason (1). Calling a plane-mirror image 'real' is the standard error — no rays actually meet behind the mirror.
Question 5
Paper 4 application style5 marks
An optical fibre has a core of refractive index $1.5$ . (a) Explain, using the critical angle, how light is kept inside the fibre. (b) Show by calculation that light striking the inside wall at $50\,\text{°}$ to the normal stays in the fibre. (c) State one use of optical fibres.
Model answer
(a) Light travels along the fibre and repeatedly hits the inside wall. If it strikes at an angle of incidence greater than the critical angle, it undergoes total internal reflection — all the light reflects back into the core with none escaping, so it is guided along the fibre.

(b) Critical angle: $\sin c = \dfrac{1}{1.5} = 0.667$ , so $c \approx 41.8\,\text{°}$ . The ray strikes at $50\,\text{°}$ , and since $50\,\text{°} > 41.8\,\text{°}$ , the angle exceeds the critical angle, so total internal reflection occurs and the light stays in the fibre.

(c) Telecommunications / transmitting data (telephone and internet signals), or medical endoscopes for looking inside the body.
Why this scores
Five marks: TIR keeps light in when the angle exceeds the critical angle (1); calculate $c \approx 41.8\,\text{°}$ (1); compare $50\,\text{°} > c$ (1); conclude TIR (1); a valid use (1).
Question 6
Paper 4 extended-response style6 marks
(a) Explain the cause of short-sightedness and how a lens corrects it. (b) Explain the cause of long-sightedness and how a lens corrects it. Refer in each case to where the image forms relative to the retina and the type of lens used.
Model answer
(a) Short-sightedness: the person cannot focus on distant objects because the eye converges the light too strongly (or the eyeball is too long), so the image of a distant object forms in front of the retina and is blurred. It is corrected with a diverging (concave) lens, which spreads the rays apart slightly before they enter the eye, so the image is pushed back to form exactly on the retina.

(b) Long-sightedness: the person cannot focus on near objects because the eye does not converge the light strongly enough (or the eyeball is too short), so the image of a near object forms behind the retina and is blurred. It is corrected with a converging (convex) lens, which converges the rays before they enter the eye, so the image is brought forward to form exactly on the retina.
Why this scores
Six marks: short sight — image in front of retina (1), diverging lens (1), diverges rays so image falls on retina (1); long sight — image behind retina (1), converging lens (1), converges rays so image falls on retina (1). Matching the correct lens to each defect is essential — the 2026 syllabus §3.2.3 requires both.

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.

Light — frequently asked questions

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

Light

Short Notes - Light

Detailed Study Notes

At a glance

What you’ll learn

Reflection at a plane mirror

Refraction and refractive index

Total internal reflection

Lenses

Correcting short-sightedness and long-sightedness

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Light

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

9Beyond 0625 — Converging lens ray diagram with magnification (object beyond 2F)

10Object between F and 2F (projector setup)

11Dispersion in a prism

12A* — Right-angled prism reflectors

13Correcting long-sightedness with a converging lens

14Correcting short-sightedness with a diverging lens

15Beyond 0625 — Magnifying glass with magnification calculation (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)

Light — frequently asked questions

Light

Short Notes - Light

Detailed Study Notes

At a glance

What you’ll learn

Reflection at a plane mirror

Refraction and refractive index

Total internal reflection

Lenses

Correcting short-sightedness and long-sightedness

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Light

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

9Beyond 0625 — Converging lens ray diagram with magnification (object beyond 2F)

10Object between F and 2F (projector setup)

11Dispersion in a prism

12A* — Right-angled prism reflectors

13Correcting long-sightedness with a converging lens

14Correcting short-sightedness with a diverging lens

15Beyond 0625 — Magnifying glass with magnification calculation (virtual image)

Snell's law

Refractive index ( $n$ )

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