302.Light

Core + ExtendedIGCSE 0625

1. Reflection of Light

Light travels in straight lines, drawn as rays — lines with arrows showing the direction the light travels.

Key terms

• Incident ray: the ray travelling towards the reflecting surface.

• Point of incidence: the point where the incident ray meets the surface.

• Normal: the line drawn at $90^{\circ}$ to the surface at the point of incidence. It is a construction line, so it is always drawn dashed.

• Angle of incidence ( $i$ ): the angle between the incident ray and the normal.

• Reflected ray: the ray travelling away from the surface.

• Angle of reflection ( $r$ ): the angle between the reflected ray and the normal.

The Law of Reflection

When light is reflected from a surface:

• The angle of incidence is equal to the angle of reflection, $i = r$ .

• The incident ray, the reflected ray and the normal all lie in the same plane.

Both angles are measured from the normal, never from the surface itself.

Experiment to demonstrate the law of reflection

Stand a plane mirror upright on a sheet of paper and draw a line along the back of the glass to mark the reflecting surface. Choose a point of incidence on that line and draw the normal at $90^{\circ}$ to it.

Direct a narrow ray from a ray box at the point of incidence. Mark the incident and reflected rays with two crosses each, or stand two optical pins upright in each ray and sight along them.

Take the mirror away, join each pair of crosses back to the point of incidence, then measure $i$ and $r$ with a protractor. Repeat for at least five different angles.

Every pair of readings gives $i = r$ to within the precision of the protractor, which confirms the law.

Plane Mirrors - Image Characteristics

Virtual image: an image formed where the reflected rays only appear to come from. The rays never actually meet there, so a virtual image cannot be caught on a screen.

The image in a plane mirror has five key properties:

• Virtual — it cannot be formed on a screen.

• Upright — the same way up as the object.

• Same size — magnification = 1.

• Laterally inverted — left and right are swapped.

• Equal distance — as far behind the mirror as the object is in front.

Locating the image by construction

Draw two rays from one point on the object to the mirror. Draw the normal at each point of incidence, then reflect each ray so that $i = r$ .

Extend both reflected rays backwards behind the mirror as dashed lines. Where they cross is the image of that point.

Measuring confirms the image is the same perpendicular distance behind the mirror as the object is in front, and that the line joining object to image crosses the mirror at $90^{\circ}$ .

Law of Reflection - Interactive Diagram

Angle of Incidence (i)35°

10°80°

Show plane mirror image formation

Law of Reflection:

Angle of Incidence (i):35°

Angle of Reflection (r):35°

i = r ✓

Remember: Angles are always measured from the normal (the dotted perpendicular line), not from the mirror surface!

Incident Ray

Incoming light ray traveling toward the mirror

Reflected Ray

Outgoing light ray bouncing off the mirror

Applications

• Periscopes — two mirrors at $45^{\circ}$ let you see over an obstacle.

• Kaleidoscopes — several mirrors produce repeated symmetrical patterns.

• Security and car mirrors — give a wide-angle view of a large area.

Worked Examples

Light strikes a mirror at 35° incidence. Find angle of reflection.

Solution

The angle of incidence is measured from the normal, so

i = 35^{\circ}

The law of reflection gives

r = i

Answer:

r = 35^{\circ}

Object is 2.5 m from mirror. Image distance?

Solution

In a plane mirror the image is the same perpendicular distance behind the mirror as the object is in front.

Answer: 2.5 m behind the mirror, so the object and its image are 5.0 m apart.

Ray reflects between parallel mirrors. If first mirror angle is 25°, what's the second?

Solution

At the first mirror the law of reflection gives

r = i = 25^{\circ}

The mirrors are parallel, so their normals are parallel too. The ray crossing between them makes equal alternate angles with the two normals, so it arrives at the second mirror with

i = 25^{\circ}

Applying the law again at the second mirror gives

r = 25^{\circ}

Answer:

25^{\circ}

— and because the ray is turned through the same angle twice, it leaves parallel to its original direction.

2. Types of Reflection

Regular (Specular) Reflection

Key Feature: Parallel incident rays remain parallel after reflection. Image is clear.

Definition: Reflection from smooth surfaces where the normal at each point is parallel.

What Happens:

• Parallel incident rays remain parallel after reflection.

• Each ray obeys the Law of Reflection (i = r) individually.

• The surface acts like a perfect mirror.

Characteristics:

• Forms clear, sharp, undistorted images.

• The image has the same size and orientation as the object.

• Virtual image forms behind the mirror.

Real-World Examples:

• Plane mirrors (bathroom, car side mirrors)

• Still water surfaces (lake, pond on calm days)

• Polished metal (aluminum, steel)

• Glass windows (reflective in dim light)

Diffuse (Irregular) Reflection

Key Feature: Parallel rays scatter in different directions. No clear image.

Definition: Reflection from rough, irregular surfaces where the normal varies at each point.

What Happens:

• The microscopically rough surface has many tiny facets at different angles.

• Each incident ray reflects according to the Law of Reflection, but the normals differ.

• Parallel incident rays scatter in different directions.

Characteristics:

• No clear image forms (light is reflected in all directions).

• Light is diffusely scattered (spread out uniformly).

• Safe viewing from any angle (no glare concentration).

Real-World Examples:

• Paper, cardboard (allows reading from any angle)

• Concrete, asphalt (road surfaces)

• Cloth, paint, wood (matte finishes)

• Skin, leaves (natural objects)

• Why we see objects: Most surfaces are rough, so they reflect light in all directions making them visible.

3. Refraction of Light

Refraction is the change in direction of a light ray when it crosses a boundary between two transparent media because its speed changes.

A medium is more optically dense if light travels more slowly in it. Glass is optically denser than air; water is optically denser than air but usually less optically dense than glass.

The normal is an imaginary line drawn at $90^{\circ}$ to the surface at the point where the ray meets the boundary. All refraction angles are measured from the normal, not from the surface.

Notation: $i$ = angle of incidence, the angle between the incident ray and the normal. $r$ = angle of refraction, the angle between the refracted ray and the normal.

Air → glass or air → water: light slows down, so the ray bends towards the normal and $r < i$ .

Glass → air or water → air: light speeds up, so the ray bends away from the normal and $r > i$ .

If the ray enters along the normal, then $i = 0^{\circ}$ and $r = 0^{\circ}$ . The ray changes speed but does not change direction.

Refraction changes the speed of the light and therefore its wavelength, because $v = f λ$ . The frequency does not change — it is fixed by the source — which is why the colour of the light stays the same.

Refractive index, $n$ , tells us how much a medium slows light down compared with air or vacuum. A larger $n$ means lower light speed and greater bending at a boundary.

For light entering a medium from air: $n = \frac{\sin i}{\sin r}$ .

Speed definition: $n = \frac{c}{v}$ , where $c = 3.0 \times 10^{8} m/s$ is the speed of light in vacuum/air and $v$ is the speed of light in the medium.

More general form: $n_{1} \sin i = n_{2} \sin r$ , where $n_{1}$ is the refractive index of the first medium and $n_{2}$ is the refractive index of the second medium.

Typical values: air $\approx 1.00$ , water $\approx 1.33$ , glass $\approx 1.50$ , diamond $\approx 2.42$ .

Interactive Refraction (Snell's Law)

Angle of Incidence (i)45°

Medium 1 (Top)

Medium 2 (Bottom)

Refractive Index Ratio (n₁/n₂):0.67

Try this: Set Medium 1 to Glass (1.5) and Medium 2 to Air (1.0). Increase the angle of incidence past 42° to see Total Internal Reflection!

Ray Diagram Rules and Exam Notation

Draw the boundary first, then draw the normal as a dashed line at $90^{\circ}$ to the boundary.

Label the incoming ray as the incident ray and the outgoing ray as the refracted ray.

Mark $i$ between the incident ray and the normal. Mark $r$ between the refracted ray and the normal.

Use arrowheads on rays to show the direction of travel.

Do not write $r < i$ or $r > i$ by memory only. First decide whether the ray slows down or speeds up.

Effects of Refraction

A swimming pool appears shallower because light from the bottom bends away from the normal as it leaves water and enters air. Your brain traces the rays back in straight lines, so the bottom appears higher than it really is.

A straw appears bent in water because rays from the underwater part change direction at the water-air boundary, while rays from the part in air do not.

Stars twinkle because starlight is refracted by moving layers of air with slightly different densities in the atmosphere.

Glass Block Experiment

Aim: show refraction through a transparent block and measure $i$ and $r$ .

Method: place a rectangular glass or Perspex block on paper, draw around it, shine a narrow ray into one face, and mark the incident and emergent rays.

At entry, air → glass: the ray slows down and bends towards the normal.

Inside the block, the ray travels in a straight line because it is travelling through one uniform medium.

At exit, glass → air: the ray speeds up and bends away from the normal.

For a rectangular block with parallel sides, the emergent ray is parallel to the incident ray, but it is shifted sideways. This shift is called lateral displacement.

For a semicircular block, a ray aimed at the centre of the curved face enters along the radius, so it meets the curved surface normally and does not refract at entry. This is useful for investigating the critical angle at the flat face.

Angle of Incidence (i)21°

Calculations (Snell's Law):

Incident angle (i):21°

Refracted angle (r):13.8°

Emergent angle:21.0°

Lateral displacement:33.5 units

Emergent ray is parallel to incident ray but laterally displaced

Using Refractive Index Equations

Use $n = \frac{\sin i}{\sin r}$ when the ray travels from air into a material and the question gives angles.

Use $n = \frac{c}{v}$ when the question gives or asks for speed of light in a material.

Use $n_{1} \sin i = n_{2} \sin r$ when both media have stated refractive indices, for example water to glass or core to cladding.

Angles must be in degrees for IGCSE calculator work unless the question clearly says otherwise.

A calculated value of $\sin r > 1$ means no refracted ray is possible. If the ray is travelling from denser to less dense medium, total internal reflection occurs.

Worked Examples

Light enters glass (n = 1.5) from air at i = 30°. Find r.

Solution

Given: i = 30°, n = 1.5

\sin r = \frac{\sin 30^{\circ}}{1.5} = 0.333

r = \sin^{- 1} (0.333) = {19.5}^{\circ}

Answer:

r = {19.5}^{\circ}

, bent towards the normal because the light slows down.

Light: i = 45°, r = 28°. Find n.

Solution

n = \frac{\sin 45^{\circ}}{\sin 28^{\circ}} = \frac{0.707}{0.469} = 1.51

Answer:

n = 1.51

Water n = 1.33. Find light speed. (c = 3.0 × 10⁸ m/s)

Solution

v = \frac{c}{n} = \frac{3.0 \times 10^{8}}{1.33} = 2.26 \times 10^{8}

m/s

Answer:

v = 2.26 \times 10^{8}

m/s

Glass (n = 1.5) to air. i = 20° in glass. Find r in air.

Solution

\sin r = 1.5 \times \sin 20^{\circ} = 1.5 \times 0.342 = 0.513

r = \sin^{- 1} (0.513) = {30.9}^{\circ}

Answer:

r = {30.9}^{\circ}

, bent away from the normal because the light speeds up on leaving the glass.

4. Total Internal Reflection (TIR)

Internal reflection is the reflection of light at a boundary while the light stays inside the denser medium, instead of passing out of it.

Total internal reflection (TIR) is the complete reflection of a light ray back into the denser medium, with no refracted ray leaving at all.

Critical angle ( $c$ ): the angle of incidence, measured in the denser medium, for which the angle of refraction in the less dense medium is exactly $90^{\circ}$ . At this angle the refracted ray travels along the boundary.

Two conditions must both be met for TIR

• The light must be travelling from an optically denser medium into a less dense one, for example glass → air or water → air.

• The angle of incidence must be greater than the critical angle, $i > c$ .

If either condition fails, the light refracts out of the medium instead. TIR is not ordinary mirror reflection: no silvering is involved, and when it happens the reflection is total, so none of the light is lost through the boundary.

The three cases at the boundary

• $i < c$ : most of the light refracts out into the air, and a weak ray is reflected back inside.

• $i = c$ : the refracted ray grazes along the boundary at $90^{\circ}$ to the normal, with a weak reflected ray.

• $i > c$ : no light escapes. All of it is reflected back into the glass, obeying $i = r$ as in any reflection.

Calculating the critical angle

For a boundary with air: $n = \frac{1}{\sin c}$ , so $\sin c = \frac{1}{n}$ and $c = \sin^{- 1} (\frac{1}{n})$ .

For two media that are not air: $\sin c = \frac{n_{2}}{n_{1}}$ , where $n_{1}$ is the denser medium and $n_{2}$ the less dense one.

Take care with notation: $c$ here is the critical angle, not the speed of light $3.0 \times 10^{8} m/s$ . Which one is meant is decided by context.

A larger refractive index gives a smaller critical angle, so total internal reflection happens more easily.

Material	Refractive index $n$	Critical angle $c$
Water	1.33	49°
Glass	1.50	42°
Diamond	2.42	24°

Refraction occurring

Incident angle (i)30°

0°critical angle 41.8°90°

i = 30° is less than c = 41.8°, so the light refracts out, bending away from the normal as it leaves the glass.

1. Optical Fibres

How they work

• An optical fibre has a central core of refractive index $n_{1}$ surrounded by cladding of slightly lower index $n_{2}$ .

• Because $n_{1} > n_{2}$ , light in the core meets the core-cladding boundary going from denser to less dense material.

• Where that angle of incidence exceeds the critical angle, the ray is totally internally reflected and stays in the core.

• Repeated TIR guides the light along the fibre, even round gentle bends, and it emerges at the far end with very little energy lost.

Uses

• Telecommunications — high-speed internet, cable television and telephone signals.

• Medical endoscopes — seeing inside the body without surgery.

• Decorative lighting.

Advantages

• Visible light and short-wavelength infrared carry very high data rates.

• No electrical interference.

• Many signals can share one fibre.

• Flexible, lightweight, and hard to tap without detection.

2. Prisms in Periscopes and Binoculars

• A $45^{\circ}$ prism is used in place of a mirror.

• Light enters a face along the normal, so it is not refracted on entry.

• It then meets the sloping face at $45^{\circ}$ , which is greater than the critical angle for glass ( $42^{\circ}$ ), so it is totally internally reflected.

• This turns the light through $90^{\circ}$ with no silvering needed, so none of the light is lost through the back of a mirror.

• Unlike a silvered mirror the surface cannot tarnish, so the image does not degrade over time, and there is no faint double image from the glass front.

• Two such prisms are used in a periscope; in binoculars they also fold a long light path into a short body and put the image the right way up.

3. Diamond Sparkle

• Diamond has a very high refractive index, $n = 2.42$ , which gives it a very small critical angle of about $24^{\circ}$ .

• Almost any ray reaching a face from inside arrives at more than $24^{\circ}$ , so it is totally internally reflected instead of escaping.

• Light bounces repeatedly inside the stone before finally leaving through the top, which is what makes a diamond look brilliant.

• The cut is designed so that those reflections send the light back out of the top face towards the viewer.

• Dispersion splits the light into colours on the way out, giving the flashes of colour known as fire.

Experimental Demonstration

Using a Semicircular Glass Block:

1. Shine a narrow ray into the curved side along a radius so that it enters without refraction.

2. Let the ray meet the flat face inside the glass. Gradually increase the angle of incidence at this flat face.

3. Below critical angle: See both refracted and reflected rays

4. At critical angle: Refracted ray runs along the surface

5. Above critical angle: Only reflected ray visible (TIR)

6. Measure the critical angle from the normal, not from the flat face.

Safety Note: Never look directly into light sources, especially lasers.

Worked Examples

A ray of light is travelling from glass into air. The refractive index of the glass is 1.5. Calculate the critical angle for the glass-air boundary.

Solution

\sin c = \frac{1}{n} = \frac{1}{1.5} = 0.667

c = \sin^{- 1} (0.667) = {41.8}^{\circ}

This gives 41.8° (any angle > 42° causes TIR)

The critical angle for light travelling from a transparent material into air is 42°. Calculate the refractive index of the material.

Solution

n = \frac{1}{\sin c} = \frac{1}{\sin 42^{\circ}} = \frac{1}{0.669} = 1.49

A ray of light travels from water into air. The refractive index of water is 1.33 and the angle of incidence in the water is 55°. Determine whether total internal reflection occurs.

Solution

Find c:

\sin c = \frac{1}{1.33} = 0.752 \to c = {48.8}^{\circ}

Compare:

i = 55^{\circ} > c = {48.8}^{\circ}

Since i > c, total internal reflection occurs.

An optical fibre has a core with refractive index 1.52 and cladding with refractive index 1.48. Calculate the critical angle at the core-cladding boundary.

Solution

\sin c = \frac{n_{2}}{n_{1}} = \frac{1.48}{1.52} = 0.974

c = \sin^{- 1} (0.974) = {76.7}^{\circ}

5. Dispersion of Light

Dispersion is the splitting of white light into a spectrum of colours by refraction in a glass prism.

White light is a mixture of visible colours. Each colour has a different frequency and wavelength, so each colour travels at a slightly different speed in glass.

How it works:

1. White light enters the prism.

2. The ray refracts towards the normal at the first surface.

3. Violet light slows down the most in glass, has the greatest refractive index, and bends the most.

4. Red light slows down the least in glass, has the smallest refractive index, and bends the least.

5. At the second surface the colours refract again and separate further.

Result: a continuous visible spectrum from red to violet.

Colour order by increasing frequency: red, orange, yellow, green, blue, indigo, violet.

Colour order by increasing wavelength is the reverse: violet, indigo, blue, green, yellow, orange, red.

Monochromatic light means visible light of a single frequency. It does not split into a spectrum because all the light has the same frequency.

Key Concept: Refraction

Different colors travel at different speeds in glass. Violet slows down the most and bends the most. Red slows down the least and bends the least.

MEMORY AIDROYGBIV

Natural Phenomena - Rainbows

A rainbow forms when sunlight is refracted, dispersed, internally reflected, and refracted again inside many water droplets.

The arc appears because droplets sending red light to your eye are seen at about 42° from the antisolar direction.

Step	What happens	Physics idea	Exam wording
1. Entry	Sunlight enters a water droplet.	Refraction and dispersion begin.	White light splits because different colours refract by different amounts.
2. Internal path	The separated colours travel through the droplet at slightly different angles.	Violet bends more; red bends less.	Different wavelengths have different refractive indices in water.
3. Reflection	Light reflects from the back inner surface of the droplet.	Internal reflection changes the direction of the rays.	In a primary rainbow, there is one internal reflection.
4. Exit	The rays leave the droplet and refract again.	Colours separate further as they emerge.	Red emerges at about 42°; violet emerges at about 40°.
5. Many droplets	Each droplet sends only one colour towards the observer's eye.	Millions of droplets together form the visible rainbow.	The curved arc is due to the constant viewing angle.

Dispersion & Rainbow Formation

Prism Dispersion

Remember ROYGBIV: Red,Orange,Yellow,Green,Blue,Indigo,Violet

Rainbow in Water Droplet

Water droplet (n ≈ 1.33)

How Rainbows Form:

1.Refraction on entry: White light enters droplet, disperses into colors
2.Reflection: Light reflects off back of droplet (one TIR)
3.Refraction on exit: Colors separate further as they leave
4.Different angles: Red exits at 42°, violet at 40° from antisolar point

Show ray pathsShow annotations

Color	Wavelength	Refractive Index (glass)	Deviation
Red	~700 nm (longest)	~1.51 (smallest)	Least
Orange	~620 nm	~1.512	↓
Yellow	~580 nm	~1.514	↓
Green	~550 nm	~1.517	↓
Blue	~470 nm	~1.52	↓
Indigo	~450 nm	~1.522	↓
Violet	~400 nm (shortest)	~1.53 (largest)	Most

Key Principle: Dispersion occurs because different wavelengths (colors) travel at different speeds in a medium, causing them to refract by different amounts. Violet slows down most (highest n) and bends most, while red slows down least (lowest n) and bends least.

A secondary rainbow is fainter because the light reflects twice inside the droplet, so more light is lost.

Feature	Primary rainbow	Secondary rainbow
Internal reflections	1 reflection	2 reflections
Brightness	Brighter and more common	Fainter
Colour order	Red on top, violet inside	Violet on top, red inside
Approximate angle	Red about 42°	Red about 51°
Exam point	Main rainbow seen most often	Two internal reflections reverse the colour order

Dispersion is useful when separating colours or wavelengths gives extra information.

Application	How dispersion is used	Key instrument / example	Exam point
Spectroscopy	Light is split into spectral lines or bands so the wavelengths can be analysed.	Star light, gas lamps, chemical samples	Each element has a unique set of spectral lines, so spectra can identify elements in stars or unknown substances.
Optical instruments	Different wavelengths are separated, measured, or compared.	Spectrometer: measures wavelengths; spectrophotometer: measures colour absorption.	Useful for accurate wavelength measurements and for analysing how much light a substance absorbs.
Chromatic aberration	Dispersion causes different colours to focus at different points in a lens.	Cameras, telescopes, microscopes	It produces coloured fringes. Achromatic compound lenses reduce the problem by cancelling much of the dispersion.
Gemstone beauty	White light disperses inside the gemstone and emerges as separated colours.	Diamond "fire"	Careful cutting increases internal reflection and dispersion, making the colours more visible.

A pure spectrum is useful when accurate wavelength or frequency comparisons are needed.

Feature	Pure spectrum	Impure spectrum
Colour separation	Colours are clearly separated with little or no overlap.	Colours overlap strongly.
Boundaries	Distinct, sharp colour regions or spectral lines.	Fuzzy boundaries between colours.
How it is produced	Use a narrow slit, collimating lens, prism, and focusing lens.	A wide beam is sent directly through a prism.
Usefulness	Best for accurate wavelength or frequency comparison.	Good for a general demonstration of dispersion, but not for accurate measurement.
Reason for difference	Each point on the slit forms a narrow image, so colours do not blur into each other much.	Many overlapping images are formed, so the colours mix and blur.

Worked Examples

Violet deviates 5.2°, red deviates 3.8°. Find angular dispersion.

Solution

Angular dispersion = δᵥ - δᵣ = 5.2° - 3.8° = 1.4°

Prism: nᵥᵢₒₗₑₜ = 1.52, nᵣₑ𝒹 = 1.48. Which deviates more?

Solution

Higher n → slower speed → more bending → greater deviation

Violet: n = 1.52 (higher) | Red: n = 1.48 (lower)

Violet deviates more (higher refractive index)

Why does rainbow appear at ~42° from antisolar point?

Solution

Process: 1. Light enters droplet → refracts | 2. Reflects inside (TIR) | 3. Exits → refracts again

Geometry: Red light: total deviation ≈ 42° | Violet: ≈ 40°

A narrow beam of white light passes through a triangular glass prism and forms a spectrum on a screen. Explain why the colours separate, and state which colour is deviated most.

Solution

White light is a mixture of colours with different wavelengths.

Each colour has a slightly different refractive index in glass, so each colour refracts by a different amount.

Violet has the greatest refractive index in glass, so it slows down and bends the most.

The spectrum forms by dispersion; violet is deviated most and red is deviated least.