P24: Light and Optics
Reflection, refraction and lenses
Light And Optics
📋 Key Definitions
Reflection: When a wave bounces off a surface. The angle of incidence equals the angle of reflection. Both angles are measured from the normal.
Refraction: The change in direction of a wave when it changes speed as it passes from one medium to another at an angle. The wave bends towards the normal when slowing down and away from the normal when speeding up.
Normal: An imaginary line perpendicular to the surface at the point of incidence. All angles are measured from the normal, NOT from the surface.
Refractive index (n): A measure of how much a material slows down light. n = speed of light in vacuum ÷ speed of light in material. Higher n = more bending.
Total internal reflection (TIR): When light travelling in a denser medium hits a boundary with a less dense medium at an angle greater than the critical angle, all the light reflects back inside.
Critical angle: The smallest angle of incidence at which total internal reflection occurs. Below this angle, some light refracts out. At this angle, light refracts along the boundary (angle of refraction = 90 degrees).
📝 Reflection
Law of reflection: The angle of incidence equals the angle of reflection. This applies to all waves, not just light.
| Type | Description | Example |
|---|---|---|
| Specular reflection | Reflection from a smooth, flat surface. All rays reflect in the same direction, producing a clear image. | Mirror, calm water |
| Diffuse reflection | Reflection from a rough surface. Rays reflect in different directions due to the uneven surface. No clear image. | Wall, paper, rough sea |
📝 Refraction
Why Refraction Happens
When light passes from one medium to another, it changes speed. If it enters a denser medium (e.g. air to glass), it slows down and bends towards the normal. If it enters a less dense medium (e.g. glass to air), it speeds up and bends away from the normal.
Remember: Light bends towards the normal when entering a more dense medium (slowing down), and away from the normal when entering a less dense medium (speeding up). If light enters along the normal, it changes speed but does NOT change direction.
| Transition | Speed change | Direction change |
|---|---|---|
| Air to Glass | Slows down | Bends towards normal |
| Glass to Air | Speeds up | Bends away from normal |
| Air to Water | Slows down | Bends towards normal |
| Water to Air | Speeds up | Bends away from normal |
| Along the normal (any) | Changes speed | No bending — passes straight through |
Snell's Law (Higher)
n1 sin theta1 = n2 sin theta2
Refractive index: n = sin i / sin r
Where:
n = refractive index (no units)
theta = angle from the normal (degrees)
i = angle of incidence
r = angle of refraction
Refractive index: n = sin i / sin r
Where:
n = refractive index (no units)
theta = angle from the normal (degrees)
i = angle of incidence
r = angle of refraction
📝 Total Internal Reflection
Two conditions needed for TIR: 1) The light must be travelling in the denser medium towards the less dense medium. 2) The angle of incidence must be greater than the critical angle for the boundary.
Critical angle: sin C = 1 / n
Where:
C = critical angle (degrees)
n = refractive index of the denser medium
Where:
C = critical angle (degrees)
n = refractive index of the denser medium
Uses of Total Internal Reflection
- Optical fibres: Light travels through the glass core by total internal reflection, carrying data over long distances with minimal signal loss. Used in internet cables, endoscopes.
- Prisms in binoculars/periscopes: Light enters the prism and undergoes TIR at the glass-air boundary, reflecting the light through 90 or 180 degrees.
📝 Lenses
| Property | Convex (Converging) Lens | Concave (Diverging) Lens |
|---|---|---|
| Shape | Thicker at the centre, thinner at edges | Thinner at the centre, thicker at edges |
| Effect on light | Converges (brings together) parallel rays to a focal point | Diverges (spreads out) parallel rays |
| Image type | Can produce real or virtual images | Always produces virtual images |
| Focal point | Real focal point where rays meet | Virtual focal point where rays appear to come from |
| Uses | Magnifying glasses, cameras, eyesight correction (long sight) | Short sight correction, peepholes |
Convex Lens Ray Diagrams
Draw three principal rays from the top of the object:
- A ray parallel to the principal axis → refracts through the focal point.
- A ray through the centre of the lens → passes straight through.
- A ray through the focal point → refracts parallel to the principal axis.
The image forms where the refracted rays cross.
Concave Lens Ray Diagrams
- A ray parallel to the principal axis → refracts so it appears to come from the focal point on the same side.
- A ray through the centre of the lens → passes straight through.
The image forms where the diverging rays appear to come from (found by extending rays back with dashed lines).
Magnification = image height / object height
Magnification has no units. A magnification greater than 1 means the image is larger than the object. A magnification less than 1 means the image is smaller.
Magnification has no units. A magnification greater than 1 means the image is larger than the object. A magnification less than 1 means the image is smaller.
🧮 Worked Examples
Example 1: Calculating refractive index
Light enters glass from air at an angle of incidence of 40 degrees. The angle of refraction is 25 degrees. Calculate the refractive index of the glass.
Solution:
n = sin i / sin r = sin(40) / sin(25) = 0.643 / 0.423 = 1.52
Example 2: Calculating critical angle
A material has a refractive index of 1.5. Calculate its critical angle.
Solution:
sin C = 1 / n = 1 / 1.5 = 0.667
C = sin⁻¹(0.667) = 41.8 degrees
Example 3: Using Snell's law
Light travels from glass (n = 1.5) into air (n = 1.0) at an angle of incidence of 30 degrees. Calculate the angle of refraction.
Solution:
n1 sin theta1 = n2 sin theta2
1.5 × sin(30) = 1.0 × sin(theta2)
1.5 × 0.5 = sin(theta2)
sin(theta2) = 0.75
theta2 = sin⁻¹(0.75) = 48.6 degrees
Example 4: Magnification calculation
An object is 2 cm tall. A convex lens produces an image that is 6 cm tall. Calculate the magnification.
Solution:
Magnification = image height / object height = 6 / 2 = 3 (image is 3 times larger)
Example 5: TIR condition check
Glass has a refractive index of 1.5 and a critical angle of 41.8 degrees. Light in the glass hits the glass-air boundary at an angle of 50 degrees. Will total internal reflection occur?
Solution:
The angle of incidence (50 degrees) is greater than the critical angle (41.8 degrees). Light is travelling from the denser medium (glass) to the less dense medium (air). Both conditions are met, so yes, total internal reflection occurs.
Example 6: Finding image height from magnification
An object is 3 cm tall. A lens produces a magnification of 0.5. Calculate the image height.
Solution:
Image height = magnification × object height = 0.5 × 3 = 1.5 cm
The image is smaller than the object (diminished).
❓ Practice Questions
Q1: Foundation State the law of reflection.
Q2: Foundation Explain the difference between specular and diffuse reflection.
Q3: Higher Light enters water from air at an angle of incidence of 35 degrees. The angle of refraction is 26 degrees. Calculate the refractive index of water.
Q4: Higher Diamond has a refractive index of 2.42. Calculate its critical angle.
Q5: Foundation An object 4 cm tall is viewed through a convex lens and produces an image 12 cm tall. Calculate the magnification.
Q6: Higher Explain how total internal reflection allows optical fibres to transmit data over long distances.
✅ Answers
- Q1: The angle of incidence equals the angle of reflection, where both angles are measured from the normal.
- Q2: Specular reflection occurs on smooth surfaces — all rays reflect in the same direction, producing a clear image (e.g. a mirror). Diffuse reflection occurs on rough surfaces — the uneven surface means each ray hits at a different angle and reflects in a different direction, so no clear image is formed (e.g. a wall).
- Q3: n = sin i / sin r = sin(35) / sin(26) = 0.574 / 0.438 = 1.31
- Q4: sin C = 1 / n = 1 / 2.42 = 0.413. C = sin⁻¹(0.413) = 24.4 degrees
- Q5: Magnification = image height / object height = 12 / 4 = 3
- Q6: Optical fibres are thin strands of glass. Light enters one end and hits the glass-air boundary at an angle greater than the critical angle, causing total internal reflection. The light reflects repeatedly along the fibre, carrying the signal with very little loss. The cladding around the core ensures the light stays within the fibre by keeping the angle above the critical angle.
🎯 Exam Tips
- Always measure angles from the normal, NOT from the surface. This is the most common error in ray diagrams.
- When drawing refraction diagrams, draw the normal first as a dashed line.
- Remember: TIR requires BOTH conditions — dense-to-less-dense AND angle greater than critical angle. Check both in your answer.
- For magnification, remember: no units, and a value between 0 and 1 means the image is diminished (smaller).
- Convex lenses produce real images when the object is beyond the focal point — the image is inverted and can be projected onto a screen.
- Concave lenses ALWAYS produce virtual, upright, diminished images — the rays diverge but appear to come from a point on the same side as the object.
- When calculating refractive index, the larger the value of n, the more the light bends and the slower the light travels in that material.
🔬 Required Practical
Required Practical: Investigating Reflection and Refraction
Aim: To investigate the reflection of light by different surfaces and the refraction of light as it passes from air into glass or water.
Method (Reflection): 1) Draw a straight line on paper and place a flat mirror along it. 2) Draw a normal line perpendicular to the mirror surface. 3) Shine a ray of light at the mirror at a set angle of incidence (measured from the normal). 4) Mark the incident and reflected rays. 5) Measure the angle of reflection from the normal. 6) Repeat for different angles of incidence. 7) Verify that angle of incidence = angle of reflection.
Method (Refraction): 1) Place a glass block on paper and trace around it. 2) Shine a ray of light into the block at an angle. 3) Mark where the ray enters and exits the block. 4) Remove the block and draw the ray path through the glass. 5) Draw normals at the entry and exit points. 6) Measure the angle of incidence and angle of refraction. 7) Calculate n = sin i / sin r.
Variables: IV: angle of incidence, DV: angle of reflection / angle of refraction, Control: same mirror / same glass block, same light source
🔢 Maths Skills
Mathematical Skills
Using Snell's law: n = sin i / sin r. You must use a calculator to find sine values of angles in degrees. To find a refractive index, divide sin(i) by sin(r). To find a critical angle, use sin C = 1/n and then use the inverse sine function. Magnification = image height / object height (no units).
Maths Example
Light enters a glass block at an angle of incidence of 50°. The angle of refraction is 30°. Calculate the refractive index of the glass. n = sin i / sin r = sin(50°) / sin(30°) = 0.766 / 0.500 = 1.53.
⚠️ Common Misconceptions
Watch Out!
1. Wrong: Angles of incidence and reflection are measured from the surface Correct: All angles in optics are measured from the normal (the perpendicular to the surface), NOT from the surface itself
2. Wrong: Light bends towards the normal when it speeds up Correct: Light bends towards the normal when it slows down (entering a denser medium) and away from the normal when it speeds up (entering a less dense medium)
3. Wrong: Total internal reflection can happen whenever light hits a boundary Correct: TIR requires BOTH conditions: light must be travelling from a denser to a less dense medium AND the angle of incidence must exceed the critical angle
✍️ 6-Mark Question
Extended Answer
6 marks: Explain how total internal reflection is used in optical fibres to transmit data over long distances. Why must the fibre be designed so that light always hits the boundary above the critical angle?
An optical fibre is a thin strand of glass with a central core surrounded by cladding of lower refractive index. Light enters one end of the fibre and travels through the core. When the light reaches the boundary between the core and the cladding, it hits at an angle greater than the critical angle for the core-cladding boundary. This means total internal reflection occurs, and all the light is reflected back into the core. The light continues to reflect repeatedly off the boundaries as it travels along the fibre, carrying the data signal with very little loss of intensity over long distances. The fibre must be designed so that light always hits the boundary above the critical angle because if the angle were below the critical angle, some light would refract out of the core into the cladding, weakening the signal. The cladding has a lower refractive index than the core, ensuring the critical angle condition is met for rays travelling along the fibre.
Mark scheme: 1 mark — light enters fibre and travels through glass core, 1 mark — light hits core-cladding boundary above critical angle, 1 mark — total internal reflection occurs repeatedly, 1 mark — signal carried with minimal loss, 1 mark — if angle were below critical angle, light would refract out and signal would weaken, 1 mark — cladding has lower refractive index to ensure TIR conditions are met
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student investigates refraction using a rectangular glass block. They measure angles of incidence and refraction and calculate n = sin i / sin r for each reading. Their results are: i=20° r=13° (n=1.52), i=30° r=19° (n=1.49), i=40° r=25° (n=1.52), i=50° r=30° (n=1.53), i=60° r=35° (n=1.51). The accepted value for the glass is 1.50.
(a) Calculate the mean refractive index from the student's results.
(b) The student's results at i = 30° give n = 1.49. Suggest why this reading might be less accurate than the others.
(c) The student repeats the experiment with a different block and gets n = 1.33. Identify the material and explain why light bends less in this material than in glass.
Answers: (a) Mean n = (1.52 + 1.49 + 1.52 + 1.53 + 1.51) ÷ 5 = 7.57 ÷ 5 = 1.514 ≈ 1.51. (b) At small angles of incidence (30°), the angle of refraction is also small (19°), so a small measurement error in the angle produces a larger percentage error in the sine value, making the calculated n less precise. (c) n = 1.33 corresponds to water. Light bends less in water than in glass because water has a lower refractive index, meaning light travels faster in water than in glass and changes speed less when entering from air.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Combined Science exams?
Get the best revision books and guides to boost your grades.
🧠 Flashcards (Spaced Repetition)
📝 Exam Questions by Topic
🎯 Target Tests (Auto-Graded)
📝 Exam Questions by Topic
📄 Past Papers for Physics (AQA)
- AQA Past Papers & Mark Schemes
- Edexcel Past Papers & Mark Schemes
- OCR Past Papers & Mark Schemes
- Eduqas Past Papers & Mark Schemes
- CCEA Past Papers & Mark Schemes
For the most accurate and up-to-date past papers, always check the official exam board websites.