P25: Light and Optics
Reflection, refraction, Snell's law, refractive index, total internal reflection, critical angle, and the behaviour of convex and concave lenses.
Reflection
The law of reflection states that the angle of incidence equals the angle of reflection. Both angles are measured from the normal, which is an imaginary line perpendicular to the surface at the point of incidence.
- Specular (mirror) reflection: smooth surface reflects all rays in the same direction
- Diffuse (scattered) reflection: rough surface reflects rays in many directions
- The image in a plane mirror is the same size as the object, virtual, upright, and laterally inverted
Refraction
Refraction is the change in direction of a wave when it crosses a boundary between two materials due to a change in wave speed. Light bends towards the normal when entering a denser medium and away from the normal when entering a less dense medium.
- When light enters a denser material (e.g. air to glass), it slows down and bends towards the normal
- When light enters a less dense material (e.g. glass to air), it speeds up and bends away from the normal
- If light enters along the normal (angle of incidence = 0°), it changes speed but does not change direction
- The frequency of the wave does not change during refraction, but the wavelength does
Remember: waves slow down in denser materials. When a wave slows down, its wavelength decreases but its frequency stays the same.
Refractive Index
The refractive index (n) of a material is a measure of how much it slows down light compared to a vacuum. A higher refractive index means light travels more slowly in that material.
Refractive Index
n = c / v
n = refractive index
c = speed of light in a vacuum (3 × 10⁸ m/s)
v = speed of light in the material (m/s)
Typical refractive indices: air ≈ 1.0, water = 1.33, glass ≈ 1.5, diamond = 2.42.
Snell's Law
Snell's law relates the angles of incidence and refraction to the refractive indices of the two materials.
Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
n₁ = refractive index of first material
n₂ = refractive index of second material
θ₁ = angle of incidence (°)
θ₂ = angle of refraction (°)
Worked Example 1: Refraction at an Air-Glass Boundary
Light travels from air (n = 1.0) into glass (n = 1.5) at an angle of incidence of 30°. Calculate the angle of refraction.
Solution:
n₁ sin θ₁ = n₂ sin θ₂
1.0 × sin 30° = 1.5 × sin θ₂
0.5 = 1.5 × sin θ₂
sin θ₂ = 0.5 / 1.5 = 0.333
θ₂ = sin⁻¹(0.333) = 19.5°
Worked Example 2: Finding Refractive Index
A ray of light strikes a glass block at 45° and is refracted to an angle of 28° inside the glass. Calculate the refractive index of the glass.
Solution:
n₁ sin θ₁ = n₂ sin θ₂
1.0 × sin 45° = n₂ × sin 28°
0.707 = n₂ × 0.469
n₂ = 0.707 / 0.469 = 1.51
Total Internal Reflection
Total internal reflection (TIR) occurs when light travelling inside a denser medium meets a boundary with a less dense medium at an angle greater than the critical angle. All the light is reflected back inside the denser medium.
Two conditions are required for TIR:
- The light must be travelling from a more dense medium towards a less dense medium
- The angle of incidence must be greater than the critical angle
Critical Angle
sin C = n₂ / n₁
C = critical angle (°)
n₁ = refractive index of the denser medium
n₂ = refractive index of the less dense medium
For light travelling from glass (n = 1.5) to air (n = 1.0):
sin C = 1.0 / 1.5 = 0.667, so C = 41.8°
Applications of Total Internal Reflection
Optical Fibres
- Used in telecommunications to transmit data as pulses of light
- Light is totally internally reflected along the fibre core
- The core has a higher refractive index than the cladding surrounding it
- Signals travel at the speed of light with minimal signal loss
- Also used in endoscopes for medical imaging
Prisms in Binoculars and Periscopes
- Right-angled prisms use TIR to reflect light through 90° or 180°
- More efficient than mirrors as no light is lost
Convex and Concave Lenses
A convex (converging) lens is thicker in the middle and causes parallel rays of light to converge at the focal point. A concave (diverging) lens is thinner in the middle and causes parallel rays to diverge.
- Focal point (F): the point where parallel rays converge after passing through a convex lens, or appear to diverge from after passing through a concave lens
- Focal length (f): the distance from the centre of the lens to the focal point
- Principal axis: the horizontal line passing through the centre of the lens
Images Formed by Convex Lenses
- Object beyond 2F: real, inverted, diminished image between F and 2F
- Object at 2F: real, inverted, same-size image at 2F
- Object between F and 2F: real, inverted, magnified image beyond 2F
- Object at F: no image (rays are parallel)
- Object between F and lens: virtual, upright, magnified image on the same side as the object (magnifying glass)
Images Formed by Concave Lenses
- Always produces a virtual, upright and diminished image regardless of object position
- The image is always on the same side of the lens as the object
Magnification
Magnification Formula
magnification = image height / object height
A magnification greater than 1 means the image is larger than the object. A magnification less than 1 means the image is smaller.
Worked Example
An object of height 2.0 cm produces an image of height 6.0 cm through a convex lens. Calculate the magnification.
Solution:
magnification = image height / object height
magnification = 6.0 / 2.0 = 3.0
Practice Questions
1. Light passes from air (n = 1.0) into water (n = 1.33) at an angle of incidence of 40°. Calculate the angle of refraction. [3 marks]
n₁ sin θ₁ = n₂ sin θ₂
1.0 × sin 40° = 1.33 × sin θ₂
0.643 = 1.33 × sin θ₂
sin θ₂ = 0.643 / 1.33 = 0.483
θ₂ = sin⁻¹(0.483) = 28.9°
2. State the two conditions required for total internal reflection to occur. [2 marks]
The light must be travelling from a more optically dense medium towards a less dense medium, and the angle of incidence must be greater than the critical angle for the boundary.
3. Calculate the critical angle for light travelling from glass (n = 1.5) into air (n = 1.0). [2 marks]
sin C = n₂ / n₁ = 1.0 / 1.5 = 0.667
C = sin⁻¹(0.667) = 41.8°
4. An object of height 3 cm produces an image of height 9 cm. Calculate the magnification. [1 mark]
magnification = 9 / 3 = 3.0
5. Describe the image formed by a concave lens. [2 marks]
A concave lens always forms a virtual, upright and diminished image on the same side of the lens as the object.
Maths Skills
Snell's Law Calculations
Key Formulae
n₁ sin θ₁ = n₂ sin θ₂ — Snell's law
n = c / v — refractive index
sin C = n₂ / n₁ — critical angle
magnification = image height / object height
Finding an Unknown Angle
Light travels from water (n = 1.33) into air (n = 1.00) at an angle of incidence of 25°. Calculate the angle of refraction.
n₁ sin θ₁ = n₂ sin θ₂
1.33 × sin 25° = 1.00 × sin θ₂
1.33 × 0.4226 = 0.562
sin θ₂ = 0.562
θ₂ = sin⁻¹(0.562) = 34.2°
Calculating Refractive Index from Critical Angle
A glass block has a critical angle of 41.8° when light travels from the glass to air. Calculate the refractive index of the glass.
sin C = n₂ / n₁ = 1.00 / n₁
sin 41.8° = 1.00 / n₁
0.667 = 1.00 / n₁
n₁ = 1.00 / 0.667 = 1.50
Magnification Calculations
- Magnification > 1: image is larger than the object
- Magnification = 1: image is the same size
- Magnification < 1: image is smaller than the object
- You can also write: magnification = image distance / object distance
Magnification Problem
A coin of diameter 2.4 cm is viewed through a magnifying glass and appears to have a diameter of 7.2 cm. Calculate the magnification.
Magnification = 7.2 / 2.4 = 3.0 (the image is three times larger than the object)
Common Misconceptions
"Light always bends towards the normal when it is refracted." Light only bends towards the normal when it enters a denser medium (slowing down). When light goes from a denser medium to a less dense medium (e.g. glass to air), it bends away from the normal because it speeds up. If light hits the surface along the normal (angle of incidence = 0°), it passes straight through without bending, even though its speed changes.
"Total internal reflection only happens in fibre optics." Total internal reflection occurs whenever light travels from a denser medium towards a less dense medium at an angle greater than the critical angle. This can happen at any such boundary: glass-to-air, water-to-air, or within prisms in binoculars and periscopes. Optical fibres simply exploit this general principle to guide light along a flexible path.
"The refractive index tells you how bright the light is after refraction." Refractive index measures how much a material slows down light compared to a vacuum. n = c / v. A higher refractive index means light travels more slowly in that material. It does not directly measure brightness, though some light is always reflected at a boundary (partial reflection).
6-Mark Extended Question
Explain how total internal reflection is used in optical fibres for communications. Discuss the properties the glass must have. [6 marks]
Optical fibres are thin strands of glass that carry data as pulses of light over long distances. The fibre has a central core with a high refractive index surrounded by cladding with a lower refractive index. When a light pulse enters the core, it strikes the boundary between the core and cladding at an angle greater than the critical angle. Because the light is travelling from a denser medium (core) to a less dense medium (cladding) and the angle of incidence exceeds the critical angle, total internal reflection occurs. The light is reflected back into the core and continues to reflect off the boundaries as it travels along the fibre, allowing it to follow the curve of the fibre even around bends. The glass must have several key properties: the core must have a higher refractive index than the cladding so that TIR can occur; the glass must be very pure and transparent to minimise absorption of light over long distances; the core must be extremely narrow to prevent light from taking multiple paths (modal dispersion), which would cause pulses to spread and overlap; the surface of the core must be very smooth to prevent light scattering at the boundary; and the cladding must be thick enough to prevent light from escaping into adjacent fibres in a bundle. Optical fibres carry far more data than copper cables and are lighter, cheaper, and more secure.
AO3: Analyse and Evaluate
A student investigates refraction by shining a ray of light into a glass block at different angles. Their results are shown below:
| Angle of incidence (°) | Angle of refraction (°) |
|---|---|
| 10 | 6.5 |
| 20 | 13.0 |
| 30 | 19.5 |
| 40 | 25.5 |
| 50 | 30.7 |
| 60 | 35.3 |
(a) Use the data for 30° incidence to calculate the refractive index of the glass.
(b) Use Snell's law to predict the angle of refraction for an angle of incidence of 45°.
(c) The student's value for 60° incidence gives n = 1.54, while the value for 10° gives n = 1.53. Explain why the result is more accurate at larger angles of incidence.
(a) n₁ sin θ₁ = n₂ sin θ₂
1.0 × sin 30° = n × sin 19.5°
0.500 = n × 0.334
n = 0.500 / 0.334 = 1.50
(b) n₁ sin θ₁ = n₂ sin θ₂
1.0 × sin 45° = 1.50 × sin θ₂
0.707 = 1.50 × sin θ₂
sin θ₂ = 0.471
θ₂ = 28.1°
(c) At larger angles of incidence, the angles of refraction are also larger. The percentage uncertainty in measuring an angle is smaller for larger angles because the absolute uncertainty (about ±1°) represents a smaller fraction of the measured value. For example, 1° uncertainty on 6.5° is a 15% error, but 1° on 35.3° is only a 2.8% error. Therefore, measurements at larger incidence angles produce more accurate calculations of refractive index.
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