Laws of reflection · Snell's Law · Optical density · Total internal reflection
📐 State and apply the law of reflection, measuring angles from the normal
🌊 Explain refraction in terms of a change in wave speed at a boundary
🔢 Use Snell's Law (n₁ sin θ₁ = n₂ sin θ₂) to calculate angles of refraction
💡 Define refractive index and relate it to optical density
🔭 Explain total internal reflection and calculate the critical angle
📡 Describe applications of total internal reflection including optical fibres
1. The Law of Reflection
When a ray of light strikes a smooth, flat surface such as a plane mirror, it bounces back in a predictable way. To describe this accurately, physicists always measure angles from the normal — an imaginary line drawn perpendicular (at 90°) to the surface at the point where the ray hits.
Normal: A construction line drawn at 90° to the reflecting (or refracting) surface at the point of incidence.
The law of reflection states two things:
The angle of incidence equals the angle of reflection: θᵢ = θᵣ
The incident ray, the reflected ray, and the normal all lie in the same plane.
θᵢ = θᵣ
Both angles are measured between the ray and the normal — not between the ray and the surface itself. A common exam mistake is to measure from the surface; always use the normal.
💡 Smooth surfaces produce specular (regular) reflection; rough surfaces scatter light in many directions — this is diffuse reflection. Both obey the law of reflection at each microscopic point, but the irregular surface redirects each ray differently.
In a plane mirror, the image formed is:
The same size as the object
Virtual (cannot be projected onto a screen)
The same distance behind the mirror as the object is in front
Laterally inverted (left–right reversed)
2. Refraction and Wave Speed
Refraction occurs when a wave crosses a boundary between two different media and changes speed. Light travels at c = 3 × 10⁸ m/s in a vacuum, but slows down when it enters a denser material such as glass or water.
Refraction: The change in direction of a wave as it passes from one medium to another, caused by a change in wave speed.
The key rules for refraction:
When light passes from a less dense medium (e.g. air) into a more dense medium (e.g. glass), it slows down and bends towards the normal (angle of refraction < angle of incidence).
When light passes from a more dense medium into a less dense medium, it speeds up and bends away from the normal (angle of refraction > angle of incidence).
If a ray hits the boundary at exactly 90° (along the normal), it passes straight through without bending.
💡 Frequency does not change when light refracts — only speed and wavelength change. Since v = fλ and f is constant, if v decreases, λ also decreases.
The refractive index (n) of a material is defined as the ratio of the speed of light in a vacuum to the speed of light in that material:
n = c ÷ v
Symbol
Quantity
Unit
n
Refractive index
dimensionless (no unit)
c
Speed of light in vacuum
m/s (3.00 × 10⁸ m/s)
v
Speed of light in the medium
m/s
Typical values: air ≈ 1.0003 (treated as 1), water ≈ 1.33, glass ≈ 1.5, diamond ≈ 2.42. A higher refractive index means light travels more slowly through that material — the material is said to be optically denser.
3. Snell's Law
Snell's Law gives a precise mathematical relationship between the angle of incidence and the angle of refraction when light crosses a boundary between two media.
n₁ sin θ₁ = n₂ sin θ₂
Symbol
Meaning
n₁
Refractive index of the first medium (incident side)
θ₁
Angle of incidence (measured from the normal), degrees or radians
n₂
Refractive index of the second medium (refracted side)
θ₂
Angle of refraction (measured from the normal)
For a ray travelling from air (n ≈ 1) into glass (n = 1.5), Snell's Law simplifies to:
sin θ₁ = 1.5 × sin θ₂
Rearranging: sin θ₂ = sin θ₁ ÷ 1.5, so θ₂ < θ₁ — the ray bends towards the normal as expected.
💡 Snell's Law is reversible: a ray refracting from glass into air at angle θ₂ emerges at angle θ₁ — exactly the reverse path.
The refractive index of a single material relative to a vacuum can also be written from Snell's Law as:
n = sin θᵢ ÷ sin θᵣ
(where the incident medium is air/vacuum with n = 1 and θᵣ is the angle in the medium).
4. Total Internal Reflection & the Critical Angle
When light travels from an optically denser medium (high n) into an optically less dense medium (low n), it bends away from the normal. As the angle of incidence increases, so does the angle of refraction. At a specific angle — the critical angle — the refracted ray travels along the boundary (angle of refraction = 90°). Beyond the critical angle, no refraction occurs: all the light is reflected back into the denser medium. This is called total internal reflection (TIR).
Critical angle (θ꜀): The angle of incidence in the denser medium at which the angle of refraction equals exactly 90°. Beyond this, TIR occurs.
Setting θ₂ = 90° in Snell's Law (sin 90° = 1) and with n₂ = 1 (air):
sin θ꜀ = n₂ ÷ n₁ = 1 ÷ n
So for glass with n = 1.5: sin θ꜀ = 1 ÷ 1.5 = 0.667, giving θ꜀ = 41.8°
Two conditions must both be met for TIR:
Light must be travelling from a denser medium into a less dense medium (n₁ > n₂).
The angle of incidence must be greater than the critical angle (θᵢ > θ꜀).
💡 A higher refractive index gives a smaller critical angle. Diamond (n = 2.42) has θ꜀ ≈ 24.4°, which is why it sparkles so brilliantly — TIR occurs easily.
Applications of TIR:
Optical fibres: Light travels along a glass or plastic fibre by repeated TIR. The core has a higher refractive index than the cladding. Used in telecommunications (internet cables) and medical endoscopes.
Periscopes and binoculars: Use glass prisms instead of mirrors — TIR gives a brighter, higher-quality reflection than silvered mirrors.
Cats' eyes on roads: Reflective road studs use TIR to return light to drivers at night.
5. Optical Density & Summary of Key Equations
Optical density is a measure of how much a medium slows down light. It is directly related to the refractive index — a material with a higher refractive index is optically denser. Note that optical density is not the same as physical density (mass per unit volume). For example, certain oils can be less physically dense than water but more optically dense.
Optically dense: A medium through which light travels more slowly, having a higher refractive index.
Equation
Use
θᵢ = θᵣ
Law of reflection
n = c ÷ v
Refractive index from wave speed
n₁ sin θ₁ = n₂ sin θ₂
Snell's Law — refraction at a boundary
n = sin θᵢ ÷ sin θᵣ
Refractive index from angles (air → medium)
sin θ꜀ = 1 ÷ n
Critical angle for medium–air boundary
The wave equation v = fλ links wave speed, frequency and wavelength. Since frequency is unchanged at a boundary, a decrease in speed means a proportional decrease in wavelength.
💡 In any refraction/TIR problem, always start by drawing the normal and measuring all angles from it, not from the surface.
Example 1: A ray of light strikes a plane mirror at an angle of 35° to the mirror surface. What is the angle of reflection? State the law of reflection.
1 Identify the angle of incidence. The angle is measured from the normal, not the surface. The angle to the surface is 35°, so the angle of incidence θᵢ = 90° − 35° = 55°.
2 Apply the law of reflection: θᵢ = θᵣ.
3 Therefore the angle of reflection θᵣ = 55°. The incident ray, reflected ray, and normal all lie in the same plane.
Angle of reflection = 55° (from the normal). The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal.
Example 2: A ray of light passes from air into a glass block. The angle of incidence is 48° and the refractive index of the glass is 1.52. Calculate the angle of refraction inside the glass.
1 Write down Snell's Law: n₁ sin θ₁ = n₂ sin θ₂. Here n₁ = 1.00 (air), θ₁ = 48°, n₂ = 1.52, θ₂ = ?
2 Rearrange for sin θ₂: sin θ₂ = (n₁ × sin θ₁) ÷ n₂ = (1.00 × sin 48°) ÷ 1.52
3 Calculate: sin 48° = 0.7431 sin θ₂ = 0.7431 ÷ 1.52 = 0.4889
4 Find the angle: θ₂ = sin⁻¹(0.4889) = 29.3°
5 Check: θ₂ (29.3°) < θ₁ (48°) ✓ — ray bends towards normal when entering denser medium.
Angle of refraction = 29.3° (to the nearest 0.1°)
Example 3: The speed of light in a type of optical glass is 1.80 × 10⁸ m/s. (a) Calculate the refractive index of the glass. (b) Calculate the critical angle for a glass–air boundary.
1Part (a) — Refractive index: Use n = c ÷ v, where c = 3.00 × 10⁸ m/s and v = 1.80 × 10⁸ m/s.
3Part (b) — Critical angle: Use sin θ꜀ = 1 ÷ n = 1 ÷ 1.67 = 0.5988
4 θ꜀ = sin⁻¹(0.5988) = 36.8°
(a) Refractive index n = 1.67 | (b) Critical angle = 36.8°
Example 4: Light travels from water (n = 1.33) into glass (n = 1.50) at an angle of incidence of 40°. Calculate the angle of refraction in the glass.
1 Both media have n > 1, so we cannot use the simplified formula. Apply full Snell's Law: n₁ sin θ₁ = n₂ sin θ₂ with n₁ = 1.33, θ₁ = 40°, n₂ = 1.50.
2 sin θ₂ = (n₁ × sin θ₁) ÷ n₂ = (1.33 × sin 40°) ÷ 1.50
3 sin 40° = 0.6428 Numerator: 1.33 × 0.6428 = 0.8549 sin θ₂ = 0.8549 ÷ 1.50 = 0.5699
4 θ₂ = sin⁻¹(0.5699) = 34.8°
5 Check: light moves from less dense (water, n=1.33) to more dense (glass, n=1.50), so it bends towards the normal. θ₂ = 34.8° < θ₁ = 40° ✓
Angle of refraction in glass = 34.8°
Question 1: A ray of light hits a flat mirror at 30° to the mirror surface. What is the angle of reflection (measured from the normal)?
Question 2: Light passes from glass (n = 1.5) into air. As the angle of incidence increases beyond the critical angle, what happens?
Question 3: The refractive index of diamond is 2.42. Calculate the speed of light in diamond. (c = 3.00 × 10⁸ m/s). Give your answer in m/s to 3 significant figures.
Question 4: Which of the following correctly describes what happens to the wavelength of light when it enters a denser medium (higher n)?
Question 5: A ray of light hits a glass–air boundary at an angle of incidence of 30°. The refractive index of the glass is 1.50. Use Snell's Law to calculate the angle of refraction in air. Give your answer in degrees to 1 decimal place.
Challenge 1 (4 marks): A student measures the angles of incidence and refraction as light passes from air into a liquid and records the following data:
Angle of incidence (°)
Angle of refraction (°)
20
14.9
40
28.9
60
40.5
(a) Use one set of data to calculate the refractive index of the liquid. (b) Calculate the critical angle for a liquid–air boundary using your value of n.
Model Answer:
(a) Using row 1: n = sin θᵢ ÷ sin θᵣ = sin 20° ÷ sin 14.9° = 0.3420 ÷ 0.2571 = 1.33 (this is the refractive index of water — the liquid is water).
You can verify: Row 2: sin40° ÷ sin28.9° = 0.6428 ÷ 0.4833 = 1.33 ✓
(b) sin θ꜀ = 1 ÷ n = 1 ÷ 1.33 = 0.7519
θ꜀ = sin⁻¹(0.7519) = 48.8°
Challenge 2 (5 marks): An optical fibre has a core refractive index of 1.55 and a cladding refractive index of 1.40. (a) Explain why cladding is needed around the core. (b) Calculate the critical angle at the core–cladding boundary. (c) A ray enters the fibre at the core–cladding boundary at an angle of incidence of 68°. State with a reason whether TIR occurs.
Model Answer:
(a) The cladding has a lower refractive index than the core, ensuring that TIR can occur at the core–cladding boundary, keeping the light signal inside the core. Without cladding, light would leak out when fibres touch each other or bend, causing signal loss. [2 marks]
(c) The angle of incidence (68°) is greater than the critical angle (64.6°), so TIR does occur. Both conditions are met: the ray is in the denser medium (core, n=1.55) travelling towards the less dense medium (cladding, n=1.40), and θᵢ > θ꜀. [1 mark]
Challenge 3 (6 marks): A ray of light travels through three layers: air → glass (n = 1.50) → water (n = 1.33) → air. The ray strikes the air–glass boundary at an angle of incidence of 55°. (a) Calculate the angle of refraction in the glass. (b) Using your answer as the angle of incidence at the glass–water boundary, calculate the angle of refraction in the water. (c) Predict and explain the angle of refraction when the ray exits back into air from the water, without calculating.
Model Answer:
(a) Air → glass: n₁ sin θ₁ = n₂ sin θ₂
1.00 × sin 55° = 1.50 × sin θ₂
sin θ₂ = sin 55° ÷ 1.50 = 0.8192 ÷ 1.50 = 0.5461
θ₂ = sin⁻¹(0.5461) = 33.1° [2 marks]
(b) Glass → water: n₁ sin θ₁ = n₂ sin θ₂
1.50 × sin 33.1° = 1.33 × sin θ₂
sin θ₂ = (1.50 × 0.5461) ÷ 1.33 = 0.8192 ÷ 1.33 = 0.6159
θ₂ = sin⁻¹(0.6159) = 38.0° [2 marks]
(c) By Snell's Law, n₁ sin θ₁ = n₂ sin θ₂. When the ray exits from water (n=1.33) into air (n=1.00) with the angle from (b) as the incident angle, the product n × sinθ must remain constant throughout: it equals 1.00 × sin55° = 0.8192 at every boundary. So the exit angle in air will be sin⁻¹(0.8192) = 55° — the same as the original angle of incidence. The ray exits parallel to its original direction. [2 marks]
Challenge 4 — Extended Response (6 marks): Explain, with reference to wave speed and refractive index, why total internal reflection can only occur when light travels from an optically denser medium into an optically less dense medium, and never the other way around. Include the concept of the critical angle in your answer.
Model Answer (mark scheme points):
• In a denser medium, light travels more slowly (higher n means lower speed: v = c÷n). [1]
• When light moves from a denser to a less dense medium, it speeds up and bends away from the normal (angle of refraction > angle of incidence). [1]
• As the angle of incidence increases, the angle of refraction increases faster, eventually reaching 90° — the refracted ray travels along the boundary. The angle of incidence at this point is the critical angle. [1]
• Beyond the critical angle, there is no possible refracted ray (it would require sin θ > 1, which is impossible), so all light is reflected back into the denser medium — this is TIR. [1]
• When light travels from a less dense to a denser medium, it bends towards the normal (angle of refraction < angle of incidence). No matter how large the angle of incidence, the angle of refraction is always less than 90°, so refraction always occurs and TIR is impossible. [2]