When light passes from one medium into another, its direction of propagation changes. This phenomenon is called refraction.
The refractive index of a medium is a quantity that describes how strongly light is refracted by that medium. It is defined as:
Because the speed of light in a vacuum is the maximum possible (c ≈ 3 × 10⁸ m/s), the refractive index of every medium is greater than 1.
| Material | Refractive Index (n) | Notes |
|---|---|---|
| Vacuum | 1.00 | Reference value |
| Air | 1.00 | Approximately 1.0003 |
| Water | 1.33 | Standard value |
| Glass | 1.50 | Typical crown glass |
| Plastic (PE) | 1.50 | Common plastic |
| Diamond | 2.42 | Very high n → brilliant sparkle |
| Direction | Change in speed | Refracted ray | Angle relation |
|---|---|---|---|
| Less dense → denser | Speed decreases | Bends toward normal | θ₂ < θ₁ |
| Denser → less dense | Speed increases | Bends away from normal | θ₂ > θ₁ |
The law of refraction has two parts:
Where:
n₁ = refractive index of the medium on the side of the incident rayn₂ = refractive index of the medium on the side of the refracted rayθ₁ = angle of incidence (between the incident ray and the normal)θ₂ = angle of refraction (between the refracted ray and the normal)| Form | Equation | Use |
|---|---|---|
| General form | n₁ sin θ₁ = n₂ sin θ₂ | Any pair of media |
| Air–medium form | sin θ₁ = n sin θ₂ | Light entering a medium from air (n₁ ≈ 1) |
| Solve for refractive index | n = sin θ₁ / sin θ₂ | Both angles known, find n |
A ray of light strikes the surface of water from air at an angle of incidence of 45°. Find the angle of refraction. (Refractive index of water n = 1.33)
Solution:
n₁ sin θ₁ = n₂ sin θ₂
1 × sin 45° = 1.33 × sin θ₂
sin θ₂ = sin 45° / 1.33 = 0.7071 / 1.33 = 0.532
θ₂ = sin⁻¹(0.532) ≈ 32.1°
✅ The angle of refraction (32.1°) is smaller than the angle of incidence (45°). The ray bends toward the normal because water is denser than air.
Light enters glass from air with an angle of incidence of 60° and an angle of refraction of 35°. Find the refractive index of the glass.
Solution:
n = sin θ₁ / sin θ₂ = sin 60° / sin 35°
n = 0.866 / 0.574 ≈ 1.51
✅ The refractive index of the glass is approximately 1.51, which matches the typical value for ordinary glass.
This means that:
When light travels from a denser medium (large n) toward a less dense medium (small n) (for example, from water to air), the angle of refraction is greater than the angle of incidence. As the angle of incidence increases, the angle of refraction also increases, until it reaches 90° — at which point the refracted ray travels along the boundary.
If the angle of incidence is increased further, the light can no longer escape into the less dense medium. Instead, it is completely reflected back into the denser medium. This is called total internal reflection (TIR).
The critical angle is the angle of incidence at which the angle of refraction is exactly 90°. At this point sin 90° = 1, so Snell's law gives:
| Medium combination | Critical angle θc | Calculation |
|---|---|---|
| Water → air | ≈ 48.6° | sin θc = 1 / 1.33 |
| Glass → air | ≈ 42.0° | sin θc = 1 / 1.50 |
| Diamond → air | ≈ 24.4° | sin θc = 1 / 2.42 |
The refractive index of water is n = 1.33. Find the critical angle for light going from water into air.
Solution:
sin θc = n₂ / n₁ = 1.00 / 1.33 = 0.7519
θc = sin⁻¹(0.7519) ≈ 48.8°
✅ The critical angle of water is about 48.8°. Any incident angle greater than this value will cause total internal reflection.
| Application | Principle | Typical uses |
|---|---|---|
| Optical fibre | Light is guided along a glass fibre by repeated total internal reflection | Telecommunications, internet, medical endoscopes |
| Periscope | Two parallel mirrors reflect light several times | Submarines, tanks, observing chemical reactions |
| Diamond sparkle | High refractive index causes light to undergo multiple total internal reflections before exiting | Gemstones, jewellery |
| Prism reflectors | Light is reflected inside a prism by total internal reflection | Binoculars, cameras |
An optical fibre has a core with a high refractive index surrounded by a cladding with a lower refractive index. Light entering the fibre is repeatedly totally internally reflected at the core–cladding boundary, allowing it to travel several kilometres without significant loss of energy.
When you look at a fish from the bank, the fish appears shallower than it really is. This happens because light leaving the water bends away from the normal, and your brain assumes the ray travelled in a straight line — leading to an incorrect estimate of depth.
For an object in water viewed from air: apparent depth = real depth / 1.33 ≈ real depth × 0.75
A swimming pool is actually 2 m deep. What is the apparent depth of a coin at the bottom when viewed from the side?
Solution:
Apparent depth = real depth / n = 2 / 1.33 ≈ 1.50 m
⚠️ The pool looks 0.5 m shallower than it really is! This is why swimming pools are always deeper than they look — never dive into unfamiliar water assuming the depth.
In the desert or on a hot road, the surface sometimes looks like a pool of water. This is actually an image of the sky, formed because light from the sky is refracted by the layers of hot air (low density, small n) near the ground and bent back upwards toward your eyes.
Sunlight enters a raindrop, is refracted, then reflected from the back surface of the drop, and refracted again as it leaves. Because the refractive index depends slightly on wavelength (violet is refracted more, red is refracted less), the colours spread into the familiar seven-colour spectrum of the rainbow.
A diamond has a very high refractive index (n = 2.42), giving it a very small critical angle of only about 24.4°. Once light enters the diamond, it undergoes many total internal reflections before finally emerging, producing the brilliant "fire" that makes diamonds sparkle.
| 中文 (Chinese) | English | Symbol |
|---|---|---|
| 折射 | Refraction | — |
| 折射率 | Refractive Index | n |
| 入射角 | Angle of Incidence | θ₁ |
| 折射角 | Angle of Refraction | θ₂ |
| 法線 | Normal | — |
| 全反射 | Total Internal Reflection | TIR |
| 臨界角 | Critical Angle | θc |
| 光纖 | Optical Fibre | — |
| 視深 | Apparent Depth | — |
A ray of light strikes a glass surface from air (n = 1) at an angle of incidence of 60°. The refractive index of the glass is 1.5. Find the angle of refraction.
n₁ sin θ₁ = n₂ sin θ₂
1 × sin 60° = 1.5 × sin θ₂
sin θ₂ = 0.866 / 1.5 = 0.577
θ₂ ≈ 35.3°
Light travels from water (n = 1.33) to air. What is the critical angle θc?
sin θc = n₂ / n₁ = 1.00 / 1.33 ≈ 0.7519
θc ≈ 48.8°
A diamond has a much higher refractive index (n = 2.42) and therefore a very small critical angle (θc ≈ 24.4°). Light entering the diamond undergoes many total internal reflections before escaping, producing the brilliant "fire" effect. Glass, with n = 1.5, has a larger critical angle (≈ 42°), so light escapes more easily and the sparkle is weaker.
A pool is 3 m deep. What is the apparent depth of a coin at the bottom when viewed from outside the water?
Apparent depth = real depth / n = 3 / 1.33 ≈ 2.26 m