Describe the passage of light through a transparent material (limited to the boundaries between two mediums only)
3.2.2 Refraction of Light
Objective
Describe how a light ray passes through a transparent material, limited to the boundary between two media.
Key Definitions
- Normal (N) – an imaginary line perpendicular to the surface at the point of incidence.
- Angle of incidence (i) – the angle between the incident ray and the normal.
- Angle of refraction (r) – the angle between the refracted ray and the normal.
- Refractive index (n) – the ratio of the speed of light in vacuum to its speed in the medium:
\$n=\dfrac{c}{v}\$
- Critical angle (c) – the angle of incidence in the denser medium for which the refracted ray in the rarer medium travels along the boundary (r = 90°).
- Internal reflection – the reflection of a ray back into the original medium when it meets a boundary.
- Total internal reflection (TIR) – internal reflection that occurs for all incidence angles greater than the critical angle.
Why Light Bends (Refraction)
When a ray passes from one transparent medium to another its speed changes. The change in speed makes the wave‑fronts pivot about the point where the ray meets the surface, producing a change in direction. Refraction occurs only at the interface; within each homogeneous medium the ray travels in a straight line.
Snell’s Law (Law of Refraction)
The quantitative relationship between the angles and the refractive indices of the two media is
\$n1\sin i = n2\sin r\$
- n1 – refractive index of the medium in which the incident ray travels.
- n2 – refractive index of the second medium.
- i – angle of incidence (measured from the normal).
- r – angle of refraction (measured from the normal).
Quick‑Recall: When the incident medium is air (n₁≈1) the law is often written as
\$n = \frac{\sin i}{\sin r}\$
where n is the refractive index of the second medium. This form is handy for laboratory determination of n.
Re‑arranged forms useful for exam questions:
- To find the angle of refraction: \$r = \sin^{-1}\!\left(\frac{n1}{n2}\sin i\right)\$
- To find the angle of incidence: \$i = \sin^{-1}\!\left(\frac{n2}{n1}\sin r\right)\$
- To find an unknown refractive index: \$n2 = n1\frac{\sin i}{\sin r}\$
Behaviour at a Single Boundary
- If n₂ > n₁ (light entering a denser medium) the ray bends towards the normal (r < i).
- If n₂ < n₁ (light entering a rarer medium) the ray bends away from the normal (r > i).
- If the incident ray is perpendicular to the surface (i = 0°) there is no bending; the ray continues straight.
Light Path Through a Transparent Slab
Critical Angle and Total Internal Reflection
When light travels from a denser medium (n₁) to a rarer medium (n₂) and the angle of incidence exceeds the critical angle, the refracted ray cannot emerge – it is reflected back into the denser medium.
From Snell’s law, setting r = 90° gives
\$\sin c = \frac{n2}{n1}\qquad (n1 > n2)\$
Numeric example (water → air)
- n₁ (water) = 1.33, n₂ (air) ≈ 1.00
- \$\sin c = \frac{1.00}{1.33}=0.752\$
- \$c = \sin^{-1}(0.752)\approx 48.8^{\circ}\$
For any incidence angle larger than 48.8°, light is totally internally reflected inside the water.
Everyday Examples of Refraction and TIR
- Optical fibres – light is guided by repeated total internal reflection.
- Diamond sparkle – high refractive index gives a small critical angle, so much light is reflected internally.
- Mirage – light from the sky is totally internally reflected in a hot, less‑dense layer of air.
- Prisms in binoculars – refraction bends the light to change direction, while TIR inside the prism keeps the beam inside.
Why Optical Fibres Work (brief explanation)
A fibre consists of a high‑index glass core surrounded by a lower‑index cladding. Light entering the core at an angle greater than the critical angle is reflected at the core–cladding boundary and travels down the fibre with very little loss.
Typical Refractive Indices
| Material | Refractive Index (n) |
|---|---|
| Air (dry, 0 °C) | 1.0003 ≈ 1.00 |
| Water (20 °C) | 1.33 |
| Glass (crown) | 1.50 – 1.52 |
| Diamond | 2.42 |
| Plastic (PMMA) | 1.49 |
Experiment: Demonstrating Refraction at Two Boundaries
- Safety note: If a laser pointer is used, never look directly into the beam and avoid pointing at anyone’s eyes.
- Set up a ray‑box (or a low‑power laser) on a flat bench.
- Place a rectangular glass slab (or clear acrylic block) on a sheet of white paper.
- Draw the normal to the first surface of the slab (use a ruler and a right‑angle triangle).
- Adjust the ray‑box so that a narrow beam strikes the first surface at a known angle of incidence i (measure with a protractor).
- Mark on the paper:
- the incident ray,
- the refracted ray inside the slab, and
- the emergent ray from the second surface.
- Measure:
- the angle of refraction r inside the slab,
- the angle of emergence i′ at the second surface.
- Record the data in a table (example below) and use Snell’s law to calculate the expected r. Compare with the measured value. Repeat for several incident angles (e.g., 20°, 40°, 60°).
| Trial | i (°) | Measured r (°) | Calculated r (°) | i′ (°) |
|---|---|---|---|---|
| 1 | 20 | – | – | – |
| 2 | 40 | – | – | – |
| 3 | 60 | – | – | – |
Observations:
- The ray is straight within each medium; bending occurs only at the two interfaces.
- When the slab is turned so that light travels from glass to air, increasing i eventually produces total internal reflection (no emergent ray).
Worked Examples
Example 1 – Find the angle of refraction
A ray in air (n₁ = 1.00) strikes a glass surface (n₂ = 1.50) at i = 30°.
\$1.00\sin30^{\circ}=1.50\sin r\$
\$\sin r=\frac{0.5}{1.50}=0.333\$
\$r=\sin^{-1}(0.333)\approx19.5^{\circ}\$
The ray bends towards the normal, reducing the angle from 30° to about 19.5°.
Example 2 – Determine an unknown refractive index
In a laboratory the measured angles are i = 45° (air) and r = 28° for an unknown transparent material. Find its refractive index.
\$n = \frac{\sin i}{\sin r}= \frac{\sin45^{\circ}}{\sin28^{\circ}}\$
\$n = \frac{0.7071}{0.4695}\approx1.51\$
The material is therefore comparable to crown glass.
Key Points for IGCSE Exams
- Define normal, angle of incidence, and angle of refraction.
- Write Snell’s law correctly, label each symbol, and be able to rearrange it for any unknown.
- Explain why the ray bends towards the normal when entering a medium of higher refractive index (speed decreases).
- Calculate the critical angle using \(\sin c = n2/n1\) and give a numeric example.
- Describe internal reflection and total internal reflection; give at least two everyday examples (optical fibre, diamond sparkle, mirage, prism in binoculars).
- State why optical fibres work – high‑index core, lower‑index cladding, TIR keeps light confined.
- Remember that refraction occurs only at the boundary; the ray travels in a straight line inside each homogeneous medium.
- When answering experimental questions, mention safety (laser eye‑hazard) and show data in a clear table.
