Light bends when it changes speed across a boundary, and that bending can be quantified. Snell's law links the angles and refractive indices, while total internal reflection explains how optical fibres and many precision devices keep light trapped inside a medium.
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How can optical fibres carry light over long distances without the light just leaking out through the sides of the cable?
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Wrong: Power is the same as energy.
Right: Power is the rate of energy transfer (energy per unit time); more power means faster energy use, not more total energy.
📚 Core Content
Refraction of light is caused by a change in speed at a boundary between media of different optical density — and this bending is one of the most direct pieces of evidence that light behaves as a wave.
When light enters a medium where it travels more slowly, it bends toward the normal. When it enters a medium where it travels faster, it bends away from the normal. The source frequency does not change at the boundary, so the wavelength changes with the speed. This wavelength change is consistent with the wave equation $v = f\lambda$: if $f$ is constant and $v$ decreases, then $\lambda$ must decrease proportionally.
The physical reason for bending can be understood by considering wavefronts. When one side of a wavefront enters the slower medium first, it slows down while the other side is still traveling faster. This causes the wavefront to pivot, changing the direction of propagation. A particle model of light cannot easily explain this directional change at a boundary without changing speed in a continuous medium.
Refractive index tells us how much a medium slows light compared with vacuum — and because it is a ratio of speeds, it is dimensionless.
The relationship $n = c/v$ means a higher refractive index corresponds to a lower light speed in that medium. This gives a compact way to compare optical density and predict bending behavior. For example, diamond has $n \approx 2.42$, which means light travels at only about $1.24 \times 10^8$ m/s inside diamond — roughly 41% of its speed in vacuum. Water has $n \approx 1.33$, so light slows to about $2.25 \times 10^8$ m/s.
The refractive index is not just a property of the material — it also depends slightly on wavelength. This is why a prism separates white light into colors: different wavelengths experience slightly different refractive indices and therefore bend by different amounts. This phenomenon, called dispersion, is responsible for rainbows and the colorful sparkle of cut diamonds.
| Medium | Refractive index (approx.) | Speed of light in medium |
|---|---|---|
| Vacuum | 1.00 | $3.00 \times 10^8$ m/s |
| Air | 1.00 | $3.00 \times 10^8$ m/s |
| Water | 1.33 | $2.25 \times 10^8$ m/s |
| Glass | 1.50 | $2.00 \times 10^8$ m/s |
| Diamond | 2.42 | $1.24 \times 10^8$ m/s |
Snell's law quantifies the bending of light at a boundary — and it does so with a simple, powerful relationship that connects the optical properties of two media.
Using $n_1 \sin \theta_1 = n_2 \sin \theta_2$, we can relate the optical properties of the two media to the incident and refracted angles. The equation works only if the angles are measured from the normal. This is the most common source of error in exam questions: students sometimes measure angles from the surface and then wonder why their answer is wrong.
The physical content of Snell's law can be understood from wavefront geometry. The ratio $\sin \theta_1 / \sin \theta_2$ equals the ratio of wave speeds $v_1/v_2$, which is also $n_2/n_1$. This means the law is not an arbitrary rule but a direct consequence of how waves change direction when their speed changes. Snell's law applies to all types of waves, not just light, which again shows the generality of wave physics.
Total internal reflection happens only when light travels from a denser medium to a less dense medium and the angle of incidence exceeds the critical angle — but both conditions must be satisfied simultaneously.
At the critical angle, the refracted ray travels along the boundary. For larger angles, no refracted ray emerges and the light is reflected entirely back into the denser medium. This is why optical fibres can trap light so effectively. The critical angle depends only on the refractive indices of the two media: $\sin i_c = n_2/n_1$ where $n_1 > n_2$. For glass to air, this simplifies to $\sin i_c = 1/n_{\text{glass}}$.
Total internal reflection is fundamentally different from ordinary reflection from a mirror. In total internal reflection, 100% of the light energy is reflected — there is no absorption by a metallic coating. This makes it extraordinarily efficient for transporting light over long distances, which is why modern telecommunications rely on optical fibres rather than metal wires for high-speed data transmission.
Total internal reflection is a practical engineering tool, not just a textbook effect — it underpins much of modern communication and medical technology.
Optical fibres use repeated total internal reflection to transmit data. The fibre consists of a core (high refractive index) surrounded by cladding (lower refractive index). Light entering the core at a shallow angle repeatedly hits the core-cladding boundary at an angle greater than the critical angle, so it is totally internally reflected and guided along the fibre even around bends. The same principle is important in endoscopes, diamonds, and prism systems in binoculars.
In endoscopes, bundles of optical fibres carry light into the body and carry images back out, allowing surgeons to perform minimally invasive procedures. In diamonds, the high refractive index means a very small critical angle, so light entering the gem tends to undergo multiple total internal reflections before exiting, creating the brilliant sparkle that makes diamonds so valuable.
A semi-circular glass block and laser or ray box can be used to test Snell's law and measure the critical angle — but careful technique is needed for reliable results.
The semi-circular shape helps arrange entry conditions so refraction is isolated cleanly at the flat face. Light enters along the radius of the curved face, so it hits the curved surface at normal incidence and does not bend there. All the bending therefore occurs at the flat face, making the geometry simple and the measurements clean. By measuring angles carefully from the normal, students can compare observed results to theoretical predictions.
To find the critical angle, rotate the block so the light hits the flat face from inside the glass at increasing angles. At some angle, the refracted ray will skim along the boundary — this is the critical angle. Beyond this angle, the light is totally internally reflected. Comparing the measured critical angle to the theoretical prediction ($\sin i_c = 1/n$) provides a direct test of the theory.
✏️ Worked Examples
Scenario: Light travels from air into glass with refractive index 1.50. The angle of incidence is 30°. Find the angle of refraction.
If the second medium had a lower refractive index than air (e.g. a hypothetical medium with $n = 0.80$), the light would bend away from the normal instead. The refracted angle would be larger than 30°, and the speed of light in that medium would exceed $c$ — which is why no known material has $n < 1$ for visible light in ordinary conditions.
Scenario: The refractive index of a glass block is 1.50. Find the critical angle for light travelling from glass into air. Then determine whether total internal reflection will occur for an angle of incidence of 50° inside the glass.
If the light were travelling from air into glass, total internal reflection could not occur at that boundary because the light would be entering the denser medium. Total internal reflection only happens when light attempts to leave the denser medium at a sufficiently large angle.
Visual Break
🏃 Activities
Write your prediction, then give the physical reason using refractive index values. Assume $n_{\text{air}} = 1.00$, $n_{\text{water}} = 1.33$, $n_{\text{glass}} = 1.50$.
Type your predictions and reasons below.
Write your predictions and reasons in your book.
Your answer should refer to both conditions for total internal reflection and explain which condition fails. Use the concept of refractive index in your explanation.
Include in your answer: the structure of the fibre (core and cladding), why the core must have a higher refractive index than the cladding, and what happens to light that hits the boundary at an angle greater than the critical angle.
Type your explanation below.
Write your explanation in your book.
Calculate the critical angle for light leaving each material into air. Then explain which material would make the best optical fibre core and which would sparkle the most when cut into a gemstone.
| Material | Refractive index | Critical angle |
|---|---|---|
| Water | 1.33 | |
| Crown glass | 1.52 | |
| Diamond | 2.42 |
Type your calculations and comparison below.
Write your calculations and comparison in your book.
Earlier you were asked how optical fibres keep light from leaking out through the sides.
The full answer: the fibre is designed so that light repeatedly hits the boundary from the denser medium side at angles greater than the critical angle. That produces total internal reflection, which keeps the light trapped and guided along the fibre. The core has a higher refractive index than the cladding, ensuring that the two conditions for total internal reflection are met at every bounce.
Now revisit your prediction. What two conditions make the fibre work?
Annotate your prediction in your book with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
✅ Check Your Understanding
1. The refractive index of a medium is given by:
2. When light enters a denser medium, it usually bends:
3. Snell's law is:
4. Total internal reflection occurs when:
5. For a medium with refractive index 2.0, the critical angle into air satisfies:
6. Optical fibres work mainly because of:
7. Explain what refractive index tells us about a medium. 3 MARKS
8. Light enters glass ($n = 1.50$) from air at an incidence angle of 30°. Find the refraction angle. 3 MARKS
9. Define total internal reflection and state the two conditions required for it to occur. 4 MARKS
Water: $\sin i_c = 1/1.33 = 0.752$ → $i_c = 48.8°$
Crown glass: $\sin i_c = 1/1.52 = 0.658$ → $i_c = 41.1°$
Diamond: $\sin i_c = 1/2.42 = 0.413$ → $i_c = 24.4°$
Best optical fibre core: Crown glass is a good choice because its moderate refractive index and critical angle allow reliable total internal reflection while being manufacturable into thin, flexible fibres.
Best gemstone: Diamond sparkles the most because its very high refractive index gives the smallest critical angle. Light entering a cut diamond is likely to undergo many total internal reflections before exiting, creating the characteristic brilliance.
1. B — refractive index is $c/v$.
2. C — entering a denser medium bends light towards the normal.
3. A — Snell's law is $n_1 \sin \theta_1 = n_2 \sin \theta_2$.
4. D — TIR needs denser-to-less-dense travel and angle above critical.
5. B — $\sin i_c = 1/2.0 = 0.5$.
6. C — optical fibres rely on repeated total internal reflection.
Q7 (3 marks): Refractive index tells us how much light is slowed in a medium compared with its speed in vacuum. A higher refractive index means a lower light speed in that medium. It is therefore a measure of optical density.
Q8 (3 marks): Use $n_1 \sin \theta_1 = n_2 \sin \theta_2$. So $1.00 \times \sin 30^\circ = 1.50 \sin \theta_2$. This gives $0.5 = 1.50 \sin \theta_2$, so $\sin \theta_2 = 0.333...$ and $\theta_2 \approx 19.5^\circ$.
Q9 (4 marks): Total internal reflection is the complete reflection of light back into a medium with no refracted ray emerging. It occurs only when light travels from a medium of higher refractive index to one of lower refractive index, and when the angle of incidence is greater than the critical angle.
Tick when you have finished the activities and checked the answers.