Total Internal Reflection and Fibre Optics Revision Notes | A Level AQA Physics

1. From Refraction to a Boundary That Reflects Everything

You already know that when light crosses a boundary between two media, it bends — that is refraction. But there is a scenario where the boundary stops transmitting light altogether and reflects every single ray back. That phenomenon, total internal reflection, is the engine behind fibre-optic broadband, endoscopes, and the sparkle of a diamond.

The key condition is direction of travel: light must be moving from a denser medium (higher refractive index $n_1$) into a less dense medium (lower refractive index $n_2$). When light travels the other way — from less dense to more dense — it always refracts inward and total internal reflection cannot occur.

Picture a ray of light inside a glass block, hitting the flat top surface. At a small angle of incidence, most light refracts out and a weak reflected ray stays inside. As you increase the angle, the refracted ray bends further from the normal. At one precise angle — the critical angle $\theta_c$ — the refracted ray runs exactly along the boundary (angle of refraction = 90°). Beyond that angle, there is no refracted ray at all: every photon bounces back inside the glass.

2. The Critical Angle and Its Formula

To derive the formula, apply Snell's law at the critical angle. At $\theta_i = \theta_c$, the refracted angle is 90°, so $\sin 90° = 1$. Snell's law gives $n_1 \sin\theta_c = n_2 \sin 90°$, which simplifies directly to the AQA specification formula:

A common special case is when the second medium is air, where $n_2 \approx 1.00$. The formula then becomes $\sin\theta_c = \frac{1}{n_1}$, which is worth memorising as a shortcut for glass-air or water-air boundaries.

Worked example: Calculating the Critical Angle for Glass

A ray of light travels from glass (refractive index $n_1 = 1.50$) into air ($n_2 = 1.00$). Calculate the critical angle. [3 marks]

1. $\sin\theta_c = \frac{n_2}{n_1}$ — [M1] Correct formula stated — this earns the method mark.
2. $\sin\theta_c = \frac{1.00}{1.50}$ — [M1] Both values substituted correctly with $n_1$ in the denominator (the denser medium).
3. $\sin\theta_c = 0.6\overline{6}$ — $1.00 \div 1.50 = 0.667$ (to 3 s.f.). Keep full precision before the inverse sine step.
4. $\theta_c = \sin^{-1}(0.667) = 41.8°$ — [A1] Final answer to 3 significant figures. Any angle of incidence greater than 41.8° inside this glass will produce total internal reflection.

Answer: $\theta_c = 41.8°$

Notice that a higher refractive index $n_1$ produces a smaller critical angle — meaning total internal reflection kicks in at shallower angles. Diamond has $n \approx 2.42$, giving $\theta_c \approx 24°$. Its facets are cut so that almost every internal ray exceeds this tiny critical angle, producing the characteristic brilliance.

3. Step-Index Optical Fibres — Structure and Function

Now that you have the physics of total internal reflection, you can see exactly why optical fibres work. A step-index optical fibre consists of two concentric cylinders: a central core made of glass or plastic with a relatively high refractive index, surrounded by a cladding layer made of a material with a lower refractive index.

The name 'step-index' refers to the refractive index profile: it does not change gradually — it steps sharply from $n_{\text{core}}$ to $n_{\text{cladding}}$ at the core-cladding boundary. This sharp boundary is what makes the total internal reflection geometry clean and predictable.

Light enters the fibre at one end and strikes the core-cladding boundary at a glancing angle — typically well above the critical angle. Because $n_{\text{core}} > n_{\text{cladding}}$, total internal reflection occurs and the ray bounces along the fibre, zig-zagging from wall to wall, until it exits at the far end. The light is effectively trapped inside the core for the entire length of the fibre.

A critical point that examiners test: without cladding, two fibres touching each other would allow light to leak from one core into the other, because there would be no controlled boundary. The cladding maintains optical isolation between fibres in a bundle.

4. Why Optical Fibres Matter — Applications and Advantages

Optical fibres carry information as pulses of light rather than as electrical signals in copper wire. Because light travels at approximately $3.0 \times 10^8 \text{ m s}^{-1}$ in a vacuum (and only slightly slower in glass), data can be transmitted over long distances at extremely high speeds with very low signal loss compared with electrical cables.

In medicine, bundles of fibres are used in endoscopes. One bundle carries light into the body; a second bundle carries the reflected image back to the observer. Each fibre in the image bundle carries one pixel of the image, so the spatial arrangement of fibres must be preserved along the entire length — this is called a coherent bundle.

• Telecommunications: high-bandwidth data transmission with low attenuation over long distances
• Medical endoscopy: visualising internal organs without surgery
• Decorative and safety lighting: directing light to precise locations
• Sensing applications: measuring temperature, pressure, or strain by monitoring changes in transmitted light

5. Worked Exam Problem — Finding the Critical Angle for a Fibre

Exam questions on optical fibres often give you the refractive indices of both the core and the cladding and ask you to find the critical angle at the core-cladding boundary. The method is identical to the glass-air calculation — the only difference is that $n_2$ is no longer 1.00. Here is a full worked example.

Worked example: Critical Angle at a Core-Cladding Boundary

An optical fibre has a core with refractive index $n_1 = 1.55$ and cladding with refractive index $n_2 = 1.45$. Calculate the critical angle at the core-cladding boundary. [3 marks]

1. $\sin\theta_c = \frac{n_2}{n_1}$ — [M1] State the formula. Note that $n_1 = 1.55$ (core, denser) goes in the denominator and $n_2 = 1.45$ (cladding, less dense) goes in the numerator.
2. $\sin\theta_c = \frac{1.45}{1.55}$ — [M1] Correct substitution. A common error is inverting the fraction — always put the medium the light is travelling INTO on top.
3. $\sin\theta_c = 0.9355...$ — $1.45 \div 1.55 = 0.9355$ (to 4 s.f.). Carry full precision into the next step.
4. $\theta_c = \sin^{-1}(0.9355) = 69.4°$ — [A1] Answer to 3 significant figures. Any ray hitting the core-cladding boundary at an angle greater than 69.4° from the normal will undergo total internal reflection and remain in the core.

Answer: $\theta_c = 69.4°$

Compare this with the glass-air result of 41.8° from the earlier example. The much smaller difference in refractive indices here (1.55 vs 1.45) produces a much larger critical angle (69.4°). This means rays must be nearly parallel to the fibre axis to undergo total internal reflection — a design consideration when coupling light into the fibre.

6. Exam Technique — Getting Every Mark

Questions on this topic follow predictable patterns. Calculation questions almost always award one mark for the formula, one for correct substitution, and one for the final answer with appropriate significant figures. If you write the formula and substitute correctly but make an arithmetic slip, you can still earn two out of three marks — so always show your working.

Describe and explain questions about optical fibres require precise language. Saying 'light bounces along the fibre' scores nothing. Saying 'light undergoes total internal reflection at the core-cladding boundary because the angle of incidence exceeds the critical angle' scores marks. The phrase 'total internal reflection' must appear explicitly.

One final point worth consolidating: the step-index fibre is the only fibre type you need for AQA A Level. The 'step' in the name refers to the abrupt change in refractive index at the core-cladding boundary — as opposed to a graded-index fibre where the index changes gradually, which is beyond the specification.