Fibre Optics in Medical Physics Revision Notes | A Level AQA Physics

1. Why Fibre Optics Changed Medicine

Before fibre optics, examining the inside of a patient's stomach meant surgery. Today, a flexible tube thinner than a finger can carry light around corners and send back a live image — no incision required. The physics that makes this possible is total internal reflection, and it is one of the most elegant applications of optics at A-level.

This lesson builds from the core physics principle — total internal reflection — through to the specific design features of medical optical fibres and the clinical applications that depend on them. Every part of the design connects back to the physics, so understanding the principle first makes everything else click.

2. Total Internal Reflection: The Core Principle

Picture a ray of light travelling through glass and hitting the glass-air boundary. At a shallow angle of incidence (measured from the normal), most light refracts out into the air and a small amount reflects back. As you increase the angle of incidence, the refracted ray bends further from the normal. At a specific angle — the critical angle — the refracted ray travels exactly along the boundary (90° to the normal). Beyond that angle, refraction is no longer possible: all the light reflects back into the glass. This is total internal reflection.

Two conditions must both be satisfied for total internal reflection to occur. The light must be travelling from a medium of higher refractive index into one of lower refractive index, and the angle of incidence must be greater than or equal to the critical angle.

The formula $\sin \theta_c = \frac{n_2}{n_1}$ comes directly from applying Snell's law ($n_1 \sin \theta_1 = n_2 \sin \theta_2$) at the critical condition where $\theta_2 = 90°$, so $\sin 90° = 1$ and the equation simplifies neatly. When the second medium is air, $n_2 \approx 1.00$, giving $\sin \theta_c = \frac{1}{n_1}$.

Worked example: Calculating the Critical Angle for Glass

A glass optical fibre core has refractive index $n_1 = 1.50$. The cladding has refractive index $n_2 = 1.44$. Calculate the critical angle at the core-cladding interface.

1. $\sin \theta_c = \frac{n_2}{n_1}$ — This is derived from Snell's law with the refracted angle set to 90°. [M1]
2. $\sin \theta_c = \frac{1.44}{1.50}$ — $n_1 = 1.50$ (core, denser) and $n_2 = 1.44$ (cladding, less dense). [M1]
3. $\sin \theta_c = 0.960$ — $1.44 \div 1.50 = 0.960$. Keep full precision before taking the inverse sine.
4. $\theta_c = \sin^{-1}(0.960) = 73.7°$ — Use inverse sine on your calculator. [A1] The critical angle is approximately 74°.
5. Check: $\sin(73.7°) \approx 0.960$ ✓ — Any ray hitting the core-cladding boundary at more than 73.7° from the normal will undergo total internal reflection and stay trapped in the core.

Answer: $\theta_c \approx 73.7°$

Notice that the critical angle here is relatively large — nearly 74°. That means light must hit the boundary at a very steep angle (nearly grazing) before total internal reflection kicks in. The closer $n_1$ and $n_2$ are in value, the larger the critical angle, and the harder it is to keep light trapped. This is why the refractive index difference between core and cladding is a key design parameter.

3. Structure of an Optical Fibre

An optical fibre is not a single piece of glass — it has a carefully engineered layered structure. The innermost region is the core, a thin strand of glass or plastic with a relatively high refractive index. Surrounding it is the cladding, a layer of glass or plastic with a slightly lower refractive index. This difference is what creates the boundary at which total internal reflection occurs.

When a ray of light enters the fibre at the correct angle, it strikes the core-cladding boundary at an angle greater than the critical angle. Total internal reflection bounces it back into the core, and it continues bouncing along the length of the fibre — even around gentle bends — without escaping. The light is effectively trapped and guided along the fibre.

A common question is: why bother with cladding at all? Could you not use a bare glass fibre in air? Technically, total internal reflection would still occur at a glass-air boundary, but in practice the bare surface would be scratched and contaminated, scattering light and causing signal loss. The cladding protects the optical surface and maintains a clean, consistent boundary. In medical devices, it also provides electrical insulation and mechanical flexibility.

4. Medical Applications: The Endoscope

The endoscope is the primary medical application of fibre optics at A-level. It is a flexible instrument inserted into the body — through the mouth to examine the oesophagus and stomach, or through other natural openings — to allow clinicians to visualise internal structures without surgery. Two distinct types of fibre bundle work together inside an endoscope, and understanding the difference between them is essential for exam questions.

Think of the coherent bundle like a grid of pixels on a screen. If you shuffle the pixels randomly, the image becomes unrecognisable. The coherent bundle preserves the spatial relationship between fibres so that the pattern of light entering at the distal (far) end is faithfully reproduced at the proximal (near) end where the clinician or camera is positioned.

The non-coherent bundle has no such requirement. Its only job is to flood the tissue with light so the coherent bundle can pick up the reflected image. Any arrangement of fibres delivers light equally well, which makes non-coherent bundles cheaper and easier to manufacture.

Modern endoscopes also incorporate a working channel — a hollow tube running alongside the fibre bundles — through which instruments such as biopsy forceps or laser fibres can be passed. This means a clinician can both image and treat a lesion during the same procedure, which is a significant clinical advantage over purely diagnostic imaging methods.

5. Properties of Optical Fibres Relevant to Medical Use

Several physical properties make optical fibres particularly well-suited to medical imaging. Each property connects directly to the underlying physics of total internal reflection and the practical demands of clinical use.

• Flexibility: glass fibres can be drawn to very small diameters (typically 10–100 micrometres), making them flexible enough to navigate the curves of the gastrointestinal tract without breaking the total internal reflection condition, provided bends are not too sharp.
• Low signal loss: high-quality glass fibres transmit light with minimal absorption or scattering, allowing images and illumination to be carried over the metre-scale lengths needed in endoscopy.
• Non-ionising radiation: optical fibres carry visible or near-infrared light, which does not ionise tissue. This makes endoscopy safer for repeated use compared with X-ray-based techniques.
• Small cross-section: a bundle of thousands of thin fibres can be packed into an endoscope narrow enough to pass through natural body openings, minimising patient discomfort.
• Electrical isolation: glass fibres carry no electrical current, reducing the risk of electrical injury to the patient during procedures.

The flexibility property deserves a closer look. When an optical fibre bends, the angle at which light strikes the core-cladding boundary changes. If the bend is too sharp, some rays will hit the boundary at an angle less than the critical angle and escape through the cladding — this is called light leakage or modal loss. Medical endoscopes are designed with a minimum bend radius to prevent this, and the refractive index difference between core and cladding is chosen to give a sufficiently large critical angle that gentle bends do not cause significant loss.

6. Comparing Fibre Optic Endoscopy with Other Imaging Methods

Fibre optic endoscopy sits alongside other medical imaging techniques — X-rays, ultrasound, MRI — and each has a different profile of advantages and limitations. At A-level, you may be asked to evaluate or compare these methods, so it is worth being precise about what endoscopy does and does not offer.

A key evaluative point is that endoscopy uses non-ionising radiation. This is a genuine advantage over X-ray techniques for patients who need repeated monitoring — for example, surveillance of Barrett's oesophagus. However, endoscopy is invasive in the sense that the instrument enters the body, carrying a small but real risk of perforation and infection. Examiners expect you to acknowledge both sides when the command word is 'evaluate'.

7. Exam Technique and Common Mistakes

Fibre optics questions at A-level tend to cluster around three areas: explaining total internal reflection with correct physics language, distinguishing coherent from non-coherent bundles, and evaluating endoscopy as a medical technique. Each area has its own set of traps.

For extended-response questions asking you to evaluate fibre optic endoscopy, the mark scheme typically rewards: a correct description of how the image is formed (total internal reflection in coherent bundle), at least one advantage with a physical or clinical reason, at least one limitation with a reason, and a concluding judgement. Answers that only list advantages without any critical evaluation will not reach the top mark band.

Fibre optic endoscopy is a direct application of total internal reflection — the same principle that governs the critical angle calculation you practised earlier. Every design feature of the endoscope, from the refractive index difference between core and cladding to the coherent arrangement of image fibres, exists to exploit or manage that single physical phenomenon.