How Fibre-Optic Cables Trap Light Through Total Internal Reflection

A light pulse can travel through a glass fibre and emerge a long way from where it entered. Yet glass is transparent. Why doesn’t the light simply escape through the sides?

Before reading on, predict which ray is easier to keep inside a fibre: one that strikes the side wall almost head-on, or one that skims along it. The second is easier to trap. The reason comes down to how light refracts at a boundary, and exactly how we measure the angle at which it arrives.

An optical fibre uses two materials to make this happen: a central core and a surrounding layer called the cladding. The core has a higher refractive index than the cladding, allowing light to be guided by repeated total internal reflections.

01How can a boundary trap light?

Imagine a light ray travelling through the core towards the core-cladding boundary. Because the cladding has a lower refractive index, light entering it would bend away from the normal.

The normal is an imaginary line perpendicular to the surface at the point where the light arrives. The angle of incidence, \(i\), is measured between the incoming ray and this normal.

This is the first detail to get right. An angle of incidence close to \(90^\circ\) means the ray is travelling almost parallel to the surface, not approaching it head-on.

As the angle of incidence increases, the refracted ray bends further away from the normal. Eventually, it reaches a limiting angle where the refracted ray travels along the boundary itself.

Increase the angle of incidence beyond that limit, and something different happens: there is no refracted ray travelling out into the cladding. Instead, the light is reflected back into the core.

This is called total internal reflection (TIR).

There are two conditions for total internal reflection:

• Light must travel from a medium with a higher refractive index into one with a lower refractive index.
• The angle of incidence must be greater than the critical angle.

Both conditions matter. A large incidence angle alone is not enough if the light is travelling into a medium with an equal or higher refractive index.

02Calculating the critical angle

The refractive index, \(n\), describes how light propagates through a material. It is defined by

\[
n=\frac{c}{v}
\]

where \(c\) is the speed of light in a vacuum and \(v\) is the speed of light in the material. Refractive index has no units.

At a boundary, the relationship between the refractive indices and the angles of incidence and refraction is given by Snell’s law:

\[
n_1\sin\theta_1=n_2\sin\theta_2
\]

For a fibre, light travels from the higher-index core into the lower-index cladding. At the critical angle, \(\theta_c\), the refracted angle is exactly \(90^\circ\).

Substituting this into Snell’s law gives

\[
n_\text{core}\sin\theta_c
=n_\text{clad}\sin90^\circ
\]

Since \(\sin90^\circ=1\),

\[
\boxed{\theta_c=
\sin^{-1}\left(
\frac{n_\text{clad}}{n_\text{core}}
\right)}
\]

This is the critical-angle equation for a core-cladding boundary, provided \(n_\text{core}>n_\text{clad}\).

The decision rule is straightforward:

• If \(i>\theta_c\), total internal reflection occurs.
• If \(i=\theta_c\), the refracted ray travels along the boundary.
• If \(i<\theta_c\), a refracted ray can enter the cladding, so total internal reflection does not occur.

Remember that the angles are measured from the normal.

Notice what the equation tells us about the materials. If the core index stays constant and the cladding index decreases, the critical angle decreases. More incidence angles can then satisfy the condition for total internal reflection.

If the two refractive indices become equal, the critical angle reaches \(90^\circ\). There is no incidence angle greater than \(90^\circ\), so total internal reflection cannot occur.

03Why the core and cladding must have different refractive indices

The core and cladding are often both made from transparent glass. They can look almost identical, but their compositions are chosen to give them different refractive indices.

The core has the higher index, and the cladding has the lower index. That relationship creates the boundary needed for total internal reflection.

Why not simply expose the core to air? Air also has a lower refractive index than glass, so a glass-air boundary can support total internal reflection. However, a practical fibre needs a controlled boundary that protects the core from its surroundings. Cladding provides that protection and helps maintain the optical conditions needed to guide light.

The index difference is also a design choice. A larger difference between the core and cladding indices generally allows a wider range of launch angles to be guided. The exact behaviour still depends on the ray’s direction and the geometry of the fibre.

One important consequence follows: a higher core refractive index does not automatically mean better light confinement. You must consider the cladding index as well.

04The light’s entry angle matters too

So far, we have considered a ray that is already travelling inside the core. In practice, light must first enter the fibre through its end face.

The angle at which light enters affects the angle at which it later meets the side boundary.

For a simple ray travelling in a plane containing the fibre’s axis, let \(r\) be the angle of the ray inside the core relative to the fibre axis. The end face is perpendicular to that axis, so \(r\) is also the refracted angle measured from the end-face normal.

At the side boundary, the normal is perpendicular to the fibre axis. The incidence angle is therefore

\[
i=90^\circ-r
\]

This gives us a useful relationship:

• A small internal angle \(r\) produces a large side-wall incidence angle \(i\), making total internal reflection more likely.
• A large internal angle \(r\) produces a smaller side-wall incidence angle \(i\), making total internal reflection less likely.

If the light enters from air at an angle \(\theta_\text{air}\) to the fibre axis, Snell’s law at the end face gives

\[
n_\text{air}\sin\theta_\text{air}
=n_\text{core}\sin r
\]

Since \(n_\text{air}\approx1.00\), we can find the internal angle \(r\), calculate \(i=90^\circ-r\), and then compare \(i\) with the critical angle.

05Why sharp bends can cause signal loss

A straight fibre can guide a ray through repeated total internal reflections. What changes when the fibre bends?

The refractive indices do not suddenly change just because the fibre curves. Instead, the geometry of the boundary changes relative to the light’s direction.

At the outside of a sufficiently tight bend, the incidence angle for some light can decrease. A ray that previously met the condition \(i>\theta_c\) may now arrive at an angle below the critical angle.

When that happens, the boundary no longer totally internally reflects that ray. Some light can enter the cladding and escape the guided path.

A gentle bend may still guide most of the light. A tight bend is more likely to cause significant loss because it can disrupt confinement for a greater amount of the light travelling through the fibre.

In real fibres, light is also described using wave optics rather than only individual rays. Bending can change how the guided optical field behaves, allowing energy to couple into light that is no longer confined to the core. The ray model gives a useful explanation of the mechanism, but it does not describe every detail of bend loss.

06How imperfections cause further losses

Even a straight fibre with a suitable core-cladding index difference is not perfectly lossless in practice.

Three effects are particularly important.

Scattering at imperfections: A rough or damaged core-cladding boundary can scatter light into different directions. Some of that scattered light may no longer satisfy the conditions needed for confinement.

Local changes in refractive index: Defects, cracks, or variations in the materials can disrupt the intended optical boundary. The light may encounter different local conditions from those predicted for a uniform fibre.

Absorption: Some optical energy is absorbed by the material and converted into other forms of energy, such as heat. This reduces the light intensity even if the remaining light continues to be guided.

These mechanisms should not be confused with ideal total internal reflection. In the ideal ray model, total internal reflection itself does not gradually consume the light at every bounce. Losses arise because real materials absorb or scatter some light, and real fibre geometries can allow light to escape.

Over a long distance, even small losses can accumulate and reduce the intensity of the signal that reaches the other end.

07A reliable method for HSC questions

When a question asks whether light will remain guided in an optical fibre, work through these checks in order.

• Check the material indices. Total internal reflection at the core-cladding boundary requires \(n_\text{core}>n_\text{clad}\).
• Calculate the critical angle. Use \(\theta_c=\sin^{-1}(n_\text{clad}/n_\text{core})\).
• Find the incidence angle. If the ray enters at an angle, use Snell’s law at the end face first. For a meridional ray in a straight fibre, use \(i=90^\circ-r\).
• Compare the angles. Total internal reflection occurs only when \(i>\theta_c\).
• Check for bends or defects. A ray that satisfies the condition in a straight, uniform fibre may not remain confined after the geometry or local material conditions change.

The most tempting mistake is to assume that any ray inside a transparent fibre will stay there. It won’t. The material combination makes total internal reflection possible, but the angle determines whether it actually occurs.

08Questions and solutions

Question 1

An ideal optical fibre has a core refractive index of \(1.49\) and a cladding refractive index of \(1.45\). A ray strikes the core-cladding boundary at \(80.0^\circ\) to the normal.

Calculate the critical angle and determine whether total internal reflection occurs.

Solution 1

Yes, total internal reflection occurs. The critical angle is approximately \(76.7^\circ\), which is smaller than the ray’s incidence angle of \(80.0^\circ\).

The critical angle is

\[
\begin{aligned}
\theta_c
&=\sin^{-1}\left(\frac{n_\text{clad}}{n_\text{core}}\right)\\
&=\sin^{-1}\left(\frac{1.45}{1.49}\right)\\
&=76.7^\circ
\end{aligned}
\]

Compare the angles:

\[
80.0^\circ>76.7^\circ
\]

The ray therefore undergoes total internal reflection and remains in the core under the ideal ray model.

The trap is to think that an angle of \(80^\circ\) is too large for reflection. In fact, because the angle is measured from the normal, a large incidence angle means the ray is approaching the boundary almost parallel to the surface.

Question 2

Two optical fibres have the same core refractive index, \(n_\text{core}=1.50\), but different cladding indices.

• Fibre X has \(n_\text{clad}=1.46\).
• Fibre Y has \(n_\text{clad}=1.42\).

A ray inside each fibre strikes the side boundary at an incidence angle of \(74^\circ\).

Calculate the critical angle for each fibre. Determine which fibre totally internally reflects the ray, and explain why the material choice changes the result.

Solution 2

Fibre Y totally internally reflects the ray, but fibre X does not. Fibre Y has a lower critical angle because its cladding has a lower refractive index relative to its core.

For fibre X,

\[
\begin{aligned}
\theta_{c,X}
&=\sin^{-1}\left(\frac{1.46}{1.50}\right)\\
&=76.7^\circ
\end{aligned}
\]

The incidence angle is \(74^\circ\), so

\[
74^\circ<76.7^\circ
\]

Total internal reflection does not occur in fibre X.

For fibre Y,

\[
\begin{aligned}
\theta_{c,Y}
&=\sin^{-1}\left(\frac{1.42}{1.50}\right)\\
&=71.2^\circ
\end{aligned}
\]

Now,

\[
74^\circ>71.2^\circ
\]

Total internal reflection occurs in fibre Y.

The lower cladding index reduces the critical angle, making the TIR condition easier to satisfy for this particular incidence angle. Both fibres have the same core index, so comparing the core indices alone would miss the reason for the different outcomes.

Question 3

A manufacturer produces two batches of optical fibre. Both have a core refractive index of \(1.48\).

• In batch A, the cladding refractive index is \(1.47\).
• In batch B, a manufacturing change makes the cladding refractive index \(1.48\).

A ray strikes the core-cladding boundary at \(85^\circ\) to the normal in each batch.

Determine whether total internal reflection occurs in each batch. Explain why a very large incidence angle is not enough by itself to guarantee confinement.

Solution 3

Total internal reflection occurs in batch A but not in batch B. In batch A, the core index is greater than the cladding index. In batch B, the indices are equal, so the core-cladding boundary no longer provides the refractive-index contrast needed for total internal reflection.

For batch A,

\[
\begin{aligned}
\theta_{c,A}
&=\sin^{-1}\left(\frac{1.47}{1.48}\right)\\
&=83.3^\circ
\end{aligned}
\]

Since

\[
85^\circ>83.3^\circ
\]

the ray undergoes total internal reflection.

For batch B,

\[
\begin{aligned}
\theta_{c,B}
&=\sin^{-1}\left(\frac{1.48}{1.48}\right)\\
&=\sin^{-1}(1)\\
&=90^\circ
\end{aligned}
\]

The ray’s incidence angle of \(85^\circ\) is less than \(90^\circ\). Therefore, total internal reflection does not occur.

When the indices are equal, an ideal boundary between the two materials produces no refractive bending or reflection due to an index difference. Light can pass into the cladding without being confined to the core.

This is an important boundary case. A large incidence angle helps only if the material combination allows a critical angle that the ray can exceed.

Question 4

An ideal fibre has a critical angle of \(77^\circ\).

In a straight section, a ray strikes the core-cladding boundary at \(84^\circ\). After the fibre passes through a tight bend, the incidence angle at the outer side of the bend decreases to \(74^\circ\).

A short section of the fibre also has a rough core-cladding interface.

Explain:

• whether total internal reflection occurs in the straight and bent sections;
• how the rough interface can cause signal loss, even if the average incidence angle remains above the critical angle elsewhere;
• whether ideal total internal reflection causes the light to lose energy at every bounce.

Solution 4

The ray undergoes total internal reflection in the straight section but can escape the guided path at the tight bend. The rough interface can cause additional scattering, while ideal total internal reflection does not progressively drain the light’s energy.

In the straight section, the incidence angle is \(84^\circ\). Since

\[
84^\circ>77^\circ
\]

total internal reflection occurs.

At the outer side of the bend, the incidence angle decreases to \(74^\circ\). Now,

\[
74^\circ<77^\circ
\]

The TIR condition is no longer satisfied. A refracted component can enter the cladding, allowing light to leave the guided path.

The rough interface introduces a separate problem. Its irregularities change the local surface orientation and can scatter light into directions that are not confined by total internal reflection. The local behaviour may differ from what would be predicted using a perfectly smooth boundary and a single average incidence angle.

Finally, an ideal total internal reflection does not consume optical energy at each bounce. In a real fibre, absorption, scattering, and leakage caused by bends or defects reduce the transmitted signal. These losses accumulate as the light travels along the fibre.

Question 5

Two ideal optical fibres have the following refractive indices.

A ray enters each fibre from air at \(20^\circ\) to the fibre axis. Treat the ray as a meridional ray, so its incidence angle at the side boundary in a straight section is \(i=90^\circ-r\), where \(r\) is its angle inside the core relative to the fibre axis.

At the outer side of a tight bend, the incidence angle decreases by exactly \(4.0^\circ\) for each ray.

Assume \(n_\text{air}=1.00\), and use the ideal ray model. If the TIR condition fails, treat the ray as no longer guided by the core.

(a) Calculate the critical angle for each fibre.

(b) Determine whether the ray is guided in each straight section, and then determine which fibre still guides the ray at the bend.

(c) The same fibres are now straight, and the launch angle is reduced to \(10^\circ\). Can you conclude that fibre A has lower real signal loss simply because it has a higher core refractive index? Justify your answer using the optical model.

Solution 5

Fibre B guides the \(20^\circ\) launch through the specified bend, whereas fibre A does not guide that ray even in the straight section. However, at a \(10^\circ\) launch angle in straight fibres, both rays satisfy the TIR condition. The higher core index of fibre A is not enough to conclude that it has lower real signal loss.

The key is to calculate the critical angle and the internal ray angle for each fibre, rather than comparing the core indices alone.

Part (a): Calculate the critical angles.

For fibre A,

\[
\begin{aligned}
\theta_{c,A}
&=\sin^{-1}\left(\frac{1.47}{1.50}\right)\\
&=78.5^\circ
\end{aligned}
\]

For fibre B,

\[
\begin{aligned}
\theta_{c,B}
&=\sin^{-1}\left(\frac{1.40}{1.48}\right)\\
&=71.1^\circ
\end{aligned}
\]

Fibre B has the lower critical angle. Its greater relative index contrast makes total internal reflection possible over a wider range of side-wall incidence angles.

Part (b): Check the \(20^\circ\) launch and the bend.

First, use Snell’s law at the end face to calculate the internal angle for fibre A.

\[
\begin{aligned}
1.00\sin20^\circ&=1.50\sin r_A\\
r_A&=\sin^{-1}\left(\frac{\sin20^\circ}{1.50}\right)\\
r_A&=13.2^\circ
\end{aligned}
\]

The side-wall incidence angle is

\[
\begin{aligned}
i_A&=90^\circ-13.2^\circ\\
&=76.8^\circ
\end{aligned}
\]

Since

\[
76.8^\circ<78.5^\circ
\]

fibre A does not totally internally reflect the ray in the straight section. At the bend, the incidence angle falls further:

\[
i_{A,\text{bend}}=76.8^\circ-4.0^\circ=72.8^\circ
\]

The ray remains outside the TIR condition.

Now calculate the internal angle for fibre B.

\[
\begin{aligned}
1.00\sin20^\circ&=1.48\sin r_B\\
r_B&=\sin^{-1}\left(\frac{\sin20^\circ}{1.48}\right)\\
r_B&=13.4^\circ
\end{aligned}
\]

The side-wall incidence angle is

\[
\begin{aligned}
i_B&=90^\circ-13.4^\circ\\
&=76.6^\circ
\end{aligned}
\]

Since

\[
76.6^\circ>71.1^\circ
\]

fibre B totally internally reflects the ray in the straight section.

At the bend,

\[
i_{B,\text{bend}}=76.6^\circ-4.0^\circ=72.6^\circ
\]

Now compare the bent incidence angle with the critical angle:

\[
72.6^\circ>71.1^\circ
\]

Fibre B still satisfies the TIR condition at the specified bend. It therefore guides the ray through the bend under the assumptions given.

The tempting prediction is that fibre A must guide light better because its core has a higher refractive index. But the critical angle depends on the ratio of the cladding index to the core index, and the launch angle determines the incidence angle at the side boundary. Fibre B’s larger relative index contrast makes it more tolerant of the specified change in incidence angle.

Part (c): Check the \(10^\circ\) launch in straight fibres.

For fibre A,

\[
\begin{aligned}
r_A&=\sin^{-1}\left(\frac{\sin10^\circ}{1.50}\right)\\
&=6.6^\circ\\
i_A&=90^\circ-6.6^\circ\\
&=83.4^\circ
\end{aligned}
\]

Since \(83.4^\circ>78.5^\circ\), total internal reflection occurs.

For fibre B,

\[
\begin{aligned}
r_B&=\sin^{-1}\left(\frac{\sin10^\circ}{1.48}\right)\\
&=6.7^\circ\\
i_B&=90^\circ-6.7^\circ\\
&=83.3^\circ
\end{aligned}
\]

Since \(83.3^\circ>71.1^\circ\), total internal reflection also occurs.

Both rays are guided in the ideal straight-fibre model. You cannot conclude that fibre A has lower real signal loss simply because its core index is higher. The calculation establishes whether the rays satisfy the confinement condition; it does not establish the total attenuation of either real fibre.

Actual signal loss also depends on absorption, scattering, material quality, launch coupling, and fibre geometry. The relative refractive indices are essential, but they are not the only factors controlling performance.

Once you can determine whether light remains guided, the next useful concept is attenuation: how absorption, scattering, and bend losses reduce the signal intensity as light travels through a fibre, even when much of the light remains confined to the core.