Total Internal Reflection - Physics for Pilot Assessments

There is a limit to how far refraction can bend a ray on its way out of a dense material, and once you reach it something abrupt happens: the light stops leaving altogether and reflects back inside instead. This lesson establishes when that switch happens, why the condition is one-directional, and what it is used for. It follows directly from Lesson 3 — the refractive index and the fact that light leaving a denser medium bends away from the normal — and needs nothing else.

Push the angle and watch the refracted ray flatten

Start with a ray inside a block of glass, heading up towards the surface almost along the normal. Most of it passes out into the air, bending away from the normal by a little, and a small fraction reflects back inside. Now tilt the ray so it strikes the surface more obliquely. The angle inside grows, and because the light is leaving a denser medium for a less dense one, the angle outside grows faster: for glass of index 1.5, an inside angle of 30° gives about 49° outside, and 40° inside gives about 75° outside. The refracted ray is being pressed down towards the surface. At the same time the reflected fraction gets steadily stronger, which you can see for yourself in a window at a shallow angle.

There is only so far that can go, because an angle measured from the normal cannot exceed 90° — at 90° the refracted ray would be running along the surface itself rather than leaving it. The angle of incidence inside the glass that puts the refracted ray exactly at 90° is the end of the road, and it has a name: the critical angle. Below it, some light leaves and some reflects. Exactly at it, the emerging ray grazes along the boundary. Above it, there is no angle outside that Snell's law could satisfy, so no light leaves at all and every bit of it reflects back into the glass. That is total internal reflection.

Three rays leave the same point inside the glass at increasing angles of incidence. The first is below the critical angle and refracts out, bending away from the normal. The second is exactly at the critical angle, so the refracted ray grazes along the boundary at 90°. The third is beyond it: nothing emerges, and the whole of the light reflects back inside, obeying the ordinary law of reflection.

The critical angle, as a number

Because the critical angle is just the special case of Snell's law where the angle outside is 90°, its value follows from the refractive index alone. Put the light in the dense material with index n and the air outside with index 1, and set the outside angle to 90°: $n \sin C = 1 \times \sin 90^\circ$ The sine of 90° is exactly 1, so n sin C = 1, and dividing both sides by n leaves the sine of the critical angle equal to one over the index.

Critical angle at a boundary with air

\sin C = \frac{1}{n}

$C$: the critical angle, measured from the normal, in degrees (°)
$n$: refractive index of the denser material — a bare ratio, no units

The shape of that relationship is more useful than any single value it produces. Because n is on the bottom, a higher refractive index gives a smaller sine and therefore a smaller critical angle — so light is trapped inside a high-index material much more easily. Diamond's index of 2.4 gives a critical angle of about 24°, which means that light entering a cut stone bounces around inside it many times before finding a face steep enough to escape from, and that is exactly what a jeweller's cut is engineered to exploit. Glass at 1.5 gives about 42°, and water at 1.33 about 49°.

Example 2 — from critical angle back to index

A transparent material is found to have a critical angle of 30° in air. What is its refractive index? This is the arrangement items favour, because it needs the relationship turned round.

Start from $\sin C = \frac{1}{n}$ and multiply both sides by n: $n \sin C = 1$
Now divide both sides by sin C, which leaves n alone: $n = \frac{1}{\sin C}$
The sine of 30° is exactly 0.5, so $n = \frac{1}{0.5} = 2$ .
Check it is possible: 2 is greater than 1, as any refractive index must be, and it sits between glass at 1.5 and diamond at 2.4 — a plausibly dense material.
Check the sense too: a small critical angle should mean a high index, and 30° is small while 2 is high. The two move in opposite directions, as the formula says they must.

It only works one way round

Total internal reflection can happen only when light is travelling from the denser medium into the less dense one — glass to air, water to air, the core of a fibre to its cladding. Going the other way it is impossible, and the reason is in the formula rather than in any extra rule. Snell's law says the sine of the angle in the denser medium is the smaller of the two, so coming from air into glass the angle inside is always smaller than the angle outside, and since the outside angle can never exceed 90°, the inside angle simply runs out of room before anything dramatic can happen — at 90° incidence in air, the angle inside glass of index 1.5 is still only about 42°. There is no angle of incidence in the less dense medium that produces a refracted angle of 90° in the denser one, so there is no critical angle in that direction at all.

Note also the word internal in the name: the reflection happens on the inside face of the denser material, with the light staying in the material it was already in. And note total, which is the genuinely unusual part. An ordinary mirror absorbs a few per cent of the light at every bounce, because reflection off a metal coating is never perfect. Total internal reflection loses nothing at the boundary itself — the entire beam turns round — which makes it a better reflector than any mirror, and is the reason it is used wherever light has to survive a great many reflections.

What it is used for

An optical fibre is a thread of very pure glass with a core of slightly higher refractive index surrounded by cladding of slightly lower index. Light injected into the core at a shallow enough angle strikes the core–cladding boundary above the critical angle, reflects totally, crosses to the other side, and reflects again, and so on for kilometres. Because no light is lost at the reflections, the only losses are absorption and scattering in the glass itself, which is why a fibre can carry a signal a very long way. Bundles of fibres do the same job for images rather than signals: a flexible endoscope carries light down one bundle to illuminate what the surgeon is looking at and an image back up another.

Light travelling along the core of an optical fibre. At the entry face it refracts toward the normal as it crosses into the denser glass; from then on it meets the core–cladding boundary only at grazing angles far beyond the critical angle, so every bounce is total internal reflection and nothing crosses into the cladding. The boundaries lose nothing — over kilometres of fibre, the only losses are absorption and scattering inside the glass itself.

Reflecting prisms are the other everyday use. A right-angled glass prism with 45° faces takes a beam entering square through one face, meets it at 45° on the sloping face — comfortably above glass's critical angle of about 42° — and turns it through exactly 90° with no coating and no losses. Two such prisms turn a beam through 180°, which is how binoculars fold a long optical path into a short body and put the image the right way up, and how a bicycle reflector throws light straight back at the car behind it. The reason these use prisms rather than mirrors is durability and efficiency: there is no silvering to tarnish and nothing to absorb.

Example 3 — deciding whether the light gets out

A ray of light inside a transparent material strikes the surface with air at 45° to the normal. Does it escape (a) if the material is glass, refractive index 1.5, and (b) if it is water, refractive index 1.33?

Every question of this shape is the same single comparison: is the angle of incidence below or above that material's critical angle?
In glass the critical angle is about 42°. The ray arrives at 45°, which is beyond it, so no light leaves at all: the surface acts as a perfect mirror and the whole beam reflects back into the glass at 45° on the other side of the normal. This is exactly the geometry a 45° reflecting prism exploits.
In water the critical angle is about 49°. The same 45° is below it, so the ray does escape, bending away from the normal as it leaves — with a weaker partial reflection returning into the water alongside it.
Same angle, opposite outcomes. The comparison is always against that material's own critical angle, never against a fixed number — which is why quoting the angle alone can never settle whether light is trapped.

Why it matters in aviation

Light held inside a denser material by repeated internal reflection is how an optical fibre carries data, and modern aircraft use fibre for high-capacity links between systems. The reasons follow from the physics rather than from fashion. A glass thread carries no current, so it cannot short or complete an unintended circuit; it adds little weight for the amount of data it moves; and because the signal is light confined by the core–cladding boundary rather than a voltage on a wire, it is unaffected by the electromagnetic interference that surrounds electrical looms, generators and radio installations. A copper run in the same location has to be screened against that interference, and a fibre does not.

The same guiding effect appears at a much smaller scale in cockpit lighting. Light fed into the edge of a transparent rod or panel travels along it by repeated total internal reflection, and by letting a controlled amount escape along its length the panel can illuminate a row of markings evenly from a single source placed out of the way. One lamp behind the panel rather than a bulb behind every legend, and nothing shining directly into the crew's eyes.

What the test asks

The commonest wrong answer is that total internal reflection can happen when light enters the denser medium — a ray going from air into glass, described as being totally reflected at some large angle of incidence. It cannot, at any angle, and the fastest way to see that is that the angle inside a denser material is always the smaller one. Expect also an option that reverses the index relationship, claiming a higher refractive index gives a larger critical angle; it gives a smaller one, because n sits on the bottom of the fraction. A third trap describes what happens beyond the critical angle as most of the light reflecting with a little still escaping — beyond C, none escapes, which is what total means. And watch for the condition being stated as a property of the material alone: every transparent material can transmit or totally reflect, and which one happens depends on the angle.