Total internal reflection and the critical angle — Physics

Before this

Reflection and refraction of waves
Refraction of light · coming soon

This leads to

Uses of each EM wave type
Lenses and ray diagrams · coming soon

A window that turns into a mirror

Swim to the bottom of a still swimming pool and look straight up. You see the sky through the surface, like looking through a window. Now look up at a slant, towards the far side of the pool. The surface changes. It shines like silver, and you see the pool floor reflected in it. The window has become a mirror.

Nothing about the water changed. Only the angle changed. So what happened to the light that should have gone out into the air?

The last lecture showed the beginning of the answer. In the explorer, when light went from glass into air at a large angle, the refracted wave disappeared, and the wave could not leave the glass at all. This lecture explains why. The same effect sends most of the world's international internet traffic through glass threads under the oceans.

Slow to fast: bending away from the normal

Recall the rule from the last lecture. A wave that speeds up as it crosses a boundary turns away from the normal. Light in glass travels more slowly than light in air. So light going from glass into air turns away from the normal, and the angle of refraction $θ_{2}$ is bigger than the angle of incidence $θ_{1}$ .

Snell's law gives the angles. For light from glass of refractive index 1.50 into air ( $n \approx 1.00$ ):

1.50 sin θ_{1} = 1.00 sin θ_{2}, so sin θ_{2} = 1.50 sin θ_{1}

$θ_{1}$ in the glass	$sin θ_{2} = 1.50 sin θ_{1}$	$θ_{2}$ in the air
10°	0.260	≈ 15.1°
20°	0.513	≈ 30.9°
30°	0.750	≈ 48.6°
40°	0.964	≈ 74.6°
45°	1.061	none

Look at the last two rows. Between 30° and 40°, the angle in the air jumps from about 49° to about 75°. The refracted ray is swinging down towards the surface. At 45°, Snell's law asks for an angle whose sine is 1.061.

No such angle exists. The sine of an angle can never be more than 1. A refracted wave at 45° would break Snell's law, and Snell's law comes from the geometry of the wavefronts. So there is no refracted wave at all.

The critical angle

Between "a refracted wave exists" and "no refracted wave" there is a boundary case. It happens when the refracted ray has turned all the way to 90°, so that it runs along the surface. The angle of incidence that does this is called the critical angle, $C$ .

Put $θ_{1} = C$ and $θ_{2} = 90°$ into Snell's law. Since $sin 90° = 1$ :

n_{1} sin C = n_{2} \times 1, so sin C = \frac{n _{2}}{n _{1}}

When the second medium is air, $n_{2} \approx 1.00$ , and this becomes $sin C = \frac{1}{n}$ . For the glass above, $sin C = \frac{1}{1.50} \approx 0.667$ , so $C \approx 41.8°$ .

Light in glass (

n = 1.50

) meeting the surface. Top: at 30°, most of the light refracts out at about 48.6°, and a faint ray reflects. Middle: at the critical angle, about 41.8°, the refracted ray runs along the surface. Bottom: at 60°, there is no refracted ray, and all the light reflects back into the glass.

Beyond the critical angle: all of it comes back

The wave still carries its energy when it reaches the boundary. That energy cannot vanish. At most boundaries it splits into two parts: a refracted part that goes on, and a reflected part that comes back. In the last lecture, light meeting a glass window head-on sent back only about 4%.

Beyond the critical angle, the refracted part cannot exist. So all the energy goes the only way left: back into the glass. This is total internal reflection. "Total", because none of the light gets out. "Internal", because it happens inside the slower medium.

The reflection still obeys the law of reflection: the angle of reflection equals the angle of incidence. The only difference is how much light comes back. At 4% you see a faint reflection. With total internal reflection you see a mirror. That is the silver surface the swimmer sees.

That is it. That is the whole idea. Total internal reflection is not a new law of physics. It is refraction pushed to the point where Snell's law has no answer.

Total internal reflection

When a wave reaches a boundary with a medium where it travels faster, there is a critical angle $C$ , given by $sin C = \frac{n _{2}}{n _{1}}$ . For light into air, $sin C = \frac{1}{n}$ .

If the angle of incidence is greater than $C$ , there is no refracted wave, and all the wave is reflected. This is total internal reflection.

It can only happen when the wave goes from a slower medium into a faster one.

Why only from slow to fast? Going into a slower medium, the wave turns towards the normal. Then $θ_{2}$ is always smaller than $θ_{1}$ , and $sin θ_{2}$ is always less than 1. Snell's law always has an answer, so a refracted wave always exists.

Explore this on the graph below. It shows $θ_{2}$ against $θ_{1}$ , together with the rays at the boundary.

Graph builder — where the refracted ray runs out

Choose a pair of media. Move the angle of incidence and watch the point climb the curve. Find where the curve stops. Then try "Air into glass", where it never stops.

Media:

Angle of incidence θ₁: 30°

Graph: across, the angle of incidence θ₁; up, the angle of refraction θ₂. Shaded: no refracted ray.

For light going from glass ( $n = 1.50$ ) into air, the angle of refraction rises faster and faster as the angle of incidence grows. It reaches 90° at the critical angle, about 41.8°, and the curve stops there.

The angle of refraction

θ_{2}

against the angle of incidence

θ_{1}

for glass into air. Beyond

C \approx 41.8°

(shaded), there is no refracted ray.

Media	Critical angle
Glass (n = 1.50) into air	≈ 41.8°
Water (n = 1.33) into air	≈ 48.8°
Diamond (n = 2.42) into air	≈ 24.4°
Fibre core (1.48) into cladding (1.46)	≈ 80.6°
Sound: air (343 m/s) into water (1500 m/s)	≈ 13.2°
Air into glass	none

What should you have seen? The curve always climbs more and more steeply, and it stops where $θ_{2}$ reaches 90°. A bigger refractive index makes it stop sooner: diamond's curve ends at about 24°. And for air into glass, the curve never reaches 90°, so there is no critical angle.

Stop and think

Many books say total internal reflection happens "when light goes from a denser medium to a less dense one". Can sound in air be totally reflected by the surface of a lake?

Show the answer

Yes, even though water is about 800 times denser than air. What matters is speed, not density. Sound travels at about 343 m/s in air and about 1500 m/s in water, so it speeds up as it enters the water. That is the slow-to-fast condition. Using the general sine law, $sin C = \frac{343}{1500} \approx 0.229$ , so $C \approx 13.2°$ . Sound from the air that meets the lake at more than about 13° to the normal cannot enter it. For light, "optically denser" just means "slower", which is why the word causes so much confusion.

Worked examples

Worked Example 1 — Why a diamond sparkles

Find the critical angle for diamond ( $n = 2.42$ ), and compare it with water ( $n = 1.33$ ).

Diamond: $sin C = \frac{1}{2.42} \approx 0.413$ , so $C \approx 24.4°$ .

Water: $sin C = \frac{1}{1.33} \approx 0.752$ , so $C \approx 48.8°$ .

Inside a diamond, any ray that meets a surface at more than about 24° to the normal is trapped. That is most rays. Diamond cutters shape the faces so that light entering the top reflects totally from the faces underneath, and then leaves back through the top towards your eye.

Check: a larger refractive index means a smaller $sin C$ , and so a smaller critical angle. 2.42 is larger than 1.33, and 24.4° is smaller than 48.8°. ✓

Worked Example 2 — A prism as a perfect mirror

A glass prism ( $n = 1.50$ ) has angles of 45°, 45° and 90°. Light enters one short face at right angles. Show that the light is totally reflected at the long face.

At the first face the light meets the boundary along the normal, so it goes straight in. The long face is at 45° to the short face, so the light meets it at 45° to its normal.

The critical angle is about 41.8°. Since 45° is greater than 41.8°, the light is totally internally reflected. It turns through 90° and leaves through the other short face, again along the normal.

Check: this works only if $C$ is less than 45°, that is, if $\frac{1}{n} < sin 45° \approx 0.707$ , so $n > 1.414$ . Glass at 1.50 passes. ✓ Prisms like this turn the light in periscopes and many binoculars. They reflect better than a silvered mirror, which absorbs some light every time.

Worked Example 3 — An angle given from the surface

A ray inside a glass block ( $n = 1.45$ ) meets the surface at 40° to the surface. Is it totally internally reflected?

First convert. Every angle in Snell's law is measured from the normal, which is at 90° to the surface. So the angle of incidence is $90° - 40° = 50°$ .

Critical angle: $sin C = \frac{1}{1.45} \approx 0.690$ , so $C \approx 43.6°$ .

50° is greater than 43.6°, so yes, the ray is totally internally reflected.

Check what would have happened without the conversion: 40° is less than 43.6°, which gives the opposite answer. One subtraction decides the whole question.

Worked Example 4 — The swimmer's window

The swimmer from the hook is 2.0 m below the surface ( $n = 1.33$ ). Light from the whole sky reaches the eye only through a circle on the surface. Find its radius.

Light from the sky can arrive at the surface from any direction, up to 90° from the normal. Inside the water, it then travels at up to the critical angle, $C \approx 48.8°$ , from the normal. So all the light from the sky reaches the eye inside a cone of half-angle 48.8°.

The cone meets the surface in a circle of radius $r = 2.0 tan 48.8° \approx 2.28$ m.

Outside that circle, the swimmer is looking up at more than the critical angle. Light coming from below reflects totally there, so the surface shows the pool, not the sky. That is the silver mirror. Check: the reasoning runs both ways. A ray from the eye going out at more than 48.8° cannot leave, so no ray from outside can come in along it. ✓

Optical fibres: light in a pipe of glass

An optical fibre is a thread of very pure glass. Light sent into one end meets the side at a large angle, reflects totally, crosses to the other side, reflects again, and so on, again and again. It cannot get out of the sides, so it follows the fibre, even around gentle bends.

A real fibre is not bare glass in air. It has a core surrounded by a layer of glass called the cladding, with a slightly lower refractive index. A common fibre for long distances has a cladding 125 µm across, about the thickness of a human hair, around a core only about 8 µm across (Corning SMF-28 data sheet).

Why not use bare glass? Total internal reflection depends on what is on the other side of the boundary. A drop of water, a fingerprint, or another fibre touching a bare surface changes the boundary, and light leaks out at that point. The cladding gives the core a boundary that is always clean and never touched.

The price is a much larger critical angle. With typical values of 1.48 for the core and 1.46 for the cladding, $sin C = \frac{1.46}{1.48} \approx 0.986$ , so $C \approx 80.6°$ . Only light travelling almost parallel to the fibre is trapped. That is fine, because the light is sent in along the fibre on purpose.

Light trapped in the core of a fibre by total internal reflection at the boundary with the cladding. The ray meets the boundary at about 84° to the normal, more than the critical angle of about 80.6°. The values 1.48 and 1.46 are typical.

The same mistake is easy to make in a calculation. Find it below.

Spot the error — will the fibre trap this ray?

A student decides whether a ray stays in a fibre. Tap the line that contains the mistake.

The working below uses the formula for glass into air, but the core is surrounded by cladding. The mistake is in line 3.

A fibre has a core of refractive index 1.48 and a cladding of refractive index 1.46.
A ray in the core meets the core–cladding boundary at 50° to the normal.
Critical angle: sin C = 1 ÷ 1.48 = 0.676, so C ≈ 42.5°. (Wrong: the outside medium is the cladding, so sin C = 1.46 ÷ 1.48 ≈ 0.986 and C ≈ 80.6°.)
50° is greater than 42.5°, so the ray is totally internally reflected.
So the light stays in the core.

In fact 50° is less than 80.6°, so the ray refracts into the cladding and is lost from the core.

Fibres carry light for very long distances. Today's best fibres lose only about 0.2 decibels per kilometre, so about 1% of the light is still there after 100 km. That is why submarine cables of glass fibre carry more than 99% of the data that travels between continents (TeleGeography, 2023). The idea is old. In 1842, Jean-Daniel Colladon in Geneva described light guided along a curved jet of water. But early glass absorbed far too much light. In 1966, Charles Kao and George Hockham argued that purer glass could bring the loss below 20 decibels per kilometre, and in 1970 researchers at Corning made a fibre that did. Kao shared the 2009 Nobel Prize in Physics for this work.

Three classic mistakes

Trap 1 — Measuring the angle from the surface

The critical angle, like every angle in Snell's law, is measured from the normal. If a question gives the angle to the surface, subtract it from 90° first. Worked Example 3 shows how this one step can reverse the answer.

Trap 2 — Looking for total internal reflection going into a slower medium

Light going from air into glass, or water into glass, slows down and turns towards the normal. It always has a refracted ray, at every angle. Total internal reflection needs the wave to speed up. For light that means going from a larger refractive index to a smaller one.

Trap 3 — Using sin C = 1/n when the outside is not air

$sin C = \frac{1}{n}$ is only the special case for air outside. In general, $sin C = \frac{n _{2}}{n _{1}}$ , with $n_{1}$ the medium the light is in. For glass (1.50) against water (1.33), $sin C \approx 0.887$ and $C \approx 62.5°$ , not 41.8°.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Window glass has $n = 1.52$ . Find its critical angle. A ray inside a window pane meets the surface at 45° to the normal. Does it get out?

Hint 1
The outside is air, so $sin C = 1/ n$ .

Hint 2
Compare 45° with $C$ . Greater than $C$ means total internal reflection.

Full answer
$sin C = \frac{1}{1.52} \approx 0.658$ , so $C \approx 41.1°$ . 45° is greater than 41.1°, so the ray is totally internally reflected and does not get out. Check with Snell's law: $sin θ_{2} = 1.52 sin 45° \approx 1.075$ , more than 1, so there is no refracted ray. ✓
A transparent plastic has a critical angle of 35° in air. Find its refractive index.

Hint 1
Rearrange $sin C = 1/ n$ for $n$ .

Hint 2
$n = 1/ sin C$ .

Full answer
$n = \frac{1}{sin 35°} \approx \frac{1}{0.574} \approx 1.74$ . Check: $\frac{1}{1.74} \approx 0.575$ , and $sin^{- 1} (0.575) \approx 35.1°$ . ✓ (The small difference is rounding.)
A glass block ( $n = 1.50$ ) sits at the bottom of a tank of water ( $n = 1.33$ ). Light inside the glass meets its top surface. Find the critical angle. Would a ray at 50° to the normal escape into the water?

Hint 1
The outside is water, not air.

Hint 2
$sin C = n_{2} / n_{1} = 1.33/1.50$ .

Full answer
$sin C = \frac{1.33}{1.50} \approx 0.887$ , so $C \approx 62.5°$ . 50° is less than 62.5°, so the ray escapes into the water. In air, the same ray would have been trapped, because 50° is greater than 41.8°. Check with Snell's law: $sin θ_{2} = \frac{1.50 sin 50°}{1.33} \approx 0.864$ , so $θ_{2} \approx 59.8°$ , a real angle. ✓
Light enters the flat end of a fibre from the air, at angle $α$ to the fibre's axis. Inside, it must meet the core–cladding boundary at more than $C$ , where $sin C = 1.46/1.48$ . Find the largest $α$ for which the light is trapped.

Hint 1
Draw it. The end face is at right angles to the axis, so its normal is the axis. The side boundary's normal is at right angles to the axis.

Hint 2
If the ray inside makes angle $β$ with the axis, it meets the side at $90° - β$ . So you need $90° - β \geq C$ , and at the end face $sin α = 1.48 sin β$ .

Full answer
$C \approx 80.6°$ , so the ray inside may make at most $β = 90° - 80.6° \approx 9.4°$ with the axis. At the end face, Snell's law gives $sin α = 1.48 sin β$ . Since $sin β = cos C = 1 - (1.46/1.48)^{2}$ , this is $sin α = 1.4 8^{2} - 1.4 6^{2} = 0.0588 \approx 0.242$ , so $α \approx 14.0°$ . Light must arrive within about 14° of the axis to be trapped. Check: $1.48 sin 9.4° \approx 0.242$ . ✓

The bigger idea

Total internal reflection shows something important about physical laws. A law can stop having an answer, and when it does, nature does not break. It does something else. Snell's law runs out at the critical angle, energy must still go somewhere, and the only place left is back. The swimmer's mirror, the diamond's sparkle and the cables under the ocean are all that one fact.

It also holds for every kind of wave, as the sound in the Stop-and-think showed. Seismic waves inside the Earth and radio waves in the upper atmosphere can be turned back in the same way, whenever they head into a region where they would travel faster.

Key concepts in this lesson

A wave going from a slower medium into a faster one turns away from the normal, and the angle of refraction reaches 90° at the critical angle $C$ .
The critical angle is given by $sin C = n_{2} / n_{1}$ , which is $sin C = 1/ n$ when the second medium is air.
Beyond the critical angle, Snell's law has no solution, so there is no refracted wave and all the wave is reflected: total internal reflection.
Total internal reflection needs the wave to speed up at the boundary; it never happens going into a slower medium.
Optical fibres trap light in a core by total internal reflection at the boundary with a cladding of slightly lower refractive index.