Total Internal Reflection & Critical Angle

Hold a diamond to the light and it blazes — not because it glows, but because it is extraordinarily good at refusing to let light out. Light that wanders into a diamond bounces around inside, trapped, escaping only through the small windows where it strikes a facet nearly head-on, and that hoarding-then-releasing is what we see as fire. The same trick sends your voice and this page as pulses of light down a hair-thin glass optical fibre for thousands of kilometres, turns a $45^\circ$ glass prism into a perfect mirror inside every pair of binoculars, and paints the shimmering false "water" of a highway mirage. The physics behind all of it is total internal reflection (TIR). It begins where Snell's law runs out of room. Going from a denser medium into a rarer one (glass to air, say), the refracted ray bends away from the normal, so as you tilt the incident ray more steeply, the refracted ray leans further and further toward the surface. At one special incidence — the critical angle $\theta_c$ — the refracted ray lies flat along the boundary, grazing at $90^\circ$ . Tilt even slightly beyond $\theta_c$ and there is simply no refracted ray at all: Snell's law would demand a sine greater than one, which is impossible, so all the light reflects back inside, obeying the ordinary law of reflection with essentially $100\%$ efficiency — a better mirror than any metal. The rule is compact: $\sin\theta_c = n_2/n_1$ , or for the common medium-to-air case $\sin\theta_c = 1/n$ . Glass gives $\theta_c = 41.8^\circ$ , water $48.6^\circ$ , and diamond a remarkably small $24.4^\circ$ — which is exactly why diamond traps light so well. This article derives the critical angle from Snell's law, explains why beyond it the light has nowhere to go, and works the numbers that power fibres, prisms and gemstones.

Key takeaways

Total internal reflection happens when light travelling in a denser medium hits the boundary with a rarer one at more than the critical angle $\theta_c$ : no light escapes — it all reflects back inside.
Two conditions, both needed: the ray must go denser → rarer ( $n_1 > n_2$ ) and strike at $\theta > \theta_c$ .
Critical angle: $\sin\theta_c = \dfrac{n_2}{n_1}$ ; for a medium–air surface $\sin\theta_c = \dfrac{1}{n}$ — glass $41.8^\circ$ , water $48.6^\circ$ , diamond $24.4^\circ$ , dense flint $38.1^\circ$ .
At exactly $\theta_c$ the refracted ray grazes at $90^\circ$ ; beyond it Snell's law would need $\sin\theta_2 > 1$ , which is impossible, so the light reflects totally.
A $45^\circ$ glass prism reflects light totally because $45^\circ > 41.8^\circ$ — the basis of the fibre, the reflecting prism, and diamond's sparkle.
Key idea: past the critical angle $\sin\theta_c = n_2/n_1$ , a denser-to-rarer ray cannot refract out at all, so it is reflected back with full intensity.

The core insight

Going from a denser to a rarer medium, the refracted ray bends away from the normal until, at the critical angle $\sin\theta_c = n_2/n_1$ , it grazes the surface at $90^\circ$ ; beyond that angle no refracted ray can exist (Snell would require $\sin\theta_2 > 1$ ), so all the light is totally internally reflected.

Where this lives today

Total internal reflection is one of the most quietly consequential effects in technology. The optical fibre — a dense glass core wrapped in lower-index cladding — traps light by TIR at every bounce and carries essentially all the world's internet and telephone traffic as pulses of infrared light, across oceans, with astonishingly little loss. Reflecting prisms at $45^\circ$ replace mirrors in binoculars, cameras, periscopes and rangefinders, because a TIR surface reflects more completely than any silver coating and never tarnishes. Endoscopes thread bundles of fibres into the body to light and image the inside. The sparkle of a cut diamond is deliberate TIR engineering: gem-cutters choose facet angles so that light entering the top bounces off the rear facets (all steeper than the $24.4^\circ$ critical angle) and returns to the eye, and the "brilliant" cut exists precisely to maximise this trapping. Even the shimmering mirage on a hot road and the way rain makes a rainbow involve the same denser-to-rarer refraction pushed toward its limit. And TIR is where the flat-surface half of ray optics ends: the chapter now curves the boundary, building refraction at spherical surfaces and then the lens, whose imaging equation $1/v - 1/u = 1/f$ (a minus, unlike the mirror's plus) closes the story that Snell's law began.

The physical picture

Start inside the glass and look outward. A ray heading from glass ( $n_1 = 1.5$ ) toward air ( $n_2 = 1$ ) is going from a denser to a rarer medium, so by Snell's law it bends away from the normal — the angle of refraction $\theta_2$ is larger than the angle of incidence $\theta_1$ . Increase $\theta_1$ a little and $\theta_2$ grows faster, leaning the escaping ray ever closer to the surface. There is clearly a limit: $\theta_2$ can be at most $90^\circ$ (flat along the boundary). The incidence that produces exactly $\theta_2 = 90^\circ$ is the critical angle $\theta_c$ . At that point the refracted ray skims along the surface and carries almost no energy outward. Now push past $\theta_c$ . Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ , would require $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} > 1$ — a demand no real angle can meet. The refracted ray does not exist. Since the light cannot cross the boundary and cannot vanish, it does the only thing left: it reflects, completely, back into the glass, obeying the familiar law of reflection. This is not the partial, dim reflection you always get at a boundary; it is total — very nearly $100\%$ of the light, with none lost to transmission — which makes TIR the best mirror in optics, used precisely because it wastes no light. Two conditions must both hold, and forgetting either is the classic error: the light must travel denser to rarer, and it must strike beyond the critical angle.

Figure

Three rays leaving a point source inside glass and striking the glass-air boundary at increasing angles — below the critical angle the ray refracts out into air with a weak partial reflection, at the critical angle it grazes along the surface at 90 degrees, and beyond the critical angle it is totally reflected back into the glass

The critical angle

One relation runs the whole topic. Set the refracted ray to graze the boundary — $\theta_2 = 90^\circ$ — in Snell's law and read off the critical incidence:

\sin\theta_c = \frac{n_2}{n_1} \qquad (1)

where: $n_1$ is the index of the denser medium the light starts in and $n_2$ the rarer medium it is trying to enter, so the ratio is less than one and a real $\theta_c$ exists only for $n_1 > n_2$ . For the very common case of a medium meeting air ( $n_2 = 1$ , $n_1 = n$ ), this simplifies to

\sin\theta_c = \frac{1}{n} \qquad (2)

where: $n$ is the index of the medium. The physical reading: a higher index gives a smaller critical angle, because denser light bends more sharply and reaches the grazing limit sooner. That single fact explains the ranking — diamond ( $n = 2.42$ ) has $\theta_c = 24.4^\circ$ , so almost any ray inside it exceeds the critical angle and stays trapped; ordinary glass ( $n = 1.5$ ) has $\theta_c = 41.8^\circ$ ; water ( $n = 4/3$ ) a generous $48.6^\circ$ , which is why water holds light in far less readily than glass or diamond. The condition for TIR then reads: light must be in the denser medium ( $n_1 > n_2$ ) and strike at $\theta_1 > \theta_c$ . If either fails — if the light is going from rarer to denser, or if it hits below $\theta_c$ — some or all of it refracts out and TIR does not occur. Below $\theta_c$ you always get partial reflection alongside the refracted ray; only beyond $\theta_c$ does reflection become total.

Critical angles at a glance

Medium (to air)	index $n$	$\sin\theta_c = 1/n$	critical angle $\theta_c$
water	$4/3 \approx 1.33$	$0.750$	$48.6^\circ$
crown glass	$1.5$	$0.6667$	$41.8^\circ$
dense flint glass	$1.62$	$0.6173$	$38.1^\circ$
diamond	$2.42$	$0.4132$	$24.4^\circ$

Figure

Light guided along an optical fibre by repeated total internal reflection — a ray enters the denser glass core and zig-zags down it, striking the core-cladding boundary each time at more than the critical angle so none escapes, carrying the signal around bends

Derivation AThe critical angle from Snell's law

The critical-angle formula is Snell's law evaluated at the one incidence that pushes the refracted ray flat against the boundary. The goal is to derive $\sin\theta_c = n_2/n_1$ (and its medium-to-air form $\sin\theta_c = 1/n$ ) by setting $\theta_2 = 90^\circ$ , and to compute $\theta_c$ for glass and diamond. The setup is a ray inside the denser medium ( $n_1$ ) meeting the boundary with the rarer one ( $n_2 < n_1$ ).

Step 1. Write Snell's law for the boundary: $n_1\sin\theta_1 = n_2\sin\theta_2$ . As the incidence $\theta_1$ increases, the refraction angle $\theta_2$ increases faster (since $n_1 > n_2$ ), heading toward its maximum possible value of $90^\circ$ .

Step 2. Impose the grazing condition. The critical angle $\theta_c$ is defined as the incidence for which $\theta_2 = 90^\circ$ . Substitute: $n_1\sin\theta_c = n_2\sin 90^\circ$ .

Step 3. Solve. Since $\sin 90^\circ = 1$ , we get $n_1\sin\theta_c = n_2$ , hence $\sin\theta_c = \dfrac{n_2}{n_1}$ . For a medium meeting air, $n_2 = 1$ and $n_1 = n$ , so $\sin\theta_c = \dfrac{1}{n}$ .

Step 4. Put numbers in. Glass, $n = 1.5$ : $\sin\theta_c = \dfrac{1}{1.5} = 0.6667$ , so $\theta_c = 41.8^\circ$ . Diamond, $n = 2.42$ : $\sin\theta_c = \dfrac{1}{2.42} = 0.4132$ , so $\theta_c = 24.4^\circ$ — much smaller, because the higher index reaches grazing sooner. A grazing check confirms the algebra: at $\theta_c$ , $n_1\sin\theta_c = 1.5 \times 0.6667 = 1.00 = n_2\sin 90^\circ$ , so the refracted ray indeed lies along the surface. Verified: $\theta_c = 41.8^\circ$ (glass), $48.6^\circ$ (water), $24.4^\circ$ (diamond), $38.1^\circ$ (flint), and the grazing identity, are recomputed in verify_hs2-total-internal-reflection.py.

Go deeper · Derivation B · 1 minWhy there is no refracted ray beyond the critical angle

Derivation BWhy there is no refracted ray beyond the critical angle

The most conceptual — and most examined — part of TIR is why the light has no choice but to reflect once past $\theta_c$ . The goal is to show that beyond the critical angle Snell's law has no solution for the refracted ray, and to read the same limit through Fermat's principle of least time (stated conceptually — the minimisation itself is calculus and belongs to the university track). Take glass to air and an incidence $\theta_1 > \theta_c$ .

Step 1. Ask Snell's law for the refracted ray. It says $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = n\sin\theta_1$ (with $n = n_1/n_2 > 1$ ). For $\theta_1$ just above $\theta_c$ , $\sin\theta_1 > \sin\theta_c = 1/n$ , so $n\sin\theta_1 > 1$ .

Step 2. Hit the impossibility. This demands $\sin\theta_2 > 1$ , but no real angle has a sine above one. There is therefore no angle $\theta_2$ — the refracted ray simply cannot exist. The mathematics is telling you the light has nowhere to go across the boundary.

Step 3. Conserve the energy. The light cannot cross and cannot disappear, so it must return into the denser medium — it reflects, and because none is transmitted, essentially all of it reflects, obeying the ordinary law of reflection ( $\theta_{\text{reflected}} = \theta_1$ ). That is total internal reflection.

Step 4. Read it through Fermat. Fermat's principle says light takes the path of least travel time; the refracted path that Snell's law selects is the least-time route across the boundary. Beyond $\theta_c$ that least-time refracted path would require the impossible $\sin\theta_2 > 1$ , so no valid transmitted path exists at all, and the least-time route that remains is the reflected one — the light stays inside. For diamond ( $\theta_c = 24.4^\circ$ ), almost every internal ray exceeds this angle, which is why light entering a cut diamond bounces many times before finding a facet to escape through, and the stone appears to blaze. Verified: the $\sin\theta_2 > 1$ impossibility beyond $\theta_c$ , and the diamond $\theta_c = 24.4^\circ$ , are recomputed in verify_hs2-total-internal-reflection.py.

Common trapTotal internal reflection needs BOTH conditions — denser to rarer AND beyond the critical angle

The commonest TIR mistakes are all about the two conditions and the direction of the formula. First, TIR can happen only going from denser to rarer ( $n_1 > n_2$ ): light travelling from air into glass can never be totally internally reflected, no matter how steep — there is no critical angle for rarer-to-denser, because the refracted ray bends toward the normal and always exists. Second, TIR needs $\theta > \theta_c$ ; at or below $\theta_c$ light still refracts out (with a partial reflection alongside), so "some light reflects" is not the same as "total reflection" — only beyond $\theta_c$ is it total. Third, the formula is $\sin\theta_c = n_2/n_1$ , rarer over denser — writing it upside-down ( $n_1/n_2 > 1$ ) asks for the arcsine of a number bigger than one and gives nonsense. Finally, a mirage is not a literal reflection off the hot road: it is light refracting smoothly through air layers whose index changes with temperature, curving until it appears to come from below — a TIR-like bending, not a bounce off a surface. Keep both conditions, the right ratio, and the honest mirage story, and TIR problems are secure.

Worked examples

Board level — critical angles for glass, water and diamond (IGCSE and CBSE pattern)

Find the critical angle for light passing from (a) crown glass ( $n = 1.5$ ), (b) water ( $n = 4/3$ ) and (c) diamond ( $n = 2.42$ ) into air.

Step 1. For a medium-to-air boundary use $\sin\theta_c = \dfrac{1}{n}$ .

Step 2. (a) Glass: $\sin\theta_c = \dfrac{1}{1.5} = 0.6667$ , so $\theta_c = \arcsin(0.6667) = \boxed{41.8^\circ}$ .

Step 3. (b) Water: $\sin\theta_c = \dfrac{1}{4/3} = 0.750$ , so $\theta_c = \arcsin(0.750) = \boxed{48.6^\circ}$ (using $n = 1.33$ gives $48.75^\circ$ ).

Step 4. (c) Diamond: $\sin\theta_c = \dfrac{1}{2.42} = 0.4132$ , so $\theta_c = \arcsin(0.4132) = \boxed{24.4^\circ}$ . The pattern is monotone: the higher the index, the smaller the critical angle, and the more readily the medium traps light. Diamond's tiny $24.4^\circ$ is the physical reason it sparkles so.

Competitive level — the reflecting prism, the fibre and dense flint (JEE Main and NEET pattern)

(a) Show that a right-angled glass prism ( $n = 1.5$ ) used at $45^\circ$ acts as a perfect mirror. (b) A dense flint glass has $n = 1.62$ ; find its critical angle. (c) Verify the grazing condition for crown glass.

Step 1. (a) The prism's critical angle is $\theta_c = 41.8^\circ$ (from part (a) above). Light entering a face of a $45^\circ{-}45^\circ{-}90^\circ$ prism strikes the hypotenuse at $45^\circ$ . Since $45^\circ > 41.8^\circ$ , the light is totally internally reflected — the prism turns the beam through $90^\circ$ with essentially no loss, which is why binoculars and periscopes use reflecting prisms rather than silvered mirrors.

Step 2. (b) Dense flint: $\sin\theta_c = \dfrac{1}{1.62} = 0.6173$ , so $\theta_c = \arcsin(0.6173) = \boxed{38.1^\circ}$ — smaller than crown glass's $41.8^\circ$ , because flint is optically denser.

Step 3. (c) Grazing check for crown glass: at $\theta_c = 41.8^\circ$ , $n_1\sin\theta_c = 1.5 \times \sin 41.8^\circ = 1.5 \times 0.6667 = \boxed{1.00}$ , which equals $n_2\sin 90^\circ = 1 \times 1$ . So the refracted ray does lie exactly along the surface at the critical angle — the algebra checks out.

Step 4. The same $\theta > \theta_c$ logic keeps light inside an optical fibre: a ray zig-zagging down the dense glass core meets the boundary with the lower-index cladding at well above the critical angle every time, so none leaks out and the signal survives kilometres of travel. Timing note: about 60 seconds. The prism part needs no calculation once you know $\theta_c = 41.8^\circ$ — just compare $45^\circ$ with it. Memorise glass $41.8^\circ$ , water $48.6^\circ$ , diamond $24.4^\circ$ and half these problems are instant.

Exam traps

Forgetting "denser to rarer". TIR occurs only when light travels from a higher-index medium to a lower-index one. Air-to-glass can never give TIR.

Only one condition. Both are required: denser → rarer and $\theta > \theta_c$ . At or below $\theta_c$ the light refracts out (with partial reflection), so it is not total.

Flipping the formula. $\sin\theta_c = n_2/n_1$ — rarer over denser (for medium–air, $1/n$ ). The reciprocal exceeds one and has no arcsine.

"Partial" versus "total". Below the critical angle some light always reflects (partial reflection); only beyond $\theta_c$ is reflection total. Do not call ordinary partial reflection "total".

Mirage as a mirror. A road mirage is gradual refraction through temperature-layered air, not a bounce off the surface — TIR-like, not a literal reflection.

Board scope. TIR and the critical angle sit on all ten boards this lane serves: CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2, IB (both SL and HL, as wave behaviour), IGCSE and A-Level. As throughout this chapter, "A-Level" means Edexcel International A-Level — CIE 9702 A-Level has no geometric optics, so a 9702 candidate does not study TIR for that paper, whereas Edexcel IAL, IGCSE and AP-2 all do.

Origins · 1 minPierre de Fermat and how this was discovered

Pierre de Fermat

1607–1665 · Toulouse

Fermat, a magistrate at Toulouse who did mathematics as a passion, gave optics its deepest organising idea: the principle of least time. In 1662 he proposed that light travels between two points along the path that takes the least time, and showed that this single principle derives Snell's law — the ray bends at a boundary by exactly the amount that minimises total travel time through the two media. The critical angle falls out of the same idea: beyond it, the least-time refracted path would require an impossible geometry ( $\sin\theta_2 > 1$ ), so no transmitted path exists and the light must reflect. The honest scope: Fermat gave the principle and its consequence, but performing the minimisation itself needs calculus and belongs to the university track — here we use the result, not the derivation. His principle also anticipated, by two centuries, the variational thinking at the heart of modern physics. The dazzling modern payoff of TIR — the light-pipe and the optical fibre — came later, from Colladon and Tyndall in the 1840s–50s and the fibre pioneers of the twentieth century.

History

1621

— Willebrord Snellius finds the refraction law whose limit defines the critical angle.

1637

— René Descartes prints the sine law of refraction in La Dioptrique.

1662

— Pierre de Fermat states the principle of least time, from which Snell's law and the critical angle follow.

1841

— Jean-Daniel Colladon guides light along a curved water jet by total internal reflection.

1854

— John Tyndall demonstrates the light-pipe, the direct ancestor of the optical fibre.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics"), total internal reflection and its applications · NCERT Science Class 10, Ch. 10, refraction · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics", critical-angle and prism problems · Cambridge IGCSE 0625, total internal reflection and optical fibres.

Provenance. Standard geometric optics; every number recomputed in verify_hs2-total-internal-reflection.py. Recompute $\sin\theta_c = 1/n$ : glass $1/1.5 = 0.6667 \to 41.8^\circ$ ; water $1/(4/3) = 0.750 \to 48.6^\circ$ (and $1/1.33 \to 48.75^\circ$ ); diamond $1/2.42 = 0.4132 \to 24.4^\circ$ ; dense flint $1/1.62 = 0.6173 \to 38.1^\circ$ . Grazing identity: $1.5 \times 0.6667 = 1.00$ . Prism check: $45^\circ > 41.8^\circ$ .

Prerequisites. Refraction of Light and Snell's Law — the bending rule whose grazing limit defines the critical angle. Leads to. Refraction at a Spherical Surface (curving the boundary toward the lens) and the UG seam Fermat's Principle of Least Time.

Go deeper (UG). Fermat's least-time principle, carried out with calculus and generalised to the whole variational formulation of optics, is developed in Fermat's Principle of Least Time in the university optics track. It is enrichment only; every result here follows from Snell's law evaluated at $\theta_2 = 90^\circ$ , with the minimisation quoted, never performed.

Why does light reflect completely instead of just partly beyond the critical angle?

Because beyond the critical angle there is no refracted ray for the light to escape into. Snell's law would require the refracted ray to have $\sin\theta_2 > 1$ , which is impossible, so transmission across the boundary is simply not allowed. The light cannot cross and it cannot vanish, so conservation of energy leaves only one option: it reflects, and since none is lost to transmission, essentially all of it reflects. Below the critical angle the situation is different — there is a valid refracted ray, so the light splits into a transmitted part and a weaker reflected part (partial reflection). It is only past $\theta_c$ , when the escape route closes, that reflection becomes total.

Why does diamond sparkle so much more than glass?

Because diamond's high refractive index ( $n = 2.42$ ) gives it a very small critical angle, $24.4^\circ$ , compared with glass's $41.8^\circ$ . That means almost any ray bouncing around inside a diamond strikes its facets at more than the critical angle and is totally internally reflected, so light entering the top is trapped and bounces many times before it finds a facet steep enough to let it out — and when it does escape, it comes back toward the viewer's eye. Gem-cutters exploit this deliberately: the angles of the "brilliant" cut are chosen so that light is caught by TIR off the back facets and returned to the top, maximising the fire and sparkle. Glass, with its larger critical angle, lets light escape far more easily, so a glass imitation looks comparatively dull.

Is a mirage really total internal reflection off the road?

Not exactly — a mirage is refraction, not a bounce off a surface. On a hot day the air just above the road is much warmer, and warmer air is slightly less dense and has a slightly lower refractive index. Light from the sky travelling downward toward the road passes through these layers of decreasing index and bends gradually, curving upward before it reaches the ground, so it appears to come from below — you see a shimmering patch that looks like water reflecting the sky. It resembles total internal reflection because the light effectively turns back, but there is no single surface and no sudden critical-angle event; it is smooth, continuous bending through a gradient of index. Calling it "TIR off the road" is a common but imprecise description.

How does an optical fibre keep the light inside around bends?

An optical fibre has a dense glass core surrounded by a lower-index cladding, so the core-cladding boundary is a denser-to-rarer interface with a critical angle. Light is launched into the core at a shallow angle so that every time it hits the boundary it does so at more than the critical angle, and is therefore totally internally reflected — it zig-zags down the fibre, losing almost nothing at each bounce. Around a gentle bend the geometry still keeps the incidence above the critical angle, so the light follows the fibre. Only if you bend the fibre too sharply does some light strike below the critical angle and leak out. Because TIR is nearly lossless, a signal can travel tens of kilometres before it needs boosting, which is what makes fibre-optic communication so powerful.

Can total internal reflection happen going from air into glass?

No — never. Total internal reflection requires light to travel from a denser medium (higher index) to a rarer one (lower index). Going from air ( $n = 1$ ) into glass ( $n = 1.5$ ) is rarer-to-denser: the refracted ray bends toward the normal and always exists, no matter how steep the incidence, so there is no critical angle and no TIR. The critical angle only exists for the denser-to-rarer direction, where the refracted ray bends away from the normal and can reach $90^\circ$ . This is why the two conditions are stated together — "denser to rarer AND beyond the critical angle" — and why forgetting the direction is such a common and costly mistake.

Is TIR on my exam, and what does A-Level mean here?

Total internal reflection and the critical angle are on every board this article serves: CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2, IB (both SL and HL, under wave behaviour), IGCSE and A-Level. But "A-Level" needs the usual caveat: Cambridge International A-Level (CIE 9702) contains no geometric optics at all, so a 9702 candidate does not study TIR for that paper. Where this lane says A-Level it means Edexcel International A-Level, which does cover refraction and total internal reflection. IGCSE 0625 and AP Physics 2 both include TIR and the critical angle. So unless you sit specifically CIE 9702, this is examinable material for you.