Total Internal Reflection

In short

When light travels from a denser medium (refractive index n_1) toward a rarer one (n_2 < n_1), Snell's law says

n_1\sin\theta_1 \;=\; n_2\sin\theta_2

As \theta_1 increases, so does \theta_2. At a special incidence angle — the critical angle \theta_c — the refracted ray skims along the surface (\theta_2 = 90°). Beyond this, Snell's law has no real solution and all the light reflects back into the denser medium. This is total internal reflection (TIR).

\boxed{\;\sin\theta_c \;=\; \frac{n_2}{n_1}\;}

For water-to-air (n_1 = 1.33, n_2 = 1.00): \theta_c = 48.6°. For glass-to-air (n_1 = 1.50): \theta_c = 41.8°. For diamond-to-air (n_1 = 2.42): \theta_c = 24.4°.

Two conditions must both hold for TIR:

Light must go from a denser medium to a rarer one.
The angle of incidence must exceed \theta_c.

Consequences.

Optical fibres (BSNL and Jio backbones) — light is trapped inside a thin glass thread by repeated TIR; signal travels hundreds of kilometres with almost no loss.
Diamond brilliance — a diamond's tiny critical angle (24.4°) means light entering the top cannot easily escape from the sides; it bounces around internally and exits through the top, bright.
Periscopes (Indian Navy's Scorpène submarines) — two 45°-angled prisms use TIR instead of silvered mirrors (no reflectivity loss).
Endoscopes (AIIMS gastroenterology) — a flexible bundle of fibres carries an image from inside the stomach to a camera outside.
Mirages on NH-4 in May — hot road surface bends light by TIR in the warm air layer; you see a "puddle" that is actually the sky.

Stand at the edge of a swimming pool in your Bandra flat complex, lean over, and look down into the water. You can see the bottom tiles. Now put on goggles and dive under. Look up. In a disk directly above your head, the sky, clouds, the diving board, and the legs of the people sitting on the edge are all visible — compressed into a bright circle. But outside that disk, you do not see the sky. You see a mirror image of the pool's walls and floor, reflected back from the water surface as if the surface were silver. There is nothing special about the water surface — it is just water and air. Yet from below, at shallow angles, it is a perfect reflector. That "mirror from below" is total internal reflection, and the disk of visible sky is Snell's window — the bright aperture through which rays from the sky can still escape to you.

This chapter explains where the cutoff between "escapes" and "reflects back" comes from, why it is exactly \sin\theta_c = n_2/n_1, and why that cutoff is the physics that lets a BSNL optical fibre carry your Instagram reel from Chennai to Singapore without a repeater for 1800 kilometres, why a diamond looks brighter than a glass imitation, and why a hot summer afternoon on National Highway 4 shows you a puddle that is not there.

Setting up the problem — light going from denser to rarer

Snell's law relates the angles on two sides of a refracting boundary:

n_1\sin\theta_1 \;=\; n_2\sin\theta_2

Why: see the refraction article for the full derivation. The physical content: the component of the wavevector along the boundary must be the same on both sides (phase continuity). Since the magnitude of the wavevector is n\omega/c in a medium of index n, the parallel component n\sin\theta is what matches.

Take the case that matters: light travels from a denser medium (higher n, like glass or water) into a rarer one (lower n, like air). So n_1 > n_2. Solving Snell's law for the refracted angle,

\sin\theta_2 \;=\; \frac{n_1}{n_2}\sin\theta_1

Since n_1/n_2 > 1, \sin\theta_2 > \sin\theta_1. The refracted ray is further from the normal than the incident ray — light bends away from the normal on entering the rarer medium.

Light travelling from water ($n_1 = 1.33$) into air ($n_2 = 1.00$). The refracted ray bends *away* from the normal. As $\theta_1$ increases, $\theta_2$ increases faster — and eventually reaches 90°.

The critical angle — where the refracted ray runs out

Because \sin\theta_2 grows faster than \sin\theta_1, there is a specific \theta_1 — call it the critical angle \theta_c — at which \sin\theta_2 = 1, i.e., \theta_2 = 90°. The refracted ray lies flat along the boundary.

Step 1. Set \theta_2 = 90° in Snell's law.

n_1\sin\theta_c \;=\; n_2\sin 90° \;=\; n_2\cdot 1

Step 2. Solve for the critical angle.

\boxed{\;\sin\theta_c \;=\; \frac{n_2}{n_1}\;}

Why: this is a simple algebraic rearrangement of Snell's law with \theta_2 = 90°. The critical angle depends only on the ratio of the two refractive indices. Note that \sin\theta_c < 1 requires n_1 > n_2 — TIR is only possible from denser to rarer medium, never the other way.

Step 3. Plug in numbers.

Denser medium	n_1	Rarer medium	n_2	\sin\theta_c	\theta_c
Water	1.33	Air	1.00	0.752	48.75°
Crown glass	1.52	Air	1.00	0.658	41.14°
Dense flint glass	1.65	Air	1.00	0.606	37.30°
Diamond	2.42	Air	1.00	0.413	24.41°
Glass fibre core	1.46	Cladding	1.43	0.979	78.26°

The denser the medium relative to the surroundings, the smaller the critical angle. Diamond's critical angle is only 24°, so light entering a cut diamond has to approach almost vertically to escape from any face — most bounces get trapped and make multiple internal reflections, producing the diamond's characteristic "fire."

What happens beyond the critical angle?

For \theta_1 > \theta_c, solving Snell's law gives \sin\theta_2 > 1. That has no real solution — \sin of a real angle is at most 1. There is no refracted ray. All the energy that Snell would have sent through the boundary stays on the incident side and reflects back.

When $\theta_1 > \theta_c$, Snell's law has no real solution for the refracted angle. *All* the incident energy reflects back into the water, obeying the law of reflection ($\theta_\text{reflected} = \theta_\text{incident}$). No light escapes into the air.

The word total means exactly what it says — the reflected beam carries 100% of the incident energy. An ordinary silvered mirror only reflects 90-95%; the rest is absorbed by the silver coating. A TIR-based reflector (glass prism at 45°) reflects nearly 100%, losing almost nothing. That is why high-quality optical instruments (periscopes, good binoculars, DSLR viewfinders) use glass prisms instead of metal mirrors.

Watch the critical angle emerge

Drag the red dot to change the incidence angle $\theta_1$ at a water-air surface (water-side, $n_1 = 1.33$). The dark dot tracks $\sin\theta_2 = (n_1/n_2)\sin\theta_1 = 1.33\sin\theta_1$. While $\sin\theta_2 < 1$, a refracted ray exists. At $\theta_1 = 48.75°$, $\sin\theta_2 = 1$ and $\theta_2 = 90°$ — the refracted ray lies flat. Beyond this angle, $\sin\theta_2 > 1$: Snell's law has no real solution and the light reflects totally.

Drag the slider past 48.75° and notice that \sin\theta_2 exceeds 1 — a mathematical signal that the refraction equation breaks down. Physics responds by refusing to let any light cross; it all reflects back.

The two conditions for TIR — both must hold

TIR is not an accident of optics. It occurs only when both conditions are met:

Direction: light must travel from a denser medium (n_1) to a rarer medium (n_2 < n_1). In the reverse direction (air to water), bending is toward the normal, and \sin\theta_2 is always less than \sin\theta_1 \le 1 — there is no angle at which the refracted ray vanishes.
Angle: the incidence angle must exceed the critical angle: \theta_1 > \theta_c.

Fail either condition and you get ordinary partial reflection plus refraction (the usual Fresnel split between a transmitted and a reflected beam).

Why the direction matters. Think of it from the geometric fact \sin\theta_c = n_2/n_1. If n_2 > n_1 (rarer to denser), this ratio exceeds 1 and no real angle satisfies the equation — there is no critical angle because there is nothing to reach. All possible incidence angles produce a real refracted ray.

Worked examples

Example 1: The BSNL optical-fibre backbone

A single-mode silica fibre has a core refractive index n_\text{core} = 1.475 and a cladding index n_\text{clad} = 1.455. Find the critical angle at the core-cladding boundary and the maximum angle a light ray can make with the fibre's axis while still being guided by TIR.

A single-mode optical fibre. The ray zigzags down the core, reflected at every boundary by TIR (because the angle inside the core exceeds the critical angle at the core-cladding surface). No light leaks into the cladding.

Step 1. Critical angle at the core-cladding boundary.

\sin\theta_c \;=\; \frac{n_\text{clad}}{n_\text{core}} \;=\; \frac{1.455}{1.475} \;=\; 0.9864

\theta_c \;=\; \arcsin(0.9864) \;=\; 80.54°

Why: the two indices are very close, so the ratio is close to 1, and the critical angle is close to 90°. This means only rays travelling almost parallel to the fibre axis can stay guided — but those rays can bounce many thousands of times per metre and still stay trapped.

Step 2. For a ray inside the core, define the angle \alpha with respect to the fibre axis. The angle with respect to the normal of the core-cladding boundary is 90° - \alpha. For TIR we need

90° - \alpha > \theta_c

\alpha < 90° - 80.54° \;=\; 9.46°

Why: the normal to the fibre wall is perpendicular to the fibre axis. If the ray makes angle \alpha with the axis, it makes 90° - \alpha with the normal. For TIR that angle must exceed \theta_c.

Step 3. The outside acceptance angle \alpha_\text{ext} — how steep the ray entering the fibre from air can be — follows from Snell at the entry face:

\sin\alpha_\text{ext} \;=\; n_\text{core}\sin\alpha_\text{max} \;=\; 1.475 \times \sin 9.46° \;=\; 1.475 \times 0.1644 \;=\; 0.2425

\alpha_\text{ext} \;=\; \arcsin(0.2425) \;=\; 14.03°

Why: the ray inside at \alpha = 9.46° came from an air-side angle \alpha_\text{ext}, related by Snell at the fibre's end face. A ray entering at a steeper angle would, after refraction into the core, exceed the TIR-safe angle and leak out. The quantity \sin\alpha_\text{ext} is called the numerical aperture (NA) of the fibre.

Result: Critical angle 80.54°; maximum internal axial angle 9.46°; external acceptance half-angle 14.03°, giving \text{NA} = 0.242.

What this shows: BSNL and Jio's single-mode fibres, strung under the Bay of Bengal and buried alongside NH-7 from Pune to Nagpur, guide light by TIR for hundreds of kilometres with losses measured in a fraction of a decibel per kilometre. The ray stays trapped because every reflection is perfect (no absorption by a metal backing), and the core-cladding index contrast is carefully chosen so \theta_c is near 90° — the fibre accepts only nearly-axial rays, which keeps dispersion small and data rate high. Your Netflix stream from a server in Singapore reaches Mumbai along exactly such a fibre.

Example 2: Why a diamond sparkles

A cut diamond has refractive index 2.42. Find its critical angle in air. Compare with glass (n = 1.50) and explain why a diamond looks brighter than a glass imitation.

Step 1. Critical angle of diamond-to-air.

\sin\theta_c \;=\; \frac{n_\text{air}}{n_\text{diamond}} \;=\; \frac{1.00}{2.42} \;=\; 0.4132

\theta_c^\text{diamond} \;=\; \arcsin(0.4132) \;=\; 24.41°

Step 2. Critical angle of glass-to-air.

\sin\theta_c^\text{glass} \;=\; \frac{1}{1.50} \;=\; 0.6667,\quad \theta_c^\text{glass} \;=\; 41.81°

Why: the denser the material, the smaller the critical angle. Diamond's critical angle is only about 24° — nearly half that of glass.

Step 3. Meaning. A light ray entering the top of a cut diamond must strike each internal face at an angle greater than 24° from the normal to escape. The geometry of a standard brilliant cut is designed so most rays make angles greater than 24° with each internal facet — so they do not escape on first try. They bounce around inside, exit only through the top (where the cut is flat and the ray meets the surface almost perpendicularly), and emerge as concentrated bright flashes.

In contrast, in a glass imitation with \theta_c = 42°, far more rays escape through side facets on first try — the stone leaks light, and it looks dull.

A ray entering a cut diamond meets each interior facet at an angle greater than $24°$, so it is totally reflected each time. After two or three bounces it emerges through the top, where it meets the surface at a smaller angle and escapes — concentrated into a bright flash.

Result: Diamond \theta_c = 24.4°; glass \theta_c = 41.8°. Diamond's much smaller critical angle traps light inside the stone and releases it only through the top — the physical basis of brilliance.

What this shows: the visible "sparkle" of a diamond is not mystical. It is TIR doing its job on a stone whose critical angle is the smallest of any common transparent material. Jewelers cut the facets at specific angles (the "pavilion angle" of 40.75° is a standard) precisely so that any ray entering the top must TIR off the lower facets before escaping through the top. A glass imitation cannot be cut to mimic this because \theta_c is too large — rays escape out the sides on first try.

Example 3: A periscope with a 45° prism

A periscope in an Indian Navy Scorpène-class submarine uses two right-angled glass prisms (index 1.50). A horizontal ray enters the upper prism, reflects off the 45° hypotenuse by TIR, travels down, and reflects again off the lower prism's hypotenuse to emerge horizontally into the observer's eye. Verify that TIR occurs at each prism.

Two right-angle prisms in a periscope. Light from above enters the upper prism, strikes the 45° hypotenuse internally, TIRs downward, travels along the tube, strikes the lower prism's hypotenuse at 45° internally, TIRs horizontally into the eye. No silvered mirrors anywhere — 100% efficient at each bounce.

Step 1. Critical angle for glass (n = 1.50).

\sin\theta_c \;=\; \frac{1.00}{1.50} \;=\; 0.667,\quad \theta_c \;=\; 41.81°

Step 2. Angle of incidence inside the prism. The ray enters the prism face perpendicularly (so it does not refract at entry). It then strikes the hypotenuse. The hypotenuse lies at 45° to the entry face, so the normal to the hypotenuse lies at 45° to the incident ray's direction.

Therefore the angle of incidence on the hypotenuse is

\theta_\text{hyp} \;=\; 45°

Step 3. Compare to critical angle.

\theta_\text{hyp} = 45° > \theta_c = 41.81°

Why: the incidence angle exceeds the critical angle, so TIR occurs at the hypotenuse. The ray reflects with the same 45° angle to the normal — which, because the hypotenuse is at 45° to vertical, means the reflected ray is perpendicular to the incident ray. It turns by exactly 90°.

Step 4. At the lower prism, the geometry is a mirror image. The downward ray hits the hypotenuse at 45°, TIRs into the horizontal direction, and exits perpendicular to the lower prism's side face (no refraction at exit).

Result: TIR occurs at both prisms because 45° > 41.81°. The submarine's captain sees the sea surface with no reflectivity loss — brighter than any silvered-mirror periscope.

What this shows: Indian Navy periscopes and classical binocular Porro prisms rely on TIR for exactly this reason. At 45° incidence in crown glass, TIR is guaranteed; the reflection is 100% efficient, needs no coating that can tarnish, and works for decades in salt-water environments. If the glass were cheap soda-lime with n < 1.41 (so \theta_c > 45°), the scheme would fail — at 45° the light would refract out of the hypotenuse, leaving only a weak partial reflection. The choice of crown glass is the physics-grade choice; lower-index glass would need silvering, and silver tarnishes underwater.

Common confusions

"TIR happens whenever light hits a water surface." Only when going from the denser side to the rarer side at an angle larger than the critical angle. Light entering a pool from above (air to water) cannot TIR — Snell's law always gives a valid refracted angle. It is the diver looking up who experiences TIR for shallow-angle rays from the pool walls.
"The reflected ray is less intense than 100% of the incident." For \theta_1 > \theta_c, every photon reflects. Energy cannot disappear — if no refracted ray carries it away, the reflected ray carries it all. Total means total.
"There is no light at all in the rarer medium when TIR happens." There is a thin evanescent wave — a field that decays exponentially within about one wavelength of the interface. It carries no net energy on its own, but can leak energy into a second medium placed very close to the surface. This is frustrated TIR and is used in optical couplers and ATR infrared spectroscopy. For most purposes (fibres, periscopes, diamonds), the evanescent wave is irrelevant.
"Snell's law is 'violated' during TIR." Snell's law is not violated — it simply has no real solution when \sin\theta_2 exceeds 1. Formally, one can extend it into complex angles and recover the evanescent wave, but the classical picture is: Snell's law predicts TIR by producing an impossible value, and physics responds by reflecting instead.
"A mirage is light actually bending upward." A mirage is TIR in a layer of hot air near the road surface. The hot layer has lower refractive index than the cooler air above. Light from the sky enters the hot layer from above at an angle larger than the layer's critical angle (measured with respect to the layer normal), TIRs back up, and reaches your eye. Your brain interprets the ray's apparent origin — below the road — as "water on the tar." The "puddle" is literally the sky, folded down by TIR.
"The critical angle depends on the colour of light." Yes — refractive index depends weakly on wavelength (dispersion), so \theta_c is slightly different for red and blue light. The difference is small (a fraction of a degree in water), but in a finely-cut diamond it splits the TIR'd white light into colours, adding "fire" (spectral flashes) to the brilliance. For most problems you can treat n as wavelength-independent without error.

You have TIR, the critical angle, and the applications. What follows is for readers who want the evanescent wave, the Fresnel equations applied to TIR, and the connection to quantum tunnelling.

The evanescent wave — light that shouldn't be there

When \theta_1 > \theta_c, the refracted ray in the rarer medium has a complex wavevector. The full electromagnetic solution still contains a field in the rarer medium, but it does not propagate — it decays exponentially away from the interface with decay length of order one wavelength.

Derivation. In the rarer medium the refracted wave has wavevector components

k_{\parallel} \;=\; n_1 k_0 \sin\theta_1 \;>\; n_2 k_0

where k_0 = \omega/c. The perpendicular component is

k_\perp^2 \;=\; (n_2 k_0)^2 - k_{\parallel}^2 \;<\; 0

so k_\perp is imaginary. Writing k_\perp = i\kappa,

E(\text{rarer},\,z) \;=\; E_0 e^{ik_{\parallel}x}\,e^{-\kappa z}

The wave travels along the interface (the x-direction) but decays exponentially with depth z into the rarer medium. The decay constant is

\kappa \;=\; k_0\sqrt{n_1^2 \sin^2\theta_1 - n_2^2}

For glass-air at \theta_1 = 60° and \lambda_0 = 500 nm: \kappa^{-1}\approx 120 nm — the evanescent field effectively vanishes within a fifth of a wavelength.

Why it carries no net energy. The time-averaged Poynting flux perpendicular to the interface is zero — all energy oscillates, nothing escapes. But if you bring a second glass surface within the decay length, the evanescent field couples to it and energy leaks through. This is frustrated TIR and is the basis of beamsplitters in laser interferometry.

Fresnel equations for TIR

The Fresnel equations for an interface give the reflection amplitudes r_s (for s-polarisation, \vec{E} perpendicular to incidence plane) and r_p (p-polarisation, \vec{E} in the plane):

r_s \;=\; \frac{n_1\cos\theta_1 - n_2\cos\theta_2}{n_1\cos\theta_1 + n_2\cos\theta_2}

r_p \;=\; \frac{n_2\cos\theta_1 - n_1\cos\theta_2}{n_2\cos\theta_1 + n_1\cos\theta_2}

For \theta_1 > \theta_c, \cos\theta_2 becomes imaginary. Plug in \cos\theta_2 = i\sqrt{\sin^2\theta_2 - 1}, and you find |r_s|^2 = |r_p|^2 = 1 — the intensity reflection is exactly 100%. That is why TIR is total, not 99%.

But: r_s and r_p acquire different phase shifts. The reflected s and p components emerge with different phases, so a linearly polarised incident wave becomes elliptically polarised on reflection. This phase-shift difference is the basis of the Fresnel rhomb — a glass parallelepiped that converts linearly to circularly polarised light using two successive TIRs.

Quantum-mechanical connection: tunnelling

Frustrated TIR is mathematically identical to quantum-mechanical tunnelling through a potential barrier. In quantum mechanics, a particle with energy less than a barrier height has a classically forbidden region in which its wavefunction decays exponentially — and if the barrier is finite in extent, some probability amplitude leaks through.

The correspondence is:

Optics	Quantum mechanics
Evanescent wave in rarer medium	Exponentially decaying wavefunction
TIR: no net energy flux	Reflection off a classically forbidden region
Frustrated TIR (thin gap)	Tunnelling through a finite barrier

So when you frustrate TIR with a closely-spaced second surface, you are doing the optical analogue of alpha-decay of a radioactive nucleus. Feynman's student days included a lecture series on exactly this analogy. It is not just an analogy — the same differential equation (Helmholtz / time-independent Schrödinger) governs both.

Numerical aperture and information capacity

For an optical fibre with core index n_1 and cladding n_2, the numerical aperture is

\text{NA} \;=\; \sqrt{n_1^2 - n_2^2}

This is the sine of the half-angle of the external acceptance cone. A BSNL single-mode fibre has NA \approx 0.14; a multi-mode fibre has NA \approx 0.27.

Higher NA means more rays can enter and be guided — more light, more modes. But more modes also means more modal dispersion: rays at steep angles take longer paths than axial rays, smearing pulses in time. For long-haul high-data-rate fibres (your Jio Fibre trunk from Mumbai to Chennai), single-mode with low NA is preferred. For short-range applications (within a building), multi-mode with higher NA is cheaper and works well.

The information capacity of a fibre is fundamentally limited by the Shannon-Hartley theorem, but the dispersion engineering that keeps a fibre useful at tens of terabits per second is essentially a game played with critical angles and mode numbers — TIR at every step.

Birefringent TIR and Nicol prisms

Some materials (calcite, quartz) are birefringent — they have two different refractive indices for two different polarisations. A Nicol prism, cut from calcite and glued with Canada balsam at a specific angle, uses selective TIR: the ordinary ray (higher index, smaller \theta_c) TIRs at the balsam layer and is absorbed, while the extraordinary ray (lower index, larger \theta_c) transmits. The emerging beam is pure linearly polarised light. This was the original polariser before Polaroid sheets were invented.

Where this leads next

Refraction of Light — Snell's Law — the precursor: ordinary refraction, the case before the critical angle.
Reflection of Light — the complementary phenomenon; TIR is reflection taken to 100% efficiency.
Prism and Dispersion — what happens when the same interface acts on different wavelengths differently.
Optical Fibres and Communication — the applied article on fibre optics, submarine cables, and why TIR underpins the internet.
The Human Eye and Defects of Vision — how the same principles apply inside the eye's lens and cornea.