Total internal reflection and its applications

Push a spoon into a glass of water at a shallow angle and, from the right viewpoint, the submerged part looks like a perfect mirror — no water is visible through it at all, only a silvery reflection. Nothing has been silvered. The light travelling inside the denser water, hitting the surface at a large enough angle, simply refuses to leave. Every last photon bounces back inside, obeying the ordinary law of reflection as if the water's surface had turned into polished metal. This is total internal reflection (TIR), and it is one of the most quietly powerful ideas in all of ray optics — not because the physics is exotic, but because a huge slice of modern technology runs on it. A diamond's fire, a mirage shimmering on a hot road, the glass fibres carrying your video call across an ocean in a fraction of a second, the periscope in a submarine, the binoculars around your neck — every one of these leans on light being trapped by TIR rather than escaping by ordinary refraction. The condition for it to happen is disarmingly simple: light must be travelling from a denser medium toward a rarer one, and it must strike the boundary at an angle steeper than a fixed threshold called the critical angle. Get the angle right and Snell's law itself breaks down — there is no refracted ray left for it to describe, only a perfect reflection. This article builds that condition from Snell's law with nothing beyond algebra and trigonometry, ties it to the exact critical-angle relation $\sin\theta_c = n_2/n_1$ , and then follows it into its flagship application: the optical fibre, whose core-cladding structure turns TIR into a near-lossless pipe for light and, with it, for the world's data.

Key takeaways

Total internal reflection (TIR) happens when light travelling from a denser medium ( $n_1$ ) toward a rarer medium ( $n_2 < n_1$ ) strikes the boundary at an angle of incidence greater than the critical angle $\theta_c$ — all the light reflects back, none refracts out.
Two conditions, both required: light must go from denser to rarer medium, AND the angle of incidence must exceed $\theta_c$ . Miss either one and TIR does not happen.
The critical-angle formula comes straight from Snell's law by setting the refraction angle to $90°$ : $\sin\theta_c = \dfrac{n_2}{n_1}$ .
At exactly $\theta_c$ , the refracted ray only grazes the surface ( $\theta_2 = 90°$ ); TIR proper needs $\theta_1$ strictly greater than $\theta_c$ .
Optical fibres use a higher-index core surrounded by a lower-index cladding so that light launched inside the core strikes the core-cladding wall above the critical angle every time, guiding it along the fibre by repeated TIR with almost no loss.
Everyday TIR: a diamond's brilliant sparkle (very small critical angle, about $24°$ ), the shimmering mirage above a hot road, totally reflecting glass prisms used in periscopes and binoculars instead of ordinary mirrors.
Key idea: when light travels from a denser to a rarer medium and strikes the interface beyond the critical angle given by sinθc = n2/n1, it cannot refract out at all and is reflected back completely — the mechanism that traps light inside optical fibres, diamonds, and totally reflecting prisms.

The core insight

Light going from a denser to a rarer medium at an angle beyond the critical angle sinθc = n2/n1 cannot refract out and is reflected back entirely, which is exactly how an optical fibre's core traps and guides light along its length.

Where it fits in the syllabus

Total internal reflection sits inside Unit IX, "Ray Optics and Optical Instruments," directly after refraction and Snell's law, and its headline application, optical fibres, is explicitly named in the CBSE 2025-26 syllabus. You are expected to state the two conditions for TIR, derive and use $\sin\theta_c = n_2/n_1$ , calculate critical angles for given media, and explain how an optical fibre's core-cladding geometry uses TIR to transmit light over long distances. The topic does not require anything beyond Snell's law algebra — no calculus, no wave treatment — so every derivation and worked example below stays at that level.

The physical picture

Consider a ray of light travelling inside an optically denser medium of refractive index $n_1$ and approaching the flat boundary with a rarer medium of refractive index $n_2$ , where $n_2 < n_1$ . Let the angle of incidence (measured from the normal, inside the denser medium) be $\theta_1$ , and the angle of refraction (inside the rarer medium) be $\theta_2$ . Snell's law connects the two media at every angle:

n_1 \sin\theta_1 = n_2 \sin\theta_2

where: $n_1, n_2$ are the refractive indices of the denser and rarer media, and $\theta_1, \theta_2$ are the angles of incidence and refraction, both measured from the normal to the surface. Because $n_1 > n_2$ , this relation forces $\sin\theta_2 > \sin\theta_1$ , so the refracted ray always bends away from the normal, into the rarer medium — a ray entering a swimming pool's water surface from below always emerges tilted farther from vertical than it went in. As $\theta_1$ is increased steadily, $\theta_2$ grows even faster, since it is being stretched away from the normal. There must therefore come a point where $\theta_2$ reaches its maximum possible value, $90°$ — the refracted ray skims along the boundary itself instead of leaving the surface. The angle of incidence at which this happens is called the critical angle, $\theta_c$ . Push $\theta_1$ beyond $\theta_c$ and Snell's law would demand $\sin\theta_2 > 1$ , which no real angle can satisfy — so a refracted ray simply ceases to exist, and all of the light energy is reflected back into the denser medium, obeying the ordinary law of reflection (angle of reflection equals angle of incidence). That is total internal reflection: not a special mirror coating, but the mathematically forced outcome of Snell's law running out of room.

Figure

A ray inside a denser medium meeting a boundary with a rarer medium at three angles of incidence, showing a small angle giving a weak reflection and strong refraction, the critical angle giving a refracted ray that grazes along the boundary, and a larger angle giving total internal reflection with no refracted ray at all

Two conditions must both hold for TIR to occur: first, the ray must be travelling from the denser medium toward the rarer one (going the other way, from rare to dense, a refracted ray exists at every angle of incidence, so TIR never happens); second, the angle of incidence must exceed the critical angle for that particular pair of media. The critical angle itself is not a universal number — it depends only on the ratio of the two refractive indices, so a glass-air boundary has a different $\theta_c$ from a glass-water boundary or a diamond-air boundary.

Derivation ARelation between the critical angle and refractive index

The goal is to derive the exact relation between the critical angle $\theta_c$ and the refractive indices $n_1$ (denser medium) and $n_2$ (rarer medium), using nothing beyond Snell's law and the definition of the critical angle as the angle of incidence at which the refracted ray grazes the boundary. This single formula is the working tool behind every critical-angle numerical in the chapter, from glass-air interfaces to optical fibre core-cladding boundaries, so it is worth deriving cleanly rather than only quoting.

Step 1. Write Snell's law for the boundary between the denser medium (refractive index $n_1$ , angle of incidence $\theta_1$ ) and the rarer medium (refractive index $n_2$ , angle of refraction $\theta_2$ ): $n_1 \sin\theta_1 = n_2 \sin\theta_2$ . This holds for every angle of incidence for which a refracted ray exists.

Step 2. Apply the defining condition for the critical angle. By definition, $\theta_c$ is the particular value of $\theta_1$ at which the refracted ray no longer bends into the rarer medium at any ordinary angle but instead skims exactly along the boundary surface, so $\theta_2 = 90°$ . Substitute $\theta_1 = \theta_c$ and $\theta_2 = 90°$ into Snell's law: $n_1 \sin\theta_c = n_2 \sin 90°$ .

Step 3. Simplify using $\sin 90° = 1$ . This gives $n_1 \sin\theta_c = n_2$ , and rearranging for $\sin\theta_c$ : \sin\theta_c = \frac{n_2}{n_1} where $n_1$ is always the refractive index of the denser (incident) medium and $n_2$ that of the rarer medium, so the ratio $n_2/n_1$ is always less than $1$ , guaranteeing that $\theta_c$ is a real, well-defined angle between $0°$ and $90°$ .

Step 4. Interpret what happens beyond $\theta_c$ . For any angle of incidence $\theta_1 > \theta_c$ , Snell's law would require $\sin\theta_2 = (n_1/n_2)\sin\theta_1 > (n_1/n_2)\sin\theta_c = 1$ , which is impossible since sine can never exceed $1$ . No refracted ray can satisfy Snell's law, so the boundary cannot transmit any light at that angle — by conservation of energy, the entire incident intensity is reflected back into the denser medium instead, following the ordinary law of reflection. This is total internal reflection, and $\sin\theta_c = n_2/n_1$ is the exact boundary between "partial refraction plus partial reflection" and "total reflection, zero refraction." Special case worth remembering: when the rarer medium is air, $n_2 \approx 1$ , so the formula simplifies to $\sin\theta_c = 1/n_1$ , or equivalently $n_1 = 1/\sin\theta_c$ — a quick way to read a medium's refractive index straight off a measured critical angle.

Applications you should recognise

Total internal reflection is not just a critical-angle numerical — it is the working principle behind several devices and everyday phenomena that the syllabus expects you to connect back to the same $\sin\theta_c = n_2/n_1$ idea.

Optical fibres. A fibre is a thin, flexible strand with a core of higher refractive index surrounded by a cladding of slightly lower refractive index. Light launched into the core at a suitable angle strikes the core-cladding wall above the critical angle every single time it reflects, so it undergoes repeated TIR and travels the whole length of the fibre — even around gentle bends — with very little loss of intensity. This is the physical basis of modern optical telecommunication (carrying enormous amounts of data as light pulses) and of medical endoscopes (carrying images out of the body along a flexible bundle of fibres).

Totally reflecting prisms. A right-angled glass prism with angles $45°$ - $45°$ - $90°$ can turn or invert a beam of light through $90°$ or $180°$ purely by TIR at its hypotenuse face, with essentially no light loss — unlike a silvered mirror, which always absorbs a small fraction of the incident light at the metal coating. Because ordinary glass has a critical angle for the glass-air interface of about $42°$ , well below the $45°$ angle at which the light strikes the hypotenuse inside such a prism, TIR is guaranteed. This trick is used inside periscopes and binoculars in place of mirrors.

Diamond's sparkle. Diamond has an unusually high refractive index (around $2.4$ ), which by $\sin\theta_c = 1/n$ gives it a very small critical angle, close to $24°$ . Light entering a cut diamond strikes its many angled internal facets well above this small critical angle almost everywhere, so it undergoes TIR repeatedly inside the stone before finally emerging — the reason a well-cut diamond throws back so much brilliant light instead of letting it leak straight through, unlike ordinary glass with its larger critical angle near $42°$ .

Mirage. On a very hot road, air just above the surface is much hotter (and hence less dense, lower refractive index) than the cooler air higher up. Light from the sky travelling down through progressively rarer layers of hot air can curve and eventually strike a layer at an angle beyond the local critical angle, undergoing TIR and bending back upward — an observer sees an apparently reflected, shimmering "pool" image of the sky on the road surface where there is no water at all.

Worked examples

Board level - critical angle for a glass-air interface

The refractive index of glass with respect to air is $1.5$ . Calculate the critical angle for a glass-air interface, and state the condition under which light travelling inside the glass undergoes total internal reflection at this interface.

Step 1. Identify the media. The denser medium is glass, $n_1 = 1.5$ ; the rarer medium is air, $n_2 = 1$ . Since light is going from glass (denser) toward air (rarer), TIR is at least geometrically possible here.

Step 2. Apply $\sin\theta_c = n_2/n_1$ : \sin\theta_c = \frac{1}{1.5} = 0.667

Step 3. Invert to find the angle: \theta_c = \sin^{-1}(0.667) \approx 41.8°

Step 4. State the condition. Total internal reflection occurs for a ray travelling from inside the glass toward the glass-air surface only if its angle of incidence at that surface exceeds $41.8°$ . For angles of incidence smaller than $41.8°$ , part of the light refracts out into the air (bent away from the normal) and part reflects partially back into the glass — TIR does not occur. Answer: $\theta_c \approx 41.8°$ ; TIR occurs only when $\theta_1 > 41.8°$ .

Competitive level - critical angle at an optical fibre's core-cladding boundary, and the effect of a sharper index difference

An optical fibre has a core of refractive index $1.55$ and a cladding of refractive index $1.45$ . (a) Find the critical angle at the core-cladding boundary. (b) A second fibre design uses a core of $1.60$ and cladding of $1.40$ instead. Find its critical angle, and explain which fibre tolerates sharper bends without losing light.

Step 1. For the first fibre, the core is denser, $n_1 = 1.55$ , and the cladding is rarer, $n_2 = 1.45$ . Apply $\sin\theta_c = n_2/n_1$ : \sin\theta_c = \frac{1.45}{1.55} = 0.935 \implies \theta_c = \sin^{-1}(0.935) \approx 69.2°

Step 2. For the second fibre, $n_1 = 1.60$ , $n_2 = 1.40$ : \sin\theta_c = \frac{1.40}{1.60} = 0.875 \implies \theta_c = \sin^{-1}(0.875) \approx 61.0°

Step 3. Compare the two critical angles. The second fibre, with the larger index difference, has the smaller critical angle ( $61.0°$ versus $69.2°$ ). A smaller $\theta_c$ means a wider range of angles of incidence (everything above $61.0°$ , rather than only above $69.2°$ ) still satisfies the TIR condition.

Step 4. Connect to bending tolerance. When a fibre bends, rays that were travelling nearly parallel to the core axis strike the core-cladding wall at a steeper angle (closer to the normal) than they would in a straight section. If that angle falls below $\theta_c$ , the ray escapes by refraction instead of reflecting, and light is lost at the bend. Because the second fibre's smaller $\theta_c$ tolerates a wider range of steep angles before losing TIR, it can survive sharper bends without significant signal loss. Answer: $\theta_c \approx 69.2°$ for the first fibre, $\theta_c \approx 61.0°$ for the second; the fibre with the larger core-cladding index difference (the second one) tolerates sharper bends. Timing note: both parts are two Snell's-law substitutions and an inverse-sine lookup — budget under two minutes total in an exam; do not attempt to derive $\sin\theta_c = n_2/n_1$ from scratch again inside a numerical, just quote and apply it.

Common traps

Thinking TIR can occur when light travels from a rarer to a denser medium — it cannot; that direction always has a real refracted ray for every angle of incidence, since $\sin\theta_2 = (n_1/n_2)\sin\theta_1$ only ever shrinks, never exceeds $1$ .
Confusing the critical angle with the actual angle of incidence in a numerical — $\theta_c$ is a fixed property of the pair of media given by $\sin\theta_c = n_2/n_1$ , while the angle of incidence is whatever the ray's geometry in the problem dictates; TIR happens only when the latter exceeds the former.
Believing that at exactly $\theta_1 = \theta_c$ the light is already totally internally reflected — at the critical angle itself the refracted ray only grazes the surface ( $\theta_2 = 90°$ ) with vanishing transmitted intensity; strict TIR (zero transmission) needs $\theta_1$ to be strictly greater than $\theta_c$ .
Writing $\sin\theta_c = n_1/n_2$ upside down — remember $n_1$ is always the denser (incident-side) medium in the denominator, so the ratio $n_2/n_1$ is always less than one and gives a sensible angle.
Assuming optical fibres work by ordinary reflection off a silvered or mirrored inner wall — the guiding mechanism is specifically total internal reflection at the core-cladding boundary, driven purely by the refractive-index step, which is why it stays essentially lossless over kilometres compared to any real metallic coating.

Common trapTIR needs both denser-to-rarer travel and an angle beyond the critical angle — either condition alone is not enough

A very common half-answer states only "the angle of incidence must be more than the critical angle" and forgets to say the ray must also be going from the denser medium into the rarer one. Both conditions are independently necessary. A ray travelling from air into glass at a large angle never undergoes TIR no matter how large that angle gets, because it is going the wrong way (rarer to denser) and a refracted ray always exists for it. Conversely, a ray travelling glass-to-air at an angle smaller than $\theta_c$ also does not undergo TIR, even though the direction is correct, because the angle condition fails. Examiners specifically reward stating both conditions together, not just the angle inequality.

You should be able to…

State the two necessary conditions for total internal reflection to occur, and explain why both are required.
Derive the relation $\sin\theta_c = n_2/n_1$ between the critical angle and the refractive indices of the two media, starting from Snell's law.
Calculate the critical angle for a given pair of media, and determine whether TIR occurs for a stated angle of incidence.
Explain, using TIR, how an optical fibre's core-cladding structure guides light along its length with minimal loss.
Identify and explain, in terms of critical angle, at least two everyday or engineered applications of TIR: totally reflecting prisms, diamond sparkle, or mirage.
Avoid the common trap of stating only the angle condition for TIR while forgetting the denser-to-rarer direction requirement.

Ask the tutor: want to work through why a diamond's small critical angle makes cut faceting angles so important to a jeweller, or see the geometry of how a $45°$ - $45°$ - $90°$ prism turns a beam through a right angle using nothing but TIR — ask the AI tutor to walk through it with you.

Why does TIR only happen going from denser to rarer, and never the other way

It comes directly from Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ . Going from denser ( $n_1$ ) to rarer ( $n_2$ ), $\sin\theta_2 = (n_1/n_2)\sin\theta_1$ can exceed $1$ once $\theta_1$ is large enough, since the multiplying factor $n_1/n_2$ is bigger than one — and that impossible situation is exactly what forces TIR. Going the other way, from rarer to denser, the multiplying factor $n_2/n_1$ is less than one, so $\sin\theta_2$ can never exceed $1$ no matter how large $\theta_1$ gets; a refracted ray always exists. That asymmetry between the two directions is built into Snell's law itself, not an extra rule you have to memorise separately.

Is the critical angle the same for every pair of media, like a universal constant

No — it depends entirely on the specific pair of refractive indices through $\sin\theta_c = n_2/n_1$ , so it is different for every combination. Glass-air gives roughly $42°$ , diamond-air gives roughly $24°$ , and a typical optical fibre's core-cladding boundary (glass-glass, with a small index difference) gives something much larger, often above $60°$ or $70°$ , because $n_2/n_1$ is close to $1$ when the two indices are close together. Whenever a numerical gives you two refractive indices, you must recompute $\theta_c$ for that specific pair rather than recalling a memorised number from a different problem.

Why does an optical fibre need a separate cladding instead of just being a bare glass strand surrounded by air

A bare strand surrounded by air would technically also support TIR, since air has a much lower refractive index than glass. The problem is practical: any contact with a surface, a smudge, dust, or even another fibre touching it, would locally change what the "rarer medium" at that spot actually is, and could let light leak out or let light from a neighbouring fibre leak in (called cross-talk). A cladding of controlled, slightly lower refractive index, permanently bonded to the core, fixes the refractive-index step precisely everywhere along the fibre's length regardless of what touches the outside of the cladding, and it also protects the light-carrying core mechanically. Fibre engineers choose the exact core-cladding index difference deliberately, since it sets the critical angle and therefore how sharply the fibre can bend without losing light.

Does total internal reflection lose any light at all, or is it genuinely 100 percent

Within the idealisation used in this syllabus (perfectly smooth boundary, single wavelength, geometric optics), TIR reflects exactly $100\%$ of the incident intensity — that is precisely why the formula $\sin\theta_c = n_2/n_1$ marks a sharp boundary between "some light refracts out" and "none does," rather than a gradual fade. In real optical fibres over long distances, some signal loss still occurs, but that comes from other physical effects, absorption and scattering by impurities in the glass, microscopic surface roughness, tiny bends, not from the TIR mechanism itself failing. For CBSE Class 12 purposes, treat TIR as a lossless, complete reflection whenever the two stated conditions are satisfied.