Total Internal Reflection and Critical Angle

This article draws a clean line through one of optics' most useful boundaries. The geometry of total internal reflection — the critical angle $\theta_c$ beyond which a ray trapped in dense glass can no longer escape, and the fibre and prism technologies built on it — is geometrical optics, and it is owned here. The dynamics — how the reflected power climbs to 100%, the ghostly evanescent field that leaks a wavelength into the forbidden medium, and the polarization-dependent phase the reflection stamps on the wave — is Fresnel's electromagnetic business, recalled by name from the companion article aq-em-brewster-angle-total-internal-reflection and not re-derived. Keep the split in mind throughout: Snell's law tells you when light is trapped; Maxwell's boundary conditions tell you what the trapped field does. That single geometric fact — light steeply enough incident from inside glass simply cannot get out — is quietly one of the most consequential in applied physics. It is why a hair-thin fibre pipes this sentence across an ocean with almost no leakage; why a cut diamond blazes, its critical angle so small that light ricochets from facet to facet, gathering colour, before escaping upward as fire; and why the best periscopes and binoculars fold their light with plain glass prisms rather than silvered mirrors — TIR reflects everything, losslessly, with no coating to tarnish. One threshold angle, fixed by a ratio of refractive indices, underwrites all three.

Key takeaways

Total internal reflection happens only dense → rare ( $n_1 > n_2$ ): beyond the critical angle $\theta_c = \arcsin(n_2/n_1)$ , Snell's law has no real transmitted angle and the whole ray is thrown back into the dense medium.
At $\theta_c$ the refracted ray grazes the interface ( $\theta_t = 90^\circ$ ); a hair steeper and it is gone. Diamond $24.4^\circ$ , crown glass $41.1^\circ$ , water $48.8^\circ$ .
An optical fibre is a TIR pipe: a high-index core in lower-index cladding traps every ray striking the wall beyond $\theta_c$ ; its acceptance cone has $\mathrm{NA} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ .
TIR is perfectly lossless in transparent media — which is why glass prisms out-reflect metal mirrors and need no coating.
Recalled from electromagnetism: beyond $\theta_c$ the reflectance is $R = 1$ , the transmitted field is an evanescent wave decaying within ~a wavelength (no net power across), and the reflection imprints a polarization-dependent phase (the Fresnel rhomb).
Key idea: past the critical angle $\theta_c = \arcsin(n_2/n_1)$ a ray in a dense medium cannot refract into a rarer one and is totally reflected — the geometry optics owns; the unit reflectance, the evanescent skin, and the reflection phase are the electromagnetic content recalled from the Fresnel treatment.

The core insight

Going from a denser medium to a rarer one, there is a critical incidence angle $\theta_c = \arcsin(n_2/n_1)$ at which the refracted ray grazes the surface. Beyond it Snell's law demands $\sin\theta_t > 1$ , which no real angle satisfies, so the light is totally reflected. The critical angle and its applications — fibres, prisms, refractometry — are geometrical optics; the unit reflectance, evanescent field, and reflection phase are electromagnetic results recalled by name.

Where this lives today

Total internal reflection is arguably the most economically consequential single piece of boundary geometry ever put to work: every optical fibre is a TIR machine, and the global internet rides on rays that strike a glass wall beyond $\theta_c$ millions of times between amplifiers. The same trick, made flexible, is the medical endoscope and the borescope inside a jet engine. Reflecting prisms fold and erect images in binoculars and periscopes by TIR at $45^\circ$ , past crown glass's $41.1^\circ$ , more efficiently than any silvered mirror and without tarnishing. The evanescent skin the electromagnetic treatment describes is now an instrument in its own right: ATR infrared spectroscopy, TIRF microscopy imaging single molecules in the $\sim 100$ nm slice above a coverslip, and optical fingerprint scanners all read the frustrated evanescent field. And a diver looking up still sees Snell's window — the whole sky crushed into a bright $97.5^\circ$ cone overhead, mirror-dark water beyond its rim.

The physical picture

Picture a ray fanning around inside a block of glass, aimed at the top surface. Strike it nearly head-on and it punches straight out into the air. Tilt it, and the refracted ray bends away from the normal — flatter and flatter as the incidence grows, because the air cannot support as steep a wave as the glass can. At one special angle the escaping ray lies flat along the surface at $90^\circ$ ; nudge the incidence past it and there is nowhere for a propagating ray to go — the surface hands all the light back, as cleanly as a perfect mirror. That handover is the critical angle, and everything here hangs off it.

Figure

Rays from dense glass at the surface: shallow (25°) refracts out, at θc (41.1°) it grazes along the surface, steep (55°) is totally reflected

Notice how asymmetric the setup is. Total internal reflection only exists on the dense side: going the other way, from air into glass, the refracted ray bends toward the normal and always finds a home, so there is no critical angle for light trying to get in — the trap is one-directional. And $\theta_c$ depends on nothing but the index ratio $n_2/n_1$ — no wavelength, no intensity, no polarization — which makes it a clean, tabulated property of any interface. The next figure turns the whole family of interfaces into a single curve.

Figure

Critical angle vs index ratio: the arcsin(n₂/n₁) curve, with diamond 24.4°, crown glass 41.1°, and water 48.8° marked on it

The curve is the whole geometry in one stroke. Left to right, $\theta_c$ rises as the index ratio climbs toward one: match the two indices and $\theta_c \to 90^\circ$ , so only a ray already skimming the surface reflects and TIR all but disappears. Right to left, $\theta_c$ collapses toward zero as the contrast grows — a very dense medium in air traps almost everything. Diamond sits far down the curve at $24.4^\circ$ because its index, $2.417$ , is so large, which is exactly why it sparkles.

The core equations

Everything begins with Snell's law and one question: how large can the incidence angle get before the refracted ray runs out of surface to lie along? Setting the refracted angle to its maximum, $90^\circ$ , converts Snell's law directly into the critical angle. The condition $n_1 > n_2$ is essential — it is what makes the threshold fall below $90^\circ$ .

\theta_c = \arcsin\!\left(\frac{n_2}{n_1}\right), \qquad n_1 > n_2 \qquad (1)

where: $\theta_c$ — critical angle, measured from the surface normal; $n_1$ — index of the incidence (denser) medium; $n_2$ — index of the transmission (rarer) medium. Because the arcsine only accepts arguments up to $1$ , a real $\theta_c$ exists only when $n_2 < n_1$ : the trap is one-way. Push the incidence past $\theta_c$ and Snell's law does not so much break as run out of real numbers. Solving it for the transmitted angle gives a sine larger than one, which no real angle can produce — so there is no propagating refracted ray at all, and the boundary reflects the entire beam.

\theta_i > \theta_c \;\Rightarrow\; \sin\theta_t = \frac{n_1}{n_2}\sin\theta_i > 1 \qquad (2)

where: $\theta_i$ — angle of incidence, now steeper than $\theta_c$ ; $\theta_t$ — the transmitted angle Snell's law would require. There is no real solution, and geometrically that is the whole content of total internal reflection: no transmitted ray means all the light comes back. (What the transmitted field does instead is the electromagnetic story below.) The most important application turns the same idea inside out. A fibre guides light by making sure every ray strikes its core–cladding wall beyond the wall's own critical angle; carrying one ray through the entrance face to the wall (Derivation B) collapses that condition into a single number — the numerical aperture — and the acceptance-cone half-angle.

\mathrm{NA} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}, \qquad \theta_a = \arcsin\!\frac{\mathrm{NA}}{n_0} \qquad (3)

where: $\mathrm{NA}$ — numerical aperture, the fibre's light-gathering power; $n_{\text{core}}, n_{\text{clad}}$ — core and cladding indices ( $n_{\text{core}} > n_{\text{clad}}$ ); $n_0$ — index at the input face (air, $n_0 = 1$ ); $\theta_a$ — acceptance half-angle. The core index cancels out of NA entirely — the acceptance cone depends only on the contrast between core and cladding.

Recalled from electromagnetism. The geometry above says nothing about how much light reflects or what the field does past the wall. That physics belongs to the companion article aq-em-brewster-angle-total-internal-reflection, and this article recalls three of its results by name rather than re-deriving them. First, beyond $\theta_c$ the Fresnel coefficients satisfy $|r_s| = |r_p| = 1$ , so the reflectance is $R = 1$ in both polarizations — all the power is reflected. Second, the transmitted field does not vanish but becomes an evanescent wave, $E \propto e^{-\kappa z}$ , decaying into the rarer medium and carrying zero time-averaged power across it. Its $1/e$ depth scale is

\delta = \frac{\lambda_0}{2\pi\sqrt{n_1^2\sin^2\theta_i - n_2^2}} \qquad (4)

where: $\delta$ — penetration depth of the evanescent field into medium 2; $\lambda_0$ — vacuum wavelength. For crown glass into air at $45^\circ$ with red light ( $633$ nm), $\delta \approx 256$ nm — a fraction of a wavelength — diverging as $\theta_i \to \theta_c^+$ . Third, total reflection is not phase-neutral: it imprints a polarization-dependent phase shift on the reflected wave ( $|r| = 1$ but $\arg r \neq 0$ ), the effect Fresnel folded into his rhomb to turn linear light circular. These three — unit reflectance, evanescent skin, reflection phase — are the electromagnetic content. Within our own track this article owns the critical-angle derivation: later articles on polarization-on-reflection and optical waveguiding recall $\theta_c = \arcsin(n_2/n_1)$ from here by name.

Derivation ACritical angle from Snell's law

Our aim is to pin down the exact incidence angle at which a ray trying to leave a dense medium runs out of room — the critical angle — and to show it is nothing more mysterious than Snell's law evaluated at its own breaking point. We push the refracted ray flatter by raising the incidence angle and ask for the incidence at which the transmitted ray lies exactly along the surface; beyond it Snell's law has no real solution, and the only place left for the light to go is back into the dense medium.

Step 1. Write Snell's law across the boundary, $n_1\sin\theta_i = n_2\sin\theta_t$ , which is really the statement that the tangential wavevector $k_x = n_1 k_0 \sin\theta_i$ is continuous across the interface. Why: the wave must stay in step with itself on both sides, so the wavevector component along the surface cannot jump — refraction is exactly this bookkeeping.

Step 2. Because $n_1 > n_2$ , the ratio $n_1/n_2 > 1$ , so as $\theta_i$ grows the required $\sin\theta_t = (n_1/n_2)\sin\theta_i$ grows faster still and reaches $1$ while $\theta_i$ is still well below $90^\circ$ ; the refracted ray flattens toward the surface. Why: a genuine threshold exists because the rarer medium cannot support as large a tangential wavenumber as the denser one — there is a ceiling to bend against.

Step 3. Set the ceiling. The largest real transmitted angle is $\theta_t = 90^\circ$ ( $\sin\theta_t = 1$ ); here the refracted ray skims the interface, the last incidence for which any propagating transmitted ray exists. Why: $90^\circ$ is the boundary between "some light escapes" and "none does", so the incidence that produces it is the dividing line.

Step 4. Solve at the ceiling: $n_1\sin\theta_c = n_2$ , hence $\sin\theta_c = n_2/n_1$ and $\theta_c = \arcsin(n_2/n_1)$ . Any steeper incidence would demand $\sin\theta_t > 1$ — no real angle — so the whole beam is totally reflected. Why: this is the owned result; every application below is just this one arcsine read off for a particular index pair. *Verified symbolically (Snell at $\theta_t = 90^\circ \Rightarrow \theta_c = \arcsin(n_2/n_1)$ ) and numerically for diamond ( $24.4^\circ$ ), crown glass ( $41.1^\circ$ ), and water ( $48.8^\circ$ ) in verify_total-internal-reflection.py.*

Go deeper · Derivation B · 1 minFibre numerical aperture and the acceptance cone

Derivation BFibre numerical aperture and the acceptance cone

A fibre traps light by arranging for every guided ray to strike the core wall beyond the critical angle. The goal here is to convert that internal condition into an external one a user can aim for: how wide a cone of incoming light, entering the flat end face, ends up trapped? Following one meridional ray through the end-face refraction and on to the core–cladding wall turns the trapping condition into a single number — the numerical aperture — and its acceptance-cone half-angle.

Step 1. A ray enters the flat end face from air ( $n_0$ ) at angle $\theta_a$ to the fibre axis and refracts to $\theta_r$ inside the core: $n_0\sin\theta_a = n_{\text{core}}\sin\theta_r$ . Why: the acceptance cone is specified in the outside medium, so we carry the ray through the entrance refraction before talking about the wall.

Step 2. That ray meets the core–cladding wall at an incidence of $90^\circ - \theta_r$ (from the wall's normal). Guiding requires TIR there: $90^\circ - \theta_r \ge \theta_{c,\text{wall}} = \arcsin(n_{\text{clad}}/n_{\text{core}})$ , i.e. $\cos\theta_r \ge n_{\text{clad}}/n_{\text{core}}$ . Why: the wall is where the trapping happens; only rays that hit it past its own critical angle stay inside.

Step 3. Take the marginal ray, $\cos\theta_r = n_{\text{clad}}/n_{\text{core}}$ , so $\sin\theta_r = \sqrt{1 - n_{\text{clad}}^2/n_{\text{core}}^2}$ . Substitute into the entrance relation: $n_0\sin\theta_a = n_{\text{core}}\sin\theta_r = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ . Why: the $n_{\text{core}}$ cancels cleanly, leaving a quantity that depends only on the two glass indices — the fibre's intrinsic light-gathering power.

Step 4. Name it and read off the cone: $\mathrm{NA} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ and acceptance half-angle $\theta_a = \arcsin(\mathrm{NA}/n_0)$ . Why: NA fixes both how much light the fibre gathers and how tightly a source must be focused in — the single fibre spec a designer reaches for first. *Verified: the $n_{\text{core}}$ cancellation and $\mathrm{NA} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ checked symbolically, and the numbers $n_{\text{core}} = 1.50$ , $n_{\text{clad}} = 1.48 \Rightarrow \mathrm{NA} = 0.244$ , $\theta_a = 14.1^\circ$ in verify_total-internal-reflection.py.*

Limiting cases

Case	Result	Meaning
$n_2 \to n_1$ (matched indices)	$\theta_c \to 90^\circ$	the trap all but vanishes: only near-grazing rays reflect
$n_1 \gg n_2$ (high contrast)	$\theta_c \to 0$	almost every internal ray is trapped — why a high-index gem holds light
diamond $\to$ air ( $n_1 = 2.417$ )	$\theta_c = 24.4^\circ$	tiny escape cone; light ricochets internally before exiting — the cut's fire
crown glass $\to$ air ( $n_1 = 1.52$ )	$\theta_c = 41.1^\circ$	below $45^\circ$ , so a $45^\circ$ prism reflects totally without a coating
water $\to$ air ( $n_1 = 1.33$ )	$\theta_c = 48.8^\circ$	Snell's window: the sky above a diver squeezed into a $97.5^\circ$ cone
$\theta_i \to \theta_c^+$	$\delta \to \infty$	the evanescent field (EM-recalled) reaches ever deeper into medium 2 at the edge

Common trap“In total internal reflection no light gets past the surface at all”

It is total in energy, not in field. Reflection sends back 100% of the power ( $R = 1$ ) and no net energy crosses the boundary — but a real, measurable field does poke through: the evanescent wave, decaying over roughly a wavelength into the rarer medium (recalled from aq-em-brewster-angle-total-internal-reflection). For crown glass into air at $45^\circ$ its $1/e$ depth is only about $256$ nm, but it is emphatically there, and you can tap it. Bring a second surface within that skin — another prism, a fingerprint ridge, a droplet — and light tunnels across the gap it supposedly could not cross: frustrated total internal reflection, the basis of beam-splitter cubes, fingerprint scanners, and near-field microscopy. Treat "totally reflected" as "no field beyond the surface" and you miss all of these — and why a TIR mirror leaks the instant something touches its face.

Worked examples

1 · Numerical aperture of a step-index fibre (confidence builder)

A step-index fibre has core index $n_{\text{core}} = 1.50$ and cladding index $n_{\text{clad}} = 1.48$ , used in air ( $n_0 = 1$ ). Find its numerical aperture and the half-angle of the cone of light it will accept and guide.

Step 1. Numerical aperture straight from Derivation B: $\mathrm{NA} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2} = \sqrt{1.50^2 - 1.48^2} = \sqrt{2.2500 - 2.1904} = \sqrt{0.0596} = \mathbf{0.244}$ .

Step 2. Acceptance half-angle: $\theta_a = \arcsin(\mathrm{NA}/n_0) = \arcsin(0.244) = \mathbf{14.1^\circ}$ — a full cone of about $28^\circ$ . A ray arriving within $\sim 14^\circ$ of the axis is captured; anything steeper leaks into the cladding within a few bounces.

Step 3. Check the bookkeeping: the marginal accepted ray refracts to $\theta_r = \arcsin(\sin 14.1^\circ / 1.50) = 9.4^\circ$ and strikes the wall at $90^\circ - 9.4^\circ = 80.6^\circ$ — exactly its critical angle $\arcsin(1.48/1.50) = 80.6^\circ$ . The widest accepted ray is the one that just barely undergoes TIR at the wall.

Step 4. The design lever: NA grows with index contrast. Dropping the cladding to $1.46$ gives $\mathrm{NA} = 0.344$ , $\theta_a = 20.1^\circ$ — more light in, but more modes and dispersion. The $0.02$ contrast of a real telecom fibre is a deliberate compromise between light collection and bandwidth. NA = 0.244, acceptance half-angle 14.1°.

2 · The diamond critical angle and why it sparkles (exam level)

A brilliant-cut diamond ( $n = 2.417$ ) sits in air. Find its critical angle, and show quantitatively why so little light escapes at each internal facet — the origin of the gem's fire.

Step 1. Critical angle: $\theta_c = \arcsin(1/2.417) = \arcsin(0.414) = \mathbf{24.4^\circ}$ . For comparison, crown glass gives $\arcsin(1/1.52) = 41.1^\circ$ : diamond's threshold is far lower.

Step 2. What a small $\theta_c$ buys: a ray can leave a facet only if it strikes within $\theta_c$ of that facet's normal — its escape cone — and the fraction of the facet's inward hemisphere inside that cone is $1 - \cos\theta_c$ . For diamond, $1 - \cos 24.4^\circ = \mathbf{0.089}$ : only about $9\%$ of internal directions escape at a facet, so roughly $91\%$ are totally reflected. For crown glass, $1 - \cos 41.1^\circ = 0.247$ — about $25\%$ escape, only $75\%$ reflect.

Step 3. Multiply it up. A ray typically strikes several pavement facets — cut so that light is thrown below $\theta_c$ at the sides and sent back upward — before finding an escape through the top, and each extra bounce is another internal path over which dispersion fans the colours apart. A shallow "brilliant" cut is engineered so that almost all light entering the crown makes two or more TIR bounces and leaves through the top as separated spectral colours: fire.

Step 4. The contrast is the whole story. A glass replica of identical shape, with its larger $41.1^\circ$ critical angle, lets a quarter of the light dribble out the sides at every bounce, so far less returns through the crown and the stone looks dull. Diamond's brilliance is not chiefly a high reflectivity per bounce — it is the small critical angle that keeps light bouncing. θc = 24.4°, escape fraction ≈ 9% per facet (versus ≈ 25% for glass).

Physics of the portrait

Total internal reflection traps light inside a transparent solid — and the man who first made a room full of people watch a beam of light bend, by trapping it inside a falling stream of water, turned it into spectacle.

Origins · 1 minJohn Tyndall and how this was discovered

John Tyndall

1820–1893 · London (Royal Institution)

John Tyndall was an Irish surveyor's son who worked on railway construction and taught school to fund a Marburg doctorate, then rose to succeed Michael Faraday as the Royal Institution's great public experimentalist. In an 1854 demonstration there he let water pour from a tank through a nozzle and shone light into the jet: the beam stayed inside the curving stream, guided bounce by bounce by total internal reflection — a "light fountain" that is the direct conceptual ancestor of the optical fibre. The effect was not his discovery; the Swiss engineer Daniel Colladon and the French physicist Jacques Babinet had guided light along water jets in the early 1840s, and Tyndall's fame long overshadowed their priority. Charles Kao's 2009 Nobel citation traces the lineage from these light pipes to the modern fibre. Beyond optics, Tyndall explained why the sky is blue (the Tyndall effect), made the first measurements of infrared absorption by water vapour and carbon dioxide — founding experimental greenhouse-gas science around 1859 — and made the first ascent of the Weisshorn in 1861. He died in 1893 of an accidental chloral overdose administered by his wife.

History

1611

— Kepler describes total internal reflection and its limiting angle in his Dioptrice

1841

— Daniel Colladon guides light along a falling water jet in Geneva; Jacques Babinet shows the same effect in the 1840s

1854

— Tyndall demonstrates light guiding by total internal reflection at the Royal Institution

1870

— Tyndall popularises the light-guiding effect in his public lectures

1966

— Kao and Hockham propose low-loss glass fibres for long-distance communication (Nobel Prize 2009)

Going further

Texts. Hecht, Optics, 5th ed., §4.7 (Total Internal Reflection) and §4.6 (Fresnel equations) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §1.2 (ray optics of fibres) and §9.1 (fibre optics) · Born & Wolf, Principles of Optics, §1.5.4 (TIR and evanescent waves) · Ghatak, Optics (numerical aperture and acceptance cone).

Provenance. The critical-angle formula $\theta_c = \arcsin(n_2/n_1)$ and its numbers are verifiable from Hecht Optics §4.7 and standard index tables; the fibre relation $\mathrm{NA} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ from Saleh & Teich §9.1; the evanescent penetration depth and imaginary- $k_z$ result (recalled here, owned by the electromagnetic article) from Saleh & Teich §6.2 and Born & Wolf §1.5.4; the Fresnel-phase and unit-reflectance results likewise recalled from Hecht §4.6–4.7. Index note (base reconciliation): this edition standardises on crown glass $n = 1.52$ ( $\theta_c = 41.1^\circ$ ) rather than the round $n = 1.5$ ( $41.8^\circ$ ), and water $n = 1.33$ ( $48.8^\circ$ ) rather than $n = 1.333$ ( $48.6^\circ$ ); each pairing is reproduced in verify_total-internal-reflection.py so the quoted angle always matches its stated index.

Prerequisites. Snell's Law of Refraction · Refractive Index and Optical Path Length · Law of Reflection · Ray Model of Light. Leads to. Dispersion and Prisms · Stops, Pupils, and Field of View.

Why is there no critical angle when light goes from a rarer into a denser medium?

Going low index to high, Snell's law gives $\sin\theta_t = (n_1/n_2)\sin\theta_i$ with $n_1/n_2 < 1$ , so $\sin\theta_t < \sin\theta_i \le 1$ always has a real solution: the refracted ray bends toward the normal and escapes at every angle, right up to grazing. Total internal reflection is intrinsically one-directional — a ray can only be trapped trying to leave the denser side — which is why $\theta_c = \arcsin(n_2/n_1)$ carries the smaller index on top, keeping the argument below one.

If the reflection is total, how can any field exist in the second medium?

Because Maxwell's boundary conditions cannot be met with zero field on one side — the tangential fields must be continuous. The resolution, recalled from the electromagnetic treatment, is the evanescent wave: a field oscillating along the surface but decaying exponentially with depth, $E \propto e^{-\kappa z}$ , over roughly a wavelength. It carries no net time-averaged power away, so energy conservation and "total" reflection hold — yet the field is genuinely there and can be tapped by anything brought within reach. "Zero average power across the boundary" and "zero field" are different statements, and the whole evanescent-sensing industry lives in the gap between them.

Why does a fibre need a glass cladding — couldn't the core just be bare in air?

Bare glass in air gives a huge numerical aperture, so on paper it traps light superbly. In practice it is a disaster: the TIR surface would be the fibre's outer surface, where dust, a fingerprint, or a supporting clip touches it — and anything within the evanescent skin frustrates the reflection and bleeds light out. The low-index cladding moves the reflecting wall inside the fibre, where it stays pristine, and sets a small, controlled index contrast that keeps the fibre single- or few-mode with low dispersion. You trade raw light-gathering for a wall you can protect.

Does bending an optical fibre make it leak light?

Yes, if you bend it too tightly. Guiding relies on every ray meeting the wall beyond $\theta_c$ ; a bend tilts the local surface, so on the outside of a curve the effective incidence angle drops back toward $\theta_c$ and rays that were safely trapped begin to refract into the cladding. The evanescent tail also reaches farther there and couples into the jacket. Manufacturers quote a minimum bend radius for exactly this reason — "macrobend loss" is the same physics as a straight fibre, only with the local incidence angle nudged the wrong way.

Why use a total-internal-reflection prism instead of an ordinary silvered mirror?

Two reasons, both from the physics here. First, TIR is lossless: beyond $\theta_c$ the reflectance is exactly $R = 1$ in transparent glass, whereas even the best metal mirror absorbs a few percent per bounce and degrades as its coating oxidises. Second, there is no coating to apply or maintain — the reflecting surface is just clean glass, which cannot tarnish or peel. That is why binoculars, SLR viewfinders, and periscopes fold their light with $45^\circ$ prisms (safely past crown glass's $41.1^\circ$ ) rather than mirrors. The catch is the warning above: the face must stay clean and dry, or the evanescent skin lets light escape.

Can I actually steal some of the light out of a totally reflected beam?

Yes — that is frustrated total internal reflection, a standard device rather than a curiosity. Bring a second high-index surface within a few hundred nanometres of the reflecting face, inside the evanescent skin, and light tunnels across; the transmitted fraction falls off roughly as $e^{-2\kappa d}$ with the gap $d$ , so you can dial the split from nearly everything to almost nothing over a sub-micron range. Beam-splitter cubes with a thin air gap, optical touch screens, and fingerprint scanners all run on it, and the mathematics is identical to quantum-mechanical tunnelling through a barrier. The upshot: "total" reflection stops being total the moment you put something in the skin.

Snell's window — what exactly does a diver see looking straight up?

Every ray from the whole sky above refracts into a cone of half-angle $\theta_c = 48.8^\circ$ around the vertical, so the diver sees the entire hemisphere of sky compressed into a bright "window" about $97.5^\circ$ wide overhead, the horizon squeezed onto its rim (darkened and colour-fringed there, as transmission drops and $\theta_c$ shifts slightly with wavelength). Beyond the rim no ray from above can reach the eye — those directions satisfy the TIR condition in reverse — so the surface acts as a mirror and the diver sees the reflected underwater scene. Surface ripples shatter the rim into the dancing bright filaments on any pool floor.