Total internal reflection and guided light

Write Snell's law for the transmitted ray, sin θ₂ = (n₁/n₂) sin θ₁, with both angles from the normal because it is the tangential component of the wavevector, k sin θ, that a flat boundary conserves. When n₁ > n₂ the prefactor exceeds one, so sin θ₂ reaches its ceiling of 1 while θ₁ is still short of grazing, at sin θ_c = n₂/n₁. Crown glass against air gives θ_c = 41.1°, water against air 48.6°, and a fibre core of 1.480 against a cladding of 1.460 gives 80.6°, only 9.4° from grazing. Past θ_c the transmitted ray does not fade; it has no solution, and the ray model assigns everything to the reflected ray. The wave treatment agrees, giving a reflection coefficient of modulus exactly 1 with only a phase shift. Reversibility reads the same inequality backwards: a ray entering from the rarer side, even at grazing, refracts to at most θ_c, so no ray that came in through the wall can ever return to it beyond θ_c. That is why total reflection runs only from higher index to lower, and why guiding needs light injected through an end face rather than through the wall.

Because θ_c is a threshold set by two indices, a measured critical angle is a measurement of index. Differentiating sin θ_c = n₂/n₁ with n₁ fixed gives cos θ_c dθ_c = dn₂/n₁, so dθ_c/dn₂ = 1/(n₁ cos θ_c) = 1/√(n₁² − n₂²). For a glass prism of index 1.750 against a water sample at 1.333, that is 0.882 rad per unit of index: a change of 0.001 in the sample moves the edge of total reflection by 0.05°, so a telescope reading to 0.01° resolves 2 × 10⁻⁴ in n₂. That is the Abbe refractometer, and it explains its dense prism: the sample must be the rarer medium, so the prism index sets the highest sample index the instrument can read. The derivative also says where the effect is sharpest. At the fibre wall, with 1.480 against 1.460, √(n₁² − n₂²) is only 0.242 and the sensitivity climbs to 4.12 rad per unit, 0.24° per 0.001: a cladding index that drifts by one part in a thousand moves the guiding condition by a quarter of a degree, which is why fibre indices are specified to four decimal places.

Look up from under still water of index 1.333. A ray arriving from zenith angle θ_z in air refracts to sin θ_w = sin θ_z/n, so the horizon itself, θ_z = 90°, appears at θ_w = θ_c = 48.6°. Every direction in the sky lies inside a cone of half-angle 48.6°, which is Snell's window, and at depth h it is a disc of diameter 2h tan θ_c = 2.27h on the surface. Outside the disc the surface is a boundary struck from below beyond θ_c, so it mirrors the bottom. The mapping is not uniform: differentiating gives dθ_w/dθ_z = cos θ_z/(n cos θ_w), which is 1/n = 0.75 at the zenith, 0.63 at 45°, and 0 at the horizon. The last 10° of sky above the horizon are squeezed into 0.98° at the rim, so the horizon and everything near it collapse into a bright ring. How much of the diver's view does the sky take? Integrate the solid angle: Ω = 2π ∫ sin θ dθ from 0 to θ_c = 2π(1 − cos θ_c) = 2.13 sr, 17% of the full 4π and 34% of the upper hemisphere. The remaining two thirds of the upward view is reflected sea floor.

A right-angle prism in air folds a beam through 90° by one total reflection at its hypotenuse, where the axial ray arrives at 45°; with n = 1.517 the critical angle is 41.2°, so the margin is only 3.8°. Now tilt the incoming beam by α at the entry face. It refracts to β = arcsin(sin α/n) inside and, on the unfavourable side, meets the hypotenuse at 45° − β. Total reflection survives while 45° − β ≥ θ_c, that is sin α ≤ n sin(45° − θ_c) = 1.517 sin 3.76° = 0.0995, so αₘₐₓ = 5.7°. A prism that must fold a ±10° field needs 45° − arcsin(sin 10°/n) ≥ arcsin(1/n), which solves to n ≥ 1.60, a dense flint; otherwise the hypotenuse is silvered and the free mirror is given up. Geometry decides which prisms can rely on the effect at all. A corner-cube retroreflector meets each of its three faces at arccos(1/√3) = 54.7°, needing only n ≥ 1.22, so it works in any glass. A pentaprism, which turns a beam through exactly 90° whatever its own tilt, meets its faces at 22.5° and would need n ≥ 2.61: no glass reaches that, so its two reflecting faces are always coated.

In a step-index fibre a meridional ray at angle θ to the axis meets the wall at 90° − θ from the normal, so it is guided when cos θ ≥ n₂/n₁. With n₁ = 1.480 and n₂ = 1.460 that is θ ≤ 9.4°. At the end face, launched from a medium n₀, n₀ sin θₐ = n₁ sin θ, and the steepest guided ray has sin θ = √(1 − n₂²/n₁²), giving n₀ sin θₐ = √(n₁² − n₂²): the numerical aperture. Here NA = √(2.1904 − 2.1316) = 0.2425, an acceptance half-angle of 14.0° in air; the small-step form n₁√(2Δ) with Δ = (n₁ − n₂)/n₁ = 0.0135 gives 0.2433, 0.3% high. The cone has a price. The axial ray covers length L in time L n₁/c; the steepest guided ray travels L/cos θ = L n₁/n₂, so it arrives later by Δt = (L n₁/c)(n₁/n₂ − 1). For 1.00 km, L n₁/c = 4.937 μs and n₁/n₂ − 1 = 0.0137, so Δt = 67.6 ns: a narrower pulse is smeared to that width, and pulses closer than about 2Δt merge, capping the rate near 7 Mbit s⁻¹ over a kilometre. Doubling the NA to gather more light roughly quadruples Δt, since Δt ∝ n₁ − n₂ ∝ NA². That trade is why graded-index and single-mode fibres exist.

The ray model keeps three promises it cannot fully justify. First, nothing crosses the wall. The wave solution puts an evanescent field in the cladding, decaying as e^(−z/d) with d = λ₀/(2π√(n₁² sin²θ − n₂²)); at 1550 nm and 84° incidence on the 1.480/1.460 wall, d = 1.3 μm. No power crosses an isolated flat wall, so the reflection is still total, but a second surface within a few d would draw power across, and the ray model has no such distance in it. Second, confinement is lossless forever. Third, a bend either guides or does not. For a core of radius a bent to radius R, the worst-placed ray parallel to the axis, the one running along the inner wall, meets the outer wall with sin φ = (R − a)/(R + a), so the ray model guides every such ray while R ≥ a(n₁ + n₂)/(n₁ − n₂): 147a, or 3.7 mm for a = 25 μm, and zero loss above that. Real fibres lose power smoothly, roughly as e^(−R/R₀), because on the outside of a bend the evanescent tail would have to outrun c/n₂ to keep pace and radiates instead; bend loss is specified at 10 to 30 mm radii, where the ray model sees nothing. The model is exact about direction and silent about everything set by a wavelength.