Refraction and Snell's law

In a linear, non-conducting medium Maxwell's equations reduce to the wave equation ∂²E/∂x² = με ∂²E/∂t², so a plane wave travels at v = 1/√(με) = c/√(εᵣ μᵣ). The refractive index n = c/v is therefore √(εᵣ μᵣ), and since μᵣ ≈ 1 for glass and water, n ≈ √εᵣ. The εᵣ that matters is the one at optical frequency. Water's static permittivity is about 80, which would give n ≈ 8.9; at 5 × 10¹⁴ Hz the molecular dipoles cannot follow the field, εᵣ falls to 1.78, and n = 1.33. So n is an optical-frequency property, not a static one, and it varies with frequency — the dispersion that a later topic quantifies. For crown glass n = 1.52 gives v = 1.97 × 10⁸ m s⁻¹. The number describes the phase speed of the wave in the medium; nothing here says a photon is slowed by being absorbed and re-emitted.

Put the boundary on the plane z = 0 with x along it. The incident wave has phase φ₁ = k₁(x sin θ₁ + z cos θ₁) − ωt and the transmitted wave φ₂ = k₂(x sin θ₂ + z cos θ₂) − ω′t. The electromagnetic boundary conditions must hold at every x on z = 0 and at every t, which is only possible if the two phases agree there up to a constant. Equate their partial derivatives on the boundary: ∂φ/∂t gives ω′ = ω, so the frequency cannot change; ∂φ/∂x gives k₁ sin θ₁ = k₂ sin θ₂, and with k = nω/c that is n₁ sin θ₁ = n₂ sin θ₂. The normal component k cos θ is the one that jumps, and that jump is the bending. Fermat's principle gave the same law in this section's opening topic by a different route; the wave argument adds two things. It says what happens to the wavelength, and it says that n sin θ is the same in every layer of a stack of parallel boundaries, so an oil film floating on water shifts a ray sideways but leaves its final direction exactly what it would have been without the film.

Write the law as sin θ₂ = (n₁/n₂) sin θ₁. Light in air meeting glass of n = 1.50 at 60.0° has sin θ₂ = 0.866/1.50 = 0.577, so θ₂ = 35.3°: it bends toward the normal by 24.7°. Reverse the ray — glass to air at 35.3° — and it returns to 60.0°, because the law is symmetric under exchanging the two sides. Differentiating gives n₁ cos θ₁ dθ₁ = n₂ cos θ₂ dθ₂, so dθ₂/dθ₁ = n₁ cos θ₁/(n₂ cos θ₂). At normal incidence this is n₁/n₂ = 0.667; at 60.0° it has fallen to cos 60.0°/(1.50 cos 35.3°) = 0.408. The refracted angle grows ever more slowly than the incident one, which is why the linear rule θ₂ = θ₁/n predicts 40.0° here, 4.7° too much. From glass into air the slope exceeds 1: at 40.0° in glass sin θ₂ = 1.50 × 0.643 = 0.964, θ₂ = 74.6°, and dθ₂/dθ₁ = 1.50 cos 40.0°/cos 74.6° = 4.33, so a 1° error in setting the incidence becomes 4.3° in the emerging ray. The slope diverges as θ₂ approaches 90°, where the law stops having a solution.

The boundary fixes f, so inside a medium λ = v/f = λ₀/n. Sodium light at λ₀ = 589.0 nm has f = 5.090 × 10¹⁴ Hz everywhere; its wavelength is 442.9 nm in water (n = 1.330) and 387.5 nm in crown glass (n = 1.520). The light in the glass is not bluer: the eye's receptors respond to frequency, the photon energy hf is unchanged, and the wavelength returns to 589.0 nm on exit. What the shorter wavelength does change is the phase accumulated per metre. A plate 1.000 mm thick of n = 1.520 holds 1.520 mm/589.0 nm = 2581 wavelengths, where the same thickness of air holds 1698, so light crossing the plate arrives 883 cycles — 883 × 2π rad — behind light beside it, delayed by L(n − 1)/c = 1.73 ps. The product nL is the optical path length, and the phase along any path is (2π/λ₀) × optical path. That bookkeeping is what interferometers, thin films and lens design all run on.

A point sits at depth d in water (n₁ = 1.33) and a viewer in air (n₂ = 1) looks down. A ray leaving the point at θ₁ to the normal meets the surface at horizontal offset d tan θ₁ and emerges at θ₂. Extrapolated back into the water, it appears to come from depth d′ = d tan θ₁/tan θ₂. Only in the near-normal limit, where tan → sin, does Snell's law turn this into d′ = d n₂/n₁ = d/1.33: a 2.00 m pool looks 1.50 m deep from above, and the sky's objects look 2.66 m away from 2.00 m below. Off the normal the exact ratio is (1/n)√[(1 − sin²θ₂)/(1 − sin²θ₂/n²)], always smaller than 1/n. At a 20° viewing angle it is 0.731 against 0.752, a 2.8% shortfall; at 40° it is 0.658, 12.5% shallower than the rule says. The image also splits into two focal lines for a bundle of rays, so a flat surface is an astigmatic imager off axis. The rule d′ = d/n is a first-order result, and the viewing angle is the small quantity it is first order in.

Every step above used four assumptions, and each fails in a specific way. The boundary was sharp compared with λ: if instead n varies continuously with height, n(z) sin θ(z) = constant layer by layer and the ray curves, which is the mirage, and why the setting Sun is lifted by about 0.5° at the horizon. The media were isotropic: in calcite n depends on direction and polarisation, and one incident ray becomes two refracted ones. They were lossless: in an absorbing medium n becomes complex, n + iκ, and refraction is accompanied by exponential decay of the field, e^{(−2πκz/λ₀)}. The light was monochromatic: since n depends on frequency, a white ray fans out, the dispersion of a later topic. And the law itself has no solution from high n to low n once sin θ₁ exceeds n₂/n₁ — the next topic takes over there. Finally, rays at all presuppose apertures large compared with λ; below that, refraction gives way to diffraction.