Boundary conditions, reflection, and refraction

Four conditions at every interface. Consider a boundary separating medium 1 and medium 2, with unit normal $\hat n$ pointing from 1 into 2. Applying Maxwell's equations to infinitesimally thin loops and pillboxes straddling the surface gives four requirements that must hold at every point of the boundary, at every instant:

• Tangential $\vec E$ is continuous, $E_{1,\text{tan}} = E_{2,\text{tan}}$, from Faraday's law on a thin loop (the flux term vanishes as the loop shrinks, leaving only the tangential-field difference).
• Tangential $\vec H$ is continuous, $H_{1,\text{tan}} = H_{2,\text{tan}}$, from the Ampère-Maxwell law by the same thin-loop argument (no free surface current).
• Normal $\vec D$ is continuous, $D_{1,\text n} = D_{2,\text n}$, from Gauss's law on a thin pillbox (no free surface charge).
• Normal $\vec B$ is continuous, $B_{1,\text n} = B_{2,\text n}$, from $\nabla\cdot\vec B = 0$ on the same pillbox.

Why reflection and refraction must exist. A single incident wave generally cannot, by itself, satisfy all four conditions simultaneously on both sides of the interface — its fields on side 1 simply do not match what side 2 needs. The resolution is that a reflected wave (staying in medium 1) and a transmitted wave (entering medium 2) must appear, with amplitudes and directions chosen so incident-plus-reflected fields on side 1 equal the transmitted field on side 2, everywhere on the boundary, at every instant. Reflection and refraction are not phenomena bolted onto electromagnetism — they are what the boundary conditions demand.

Phase matching along the interface. Write the interface as the plane $z = 0$ and each of the three waves — incident, reflected, transmitted — as $\cos(\vec k\cdot\vec r - \omega t)$. For the boundary condition to hold at every point $(x,y)$ on the interface and every time $t$, the phase factors of all three waves must agree there. This forces two things: all three waves share the same frequency $\omega$, and their wavevector components tangential to the interface are equal,

where: $k_{i,\parallel}$, $k_{r,\parallel}$, $k_{t,\parallel}$ are the components of the incident, reflected, and transmitted wavevectors lying in the interface plane. Combined with $|\vec k| = n\omega/c$ in a medium of refractive index $n$, this single matching condition yields both the law of reflection, $\theta_i = \theta_r$, and Snell's law of refraction — worked out in the derivation below.

A light ray travelling in air ($n_1 = 1.00$) strikes the flat surface of a glass slab ($n_2 = 1.50$) at an angle of incidence $\theta_i = 40^\circ$ from the normal. Find the angle of reflection and the angle of refraction.

Step 1. Angle of reflection. By the law of reflection derived above from phase matching, $\theta_r = \theta_i = 40^\circ$ — no calculation needed beyond recognizing that both wavevectors share the same magnitude and tangential component in medium 1.

Step 2. Set up Snell's law for the refracted ray. $n_1\sin\theta_i = n_2\sin\theta_t \implies \sin\theta_t = \dfrac{n_1}{n_2}\sin\theta_i = \dfrac{1.00}{1.50}\sin(40^\circ)$.

Step 3. Evaluate. $\sin(40^\circ) \approx 0.643$, so $\sin\theta_t = \dfrac{0.643}{1.50} \approx 0.4287$, giving $\theta_t = \sin^{-1}(0.4287) \approx 25.4^\circ$. The reflected ray leaves at $40^\circ$ and the refracted ray bends toward the normal, entering the glass at about $25.4^\circ$ — exactly as expected since $n_2 > n_1$ makes light bend toward the normal on entering the denser medium.

Competitive stretch

Now send a wave the other way: from water ($n_1 = 1.33$) toward air ($n_2 = 1.00$). Find the critical angle for total internal reflection. At the critical angle $\theta_c$, the refracted ray grazes the interface, so $\theta_t = 90^\circ$ and $\sin\theta_t = 1$. Snell's law gives $n_1 \sin\theta_c = n_2 \sin(90^\circ) = n_2$, so $\sin\theta_c = n_2/n_1 = 1.00/1.33 = 0.7519$, and $\theta_c = \sin^{-1}(0.7519) \approx 48.75^\circ$. Beyond this angle, the tangential-matching condition $\sin\theta_t = (n_1/n_2)\sin\theta_i$ would require $\sin\theta_t > 1$, which has no real solution — a propagating transmitted wave cannot satisfy the boundary condition, so all the energy reflects (total internal reflection), while an evanescent, non-propagating field still exists just past the boundary to keep the fields continuous.

• Treating Snell's law as a memorized empirical rule disconnected from Maxwell's equations, instead of a direct consequence of phase-matching along the interface.
• Mixing up which components are continuous: it is tangential $E$ and $H$, and normal $D$ and $B$ — swapping tangential and normal is a frequent slip.
• Trying to compute the full Fresnel reflection and transmission amplitudes or Brewster's angle here — that needs all four conditions applied quantitatively and is out of scope; this topic stops at existence of the two waves and Snell's law.
• Assuming total internal reflection means zero field beyond the interface — an evanescent field exists just past the boundary even though no energy escapes on average.

Ask the tutor: if you would like to see the full algebraic derivation of the Fresnel reflection and transmission amplitudes, or a walkthrough of the evanescent wave beyond a total-internal-reflection boundary.