Reflection, Refraction and Fresnel Equations

Statement

For a monochromatic plane electromagnetic wave incident from a linear, isotropic, non-absorbing dielectric of refractive index \( n_1 \) onto a sharp planar interface with a second such dielectric of index \( n_2 \), the electromagnetic boundary conditions force (i) the reflected and transmitted waves to share the incident frequency and tangential wavevector, giving the law of reflection \( \theta_r = \theta_i \) and Snell's law \( n_1 \sin\theta_i = n_2 \sin\theta_t \); and (ii) the amplitude ratios to take the Fresnel forms, for s-polarization (electric field perpendicular to the plane of incidence) \( r_s = \dfrac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t} \), \( t_s = \dfrac{2 n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t} \), and for p-polarization (electric field in the plane of incidence) \( r_p = \dfrac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t} \), \( t_p = \dfrac{2 n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t} \), for non-magnetic media (\( \mu_1 = \mu_2 = \mu_0 \)).

Why it matters

Everything optics does at an interface — the 4% glare off a window, the polarizing action of a lake surface, anti-reflection coatings, fibre-optic confinement, ellipsometry of nanometre films — is quantitatively contained in this one boundary-value problem. It is the first place in electromagnetism where Maxwell's equations produce not just the existence of a reflected ray (which geometry alone could guess) but its exact amplitude and phase, including the sign flips and the zero at Brewster's angle that no ray theory can supply.

Methodologically it is the archetype of a matching calculation: bulk solutions on either side, stitched together by continuity conditions, with the kinematics (Snell) coming from phase matching and the dynamics (Fresnel coefficients) from amplitude matching. The same two-stage logic recurs in quantum scattering off a potential step, acoustic impedance mismatch, and transmission-line junctions.

Assumptions

Derivation

Geometry: the interface is the plane \( z = 0 \), medium 1 fills \( z < 0 \), medium 2 fills \( z > 0 \), and the plane of incidence is the \( xz \)-plane. Angles are measured from the interface normal \( \hat{\mathbf{z}} \). Complex amplitudes carry tildes.

Result

Limiting cases

• Normal incidence (\( \theta_i = 0 \)): \( r_s = -r_p \to \frac{n_1 - n_2}{n_1 + n_2} \) in the s-convention; reflectance \( R = \left( \frac{n_1 - n_2}{n_1 + n_2} \right)^2 \) — 4.0% for air–glass. The s/p distinction disappears physically (no plane of incidence); the apparent sign difference is pure convention.
• Matched media (\( n_1 = n_2 \)): \( \theta_t = \theta_i \), \( r_s = r_p = 0 \), \( t_s = t_p = 1 \) — the interface is invisible, as it must be.
• Grazing incidence (\( \theta_i \to \pi/2 \)): \( \cos\theta_i \to 0 \), so \( r_s \to -1 \) and \( r_p \to -1 \), hence \( R \to 1 \) for both polarizations: any interface becomes a perfect mirror at grazing angles — why wet roads glare at sunset and X-ray telescopes use grazing optics.
• Brewster's angle: \( r_p = 0 \) when \( n_2\cos\theta_i = n_1\cos\theta_t \), which with Snell gives \( \theta_i + \theta_t = \pi/2 \) and \( \tan\theta_B = n_2/n_1 \); reflected light is then perfectly s-polarized.
• Critical angle (\( n_1 > n_2 \)): at \( \sin\theta_c = n_2/n_1 \), \( \theta_t \to \pi/2 \) and \( t \) does not vanish — the transmitted wave skims the surface; beyond \( \theta_c \), \( \cos\theta_t = i\sqrt{(n_1/n_2)^2\sin^2\theta_i - 1} \) is imaginary, \( |r_s| = |r_p| = 1 \) (total internal reflection) and the transmitted field becomes evanescent.
• Quantum analogue: at normal incidence \( r = \frac{k_1 - k_2}{k_1 + k_2} \) — identical in form to reflection of a matter wave off a potential step, with \( k = n\omega/c \) playing the wavenumber role.

Breaks when

Failure modes

• Squaring \( t \) for the transmittance. \( T = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2 \), not \( |t|^2 \). At normal incidence air–glass, \( t = 0.8 \) so \( |t|^2 = 0.64 \), yet \( T = 0.96 \): the flux factor is not optional. Note \( t \) can even exceed 1 (going into the rarer medium) without violating anything.
• Treating the sign of \( r_p \) at normal incidence as physics. \( r_s \to -0.2 \) and \( r_p \to +0.2 \) for air–glass describe the same reflected field; the sign difference records the convention for the positive p-direction, not a real \( \pi \) phase difference between polarizations.
• Measuring angles from the surface instead of the normal. Snell's law and every cosine above use the normal; the classic symptom is a "critical angle" of \( 48^\circ \) becoming \( 42^\circ \) or reflectances at "\( 30^\circ \)" that match no table.
• Pushing real \( \theta_t \) past the critical angle. Beyond \( \theta_c \), \( \sin\theta_t > 1 \) and a calculator returns an error or garbage; the correct move is \( \cos\theta_t \to i\sqrt{\sin^2\theta_t - 1} \), giving unimodular complex \( r \) and the total-internal-reflection phase shifts.
• Confusing s and p. "Parallel" (p) means parallel to the plane of incidence, not to the interface; the polarization whose \( \mathbf{E} \) is parallel to the surface is s. Mislabeling swaps Brewster behaviour between polarizations.
• Using \( \tan\theta_B = n_2/n_1 \) for magnetic media or for internal reflection without care. The formula assumes \( \mu_1 = \mu_2 \); and for \( n_1 > n_2 \) Brewster's angle still exists (e.g. glass-to-air \( \theta_B \approx 33.7^\circ \)) but lies below the critical angle — students often assume TIR wipes it out.

Discussion

The derivation splits cleanly into kinematics and dynamics, and the split is a symmetry statement. The interface preserves translational invariance in \( x, y \) and in \( t \); the corresponding conserved quantities are the tangential wavevector and the frequency. Snell's law is therefore conservation of the photon's tangential momentum \( \hbar k_x \) at a boundary that can absorb normal momentum but not tangential momentum — which is also why radiation pressure on a transparent interface has a normal component only. Any wave theory with a linear dispersive interface obeys "Snell's law" in this sense: seismic waves, electron waves at heterojunctions, sound at a thermocline.

The Fresnel coefficients are most naturally read as impedance-matching statements. Writing \( Z = \sqrt{\mu/\varepsilon} \) and defining angle-dependent effective impedances \( Z_s = Z/\cos\theta \), \( Z_p = Z\cos\theta \), both results collapse to the single transmission-line form \( r = (Z_2^{\mathrm{eff}} - Z_1^{\mathrm{eff}})/(Z_2^{\mathrm{eff}} + Z_1^{\mathrm{eff}}) \). Brewster's angle is then nothing mysterious: it is the angle at which two media of different \( n \) happen to present equal p-impedances. The microscopic picture agrees — at Brewster incidence the transmitted ray is perpendicular to the would-be reflected ray, and the induced dipoles in medium 2, oscillating along the transmitted \( \mathbf{E} \), cannot radiate along their own axis, so nothing is available to build the reflected wave.

Deeper still, the reflected wave is not "bounced off the surface" at all. Ewald–Oseen extinction shows that the incident wave propagates undisturbed at speed \( c \) everywhere, while the polarized molecules of medium 2 radiate a secondary field that (i) exactly cancels the incident wave inside medium 2, (ii) rebuilds it as a slower wave along the refracted direction, and (iii) sends the reflected wave back into medium 1. The Fresnel coefficients thus summarize a bulk interference effect; the sharp-interface boundary-condition calculation is a shortcut that gets the same answer because the boundary conditions themselves encode the collective molecular response.

Analyticity and unitarity organize the extensions. As functions of \( \cos\theta_t \), the coefficients continue smoothly into the total-internal-reflection regime, where \( r_s = e^{-i\delta_s} \), \( r_p = e^{-i\delta_p} \) with \( \tan\frac{\delta_s}{2} = \frac{\sqrt{\sin^2\theta_i - (n_2/n_1)^2}}{\cos\theta_i} \) and a distinct \( \delta_p \); the differential phase \( \delta_p - \delta_s \) is what a Fresnel rhomb exploits to make circular polarization without birefringence, and the \( \theta \)-derivative of the phase gives the Goos–Hänchen lateral beam shift. The pair \( (r, t) \) together with the reversed-side coefficients \( (r', t') \) form a \( 2\times 2 \) scattering matrix; losslessness makes it unitary (with the flux-normalized amplitudes) and time-reversal invariance yields the Stokes relations \( r' = -r \), \( t t' + r^2 = 1 \) — the same constraints that govern beam splitters in quantum optics, where they enforce the commutation relations of the output photon modes.

Common misconceptions. "Some light is reflected because the surface is shiny" — reflectance is a bulk-contrast effect, present for perfectly smooth, perfectly transparent media, and vanishing whenever \( n_1 = n_2 \) however the surface is finished (index-matching fluid makes glass disappear). "Total internal reflection means no field beyond the interface" — the evanescent field penetrates a distance of order \( \lambda \) and carries no time-averaged normal flux, but it can be tapped (frustrated TIR, near-field microscopy). "At Brewster's angle nothing is reflected" — only p-polarized light vanishes; unpolarized light still reflects (s-only), which is precisely why the reflection is useful as a polarizer.

Worked examples

Example 1 — air to glass at \( 45^\circ \): both polarizations, with energy check. Light of any wavelength in air (\( n_1 = 1.00 \)) strikes crown glass (\( n_2 = 1.50 \)) at \( \theta_i = 45.0^\circ \). Find \( r_s, r_p \), the reflectances, and verify \( R_s + T_s = 1 \).

Example 2 — Brewster's angle for an air–glass surface and the residual s-reflectance. Find \( \theta_B \) for \( n_1 = 1.00 \to n_2 = 1.50 \), confirm the geometry, and compute what fraction of unpolarized light is reflected there.

Problems

1. Light strikes a calm water surface (\( n = 1.33 \)) from air at normal incidence. Compute the reflectance and transmittance and verify they sum to 1.
   Solution

   At normal incidence \( r = \frac{n_1 - n_2}{n_1 + n_2} = \frac{1 - 1.33}{1 + 1.33} = \frac{-0.33}{2.33} = -0.1416 \), so \( R = r^2 = 0.0201 \) — about 2.0%. The transmission amplitude is \( t = \frac{2n_1}{n_1 + n_2} = \frac{2}{2.33} = 0.8584 \), and \( T = \frac{n_2}{n_1} t^2 = 1.33 \times 0.7368 = 0.9799 \). Check: \( R + T = 0.0201 + 0.9799 = 1.0000 \). Note \( t^2 = 0.737 \) alone is not the transmittance; the impedance factor \( n_2/n_1 \) supplies the rest.
2. Find the critical angle for (a) crown glass (\( n = 1.52 \)) to air, and (b) water (\( n = 1.33 \)) to air. Explain why no critical angle exists going from air into either medium.
   Solution

   (a) \( \sin\theta_c = n_2/n_1 = 1/1.52 = 0.6579 \Rightarrow \theta_c = 41.1^\circ \). (b) \( \sin\theta_c = 1/1.33 = 0.7519 \Rightarrow \theta_c = 48.8^\circ \). Going from air (\( n_1 = 1 \)) into a denser medium, Snell's law gives \( \sin\theta_t = \sin\theta_i / n_2 < \sin\theta_i \le 1 \): a real transmitted angle exists for every incidence angle, so total internal reflection requires \( n_1 > n_2 \) — light must be moving toward the optically rarer medium.
3. Unpolarized light hits an air–glass interface (\( n_2 = 1.50 \)) at \( \theta_i = 60.0^\circ \). Find \( r_s \), \( r_p \), \( R_s \), \( R_p \), the reflectance for unpolarized light, and the degree of polarization of the reflected beam, \( P = \frac{R_s - R_p}{R_s + R_p} \).
   Solution

   Snell: \( \sin\theta_t = \sin 60^\circ / 1.5 = 0.8660/1.5 = 0.5774 \Rightarrow \theta_t = 35.26^\circ \), \( \cos\theta_t = 0.8165 \); \( \cos 60^\circ = 0.5000 \). \( r_s = \frac{0.5000 - 1.5(0.8165)}{0.5000 + 1.5(0.8165)} = \frac{-0.7247}{1.7247} = -0.4202 \Rightarrow R_s = 0.1766 \). \( r_p = \frac{1.5(0.5000) - 0.8165}{1.5(0.5000) + 0.8165} = \frac{-0.0665}{1.5665} = -0.0425 \Rightarrow R_p = 0.0018 \). Unpolarized reflectance: \( R = \frac{1}{2}(0.1766 + 0.0018) = 0.0892 \), i.e. 8.9%. Degree of polarization: \( P = \frac{0.1766 - 0.0018}{0.1766 + 0.0018} = \frac{0.1748}{0.1784} = 0.980 \) — the reflected beam is 98% polarized, since \( 60^\circ \) lies just past Brewster's angle (\( 56.3^\circ \)). Note \( r_p \) has changed sign relative to its value below \( \theta_B \): the p-amplitude passes through zero at Brewster and re-emerges with a \( \pi \) phase change.
4. Diamond has \( n = 2.417 \). Find (a) its Brewster angle in air, (b) its normal-incidence reflectance, and (c) its critical angle. Use (b) and (c) together to explain the brilliance of a cut diamond.
   Solution

   (a) \( \theta_B = \arctan(2.417) = 67.5^\circ \). (b) \( R = \left( \frac{n-1}{n+1} \right)^2 = \left( \frac{1.417}{3.417} \right)^2 = (0.4147)^2 = 0.172 \) — about 17.2% per surface, more than four times the 4% of glass. (c) \( \theta_c = \arcsin(1/2.417) = \arcsin(0.4137) = 24.4^\circ \). Brilliance: the tiny critical angle means light entering the crown is very easily trapped by total internal reflection off the pavilion facets (any internal incidence beyond \( 24.4^\circ \) reflects with \( |r| = 1 \)), while the large index gives strong first-surface sparkle (17% flash per facet). A well-cut stone routes trapped light back out through the crown; the same geometry cut in glass (\( \theta_c = 41.8^\circ \)) leaks light through the pavilion and looks dull.
5. A helium–neon beam (\( \lambda_0 = 633\ \mathrm{nm} \)) inside glass (\( n_1 = 1.50 \)) strikes the glass–air interface at \( \theta_i = 45.0^\circ \). (a) Confirm this exceeds the critical angle. (b) Show the transmitted field has the evanescent form \( E \propto e^{-\kappa z} \) and find the penetration depth \( d = 1/\kappa \). (c) Compute \( |r_s| \) and the total-internal-reflection phase shift \( \delta_s \), where \( r_s = e^{-i\delta_s} \), \( \tan\frac{\delta_s}{2} = \frac{\sqrt{n_1^2\sin^2\theta_i - n_2^2}}{n_1\cos\theta_i} \).
   Solution

   (a) \( \theta_c = \arcsin(1/1.50) = 41.8^\circ < 45.0^\circ \): total internal reflection occurs. (b) Phase matching fixes \( k_x = n_1 \frac{\omega}{c} \sin\theta_i \) in both media. In air, \( k_x^2 + k_z^2 = (\omega/c)^2 \), so \( k_z^2 = \frac{\omega^2}{c^2}\left( 1 - n_1^2\sin^2\theta_i \right) \). With \( n_1^2\sin^2\theta_i = 2.25 \times 0.500 = 1.125 > 1 \), \( k_z = i\kappa \) with \( \kappa = \frac{2\pi}{\lambda_0}\sqrt{n_1^2\sin^2\theta_i - 1} = \frac{2\pi}{633\ \mathrm{nm}}\sqrt{0.125} = \frac{6.2832 \times 0.3536}{633\ \mathrm{nm}} = 3.509 \times 10^{-3}\ \mathrm{nm}^{-1} \). The transmitted field is \( E_T \propto e^{i k_x x} e^{-\kappa z} \): propagating along the surface, exponentially decaying into the air. Penetration depth: \( d = 1/\kappa = 285\ \mathrm{nm} \approx 0.45\,\lambda_0 \) — a second glass surface brought within a few hundred nanometres would frustrate the TIR and couple light across. (c) \( \sqrt{n_1^2\sin^2\theta_i - n_2^2} = \sqrt{1.125 - 1} = 0.3536 \); \( n_1\cos\theta_i = 1.50 \times 0.7071 = 1.0607 \). So \( \tan\frac{\delta_s}{2} = 0.3536/1.0607 = 0.3334 \Rightarrow \frac{\delta_s}{2} = 18.44^\circ \Rightarrow \delta_s = 36.9^\circ \). Since \( r_s = \frac{n_1\cos\theta_i - i\sqrt{\cdots}}{n_1\cos\theta_i + i\sqrt{\cdots}} \) is a ratio of complex conjugates, \( |r_s| = 1 \) exactly: all the energy returns, but delayed in phase by \( 36.9^\circ \) — the phase shift a Fresnel rhomb stacks up (using \( \delta_p - \delta_s \) over two bounces) to convert linear to circular polarization.