Interfaces & Fresnel amplitudes

Place a planar interface at z=0. Medium 1 occupies z<0 and medium 2 occupies z>0; the unit normal n=+z points from 1 to 2. The plane of incidence is the x–z plane. A monochromatic wave arrives from medium 1 with positive +z component, reflects with negative z component and transmits with positive z component.

Assume both media are homogeneous, isotropic, linear and lossless with real positive ε_j and μ_j. Define k_j=ω√(μ_jε_j) and η_j=√(μ_j/ε_j). The s or transverse-electric (TE) unit vector is +y, perpendicular to the incidence plane. AoN defines the p or transverse-magnetic (TM) unit vector separately for EACH wave as p=s×k̂. This right-handed basis convention matters: the reflected p unit vector reverses its tangential x direction at normal incidence.

All Fresnel coefficients in this chapter are complex FIELD-amplitude ratios in these explicitly declared bases, not power ratios.

Apply Faraday's and Ampère–Maxwell's integral laws to an infinitesimal rectangular loop straddling the interface. In the zero-height limit, finite flux terms vanish and the tangential electric field is continuous unless an ideal magnetic surface-current model is introduced; the tangential magnetic field jumps by the free surface current density K_f.

Applying Gauss's laws to an infinitesimal pillbox gives the normal conditions. The normal displacement D jumps by free surface charge σ_f, while the normal B is continuous because magnetic monopoles are absent in Maxwell electromagnetism.

For the simple dielectric interface used to derive Fresnel amplitudes, take σ_f=0 and K_f=0. Notice what is NOT implied: the normal E component need not be continuous when ε changes, and the normal H component need not be continuous when μ changes.

At z=0 the boundary equations must hold for every tangential coordinate x and every time t. Incident, reflected and transmitted exponentials can therefore cancel each other for all x,t only if their frequencies and tangential wavevector components match.

Because the reflected wave remains in medium 1, equality of tangential components plus equal magnitudes k_i=k_r gives θ_r=θ_i. Between media, k₁sinθ_i=k₂sinθ_t. For ordinary lossless positive-index media define n_j=ck_j/ω=√(ε_rj μ_rj), giving Snell's law n₁sinθ_i=n₂sinθ_t.

This argument is a phase-matching statement, not a ray-optics postulate. It also explains why translational symmetry parallel to the interface conserves tangential momentum.

For s polarization, E is along +y for incident, reflected and transmitted waves. Write E_r=r_s E_i and E_t=t_s E_i. Tangential electric continuity gives 1+r_s=t_s.

Use H=(1/η)k̂×E. The incident tangential H_x is −E_i cosθ_i/η₁; the reflected H_x is +E_r cosθ_i/η₁; the transmitted H_x is −E_t cosθ_t/η₂. Continuity of H_x therefore gives (1−r_s)cosθ_i/η₁=t_s cosθ_t/η₂.

Solving the two equations yields the s-polarized Fresnel FIELD-amplitude coefficients. In nonmagnetic media η_j=η₀/n_j, recovering the familiar n cosθ form.

For p polarization define p_a=s×k̂_a for each wave. Then p_i=cosθ_i x̂−sinθ_i ẑ, p_r=−cosθ_i x̂−sinθ_i ẑ and p_t=cosθ_t x̂−sinθ_t ẑ. The reflected p basis therefore has a NEGATIVE tangential x component.

With E_r=r_pE_i along p_r and E_t=t_pE_i along p_t, tangential electric continuity gives cosθ_i(1−r_p)=t_p cosθ_t. Meanwhile H points +y for all three p-basis waves under this definition, giving (1+r_p)/η₁=t_p/η₂.

Solving gives the standard p-basis Fresnel coefficient. At normal incidence r_p=−r_s under this basis convention; this does NOT mean the physical reflected tangential electric field differs. Because p_r=−x̂ at normal incidence, its LAB-FRAME tangential amplitude is −r_p, which equals r_s. Comparing formulas without comparing basis vectors is a sign error.

At normal incidence θ_i=θ_t=0. The two physical tangential polarizations are equivalent by rotational symmetry. In AoN's chosen p basis, r_p=−r_s only because p_r reverses relative to the laboratory x axis; the actual reflected lab-frame tangential field has the same coefficient as s.

For nonmagnetic media, the p-basis numerator vanishes when n₂cosθ_i=n₁cosθ_t. Combined with Snell's law this gives tanθ_B=n₂/n₁: the Brewster angle. The reflected p-basis field vanishes there.

If n₁>n₂, real transmitted angles cease to exist above θ_c=asin(n₂/n₁). The transmitted normal wavevector becomes imaginary, producing an evanescent field in medium 2. For ideal lossless total internal reflection |r_s|=|r_p|=1, but their phases generally differ; chapter 05 will connect this to energy flow.

The algebra above assumes real positive ε, μ, real propagation angles and scalar wave impedances. In a lossy isotropic medium, ε and possibly μ become complex functions of frequency, the propagation constant is complex, and choosing the physically decaying transmitted branch is part of the solution.

In anisotropic media, D need not be parallel to E, phase velocity and energy-flow direction may differ, and s/p may cease to diagonalize the interface problem. A conductive sheet such as graphene additionally introduces a surface current K_f=σ_s E_t, so tangential H is no longer continuous.

Those cases are not obtained safely by replacing n with one complex number in every real-angle formula without tracking branch choices, tensor response and sheet currents.

Take nonmagnetic medium 1 with n₁=1.00 and medium 2 with n₂=1.50, incident at θ_i=45°. Snell gives sinθ_t=(1/1.5)sin45°, hence θ_t≈28.13°.

Using the nonmagnetic Fresnel forms gives r_s≈−0.3033 and t_s≈0.6967. With AoN's p=s×k̂ basis, r_p≈+0.0920 and t_p≈0.7280. The opposite signs of r_s and r_p do not by themselves state opposite physical phase reversals in a common lab axis because the reflected p unit vector is oriented differently.

At this stage these are FIELD amplitudes only. Squaring |r| or |t| blindly is not yet a transmitted-power calculation because wave impedance and the normal direction cosine enter the flux. Chapter 05 derives the correct power factors.

1. Starting from the integral Faraday law, explain why tangential E is continuous across the ideal interface used here.
2. Use phase matching at z=0 to derive θ_r=θ_i and n₁sinθ_i=n₂sinθ_t.
3. For s polarization, rederive the signs of H_i,x, H_r,x and H_t,x from H=(1/η)k̂×E before solving for r_s.
4. Derive r_s and t_s from the two boundary equations without quoting the Fresnel formulas.
5. Using p=s×k̂, calculate p_i,p_r,p_t and explain why p_r has negative x component.
6. Derive r_p and t_p with AoN's p basis. Show explicitly why t_p≠1+r_p in general.
7. At normal incidence from air to glass n₂=1.5, calculate r_s and r_p in AoN's bases, then calculate the reflected LAB-x field coefficient for a p-polarized incident field.
8. For air to glass n₂/n₁=1.5, derive and calculate the Brewster angle.
9. For glass n₁=1.5 to air n₂=1.0, calculate the critical angle and state what becomes non-propagating above it.
10. At θ_i=45°, n₁=1,n₂=1.5, reproduce θ_t,r_s,t_s,r_p,t_p and explain why these numbers alone are insufficient to state transmitted power.

Chapter 05 will close EM-04 by deriving time-averaged Poynting flux, power reflection/transmission coefficients, wave impedance in media, energy conservation at lossless interfaces and the modifications introduced by absorption.