Fresnel Equations at an Interface

Derivation

Fresnel Equations at an Interface

D-292 Home PU-305 Threads waves · light · matter Depends on Electromagnetic Wave Equation in Vacuum, Bound Charge and the Displacement Field

Statement

For a monochromatic plane wave incident from a lossless dielectric of refractive index \(n_1\) onto a planar interface with a second lossless dielectric \(n_2\), enforcing continuity of the tangential components of \(\vec{E}\) and \(\vec{H}\) across the boundary yields the Fresnel amplitude coefficients. For the two independent linear polarizations these are \[ r_s=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t},\qquad r_p=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t}, \] with the transmission amplitudes \(t_s=1+r_s\) and \(t_p\) following, and the reflected \(p\)-amplitude vanishing at Brewster's angle \(\tan\theta_B=n_2/n_1\).

Why it matters

The Fresnel equations are the quantitative bridge between Maxwell's equations and everything we see: the glare off water, the 4% loss at each face of a lens, the operation of anti-reflection coatings, beam splitters, and optical fibres. They fix not only how much light reflects but its phase and polarization state, which is why they underpin ellipsometry, thin-film metrology, and the design of every polarizing optic.

Brewster's angle, an immediate corollary, is why polarized sunglasses cut horizontal glare and why laser cavities use Brewster windows to enforce a polarization with zero reflection loss.

Assumptions

Both media are linear, isotropic, homogeneous dielectrics.If dropped, \(\varepsilon\) becomes a tensor (anisotropy) or spatially varying; the scalar coefficients are replaced by a mode-coupling matrix and ordinary/extraordinary rays split.

Both media are non-magnetic, \(\mu_1=\mu_2=\mu_0\).If dropped, the refractive index no longer sets the wave impedance alone; every \(n\) must be replaced by an impedance \(Z=\sqrt{\mu/\varepsilon}\) and the coefficients rewritten in terms of \(Z\).

Both media are lossless, so \(n_1,n_2\) are real.If dropped, \(n\to n+i\kappa\) is complex, the refracted angle becomes complex, and \(r\) acquires a magnitude and a phase describing absorption and evanescent penetration; Brewster's angle degrades into a pseudo-Brewster minimum.

A single monochromatic plane wave of frequency \(\omega\).If dropped, dispersion \(n(\omega)\) makes each spectral component reflect differently; a pulse or beam must be decomposed and superposed.

The interface is atomically sharp compared with the wavelength.If dropped (a graded index over a distance \(\sim\lambda\)), the single boundary condition is replaced by a transfer-matrix integration through the transition layer and reflection is suppressed.

Derivation

Take the interface as the plane \(z=0\), with medium 1 in \(z<0\) and medium 2 in \(z>0\). The plane of incidence is the \(xz\)-plane. Incident, reflected and transmitted wavevectors are

\[ \vec{k}_i=k_1(\sin\theta_i,0,\cos\theta_i),\quad \vec{k}_r=k_1(\sin\theta_i,0,-\cos\theta_i),\quad \vec{k}_t=k_2(\sin\theta_t,0,\cos\theta_t), \]

with \(k_j=n_j\omega/c\). We treat the \(s\)-polarization (\(\vec{E}\parallel\hat{y}\), transverse-electric) in full, then quote the \(p\)-result and derive Brewster's angle.

\[ k_{ix}=k_{rx}=k_{tx}\ \Rightarrow\ k_1\sin\theta_i=k_1\sin\theta_r=k_2\sin\theta_t \]

The total field must satisfy the same boundary condition at every point of \(z=0\) and all times; this forces the tangential phase factor \(e^{i(k_x x-\omega t)}\) to be common to all three waves. Hence \(\theta_r=\theta_i\) and Snell's law \(n_1\sin\theta_i=n_2\sin\theta_t\). A

\[ E_{0i}+E_{0r}=E_{0t} \]

For \(s\)-polarization \(\vec{E}=E_0\,\hat{y}\) is purely tangential to the interface. With no free surface charge, the tangential component of \(\vec{E}\) is continuous, so the \(y\)-amplitudes add on the incident side and equal the transmitted amplitude. A

\[ \vec{B}=\frac{1}{\omega}\,\vec{k}\times\vec{E}\quad\Rightarrow\quad B_x=-\frac{k_z}{\omega}E_y \]

Faraday's law for a plane wave, \(\nabla\times\vec{E}=-\partial_t\vec{B}\), gives \(\vec{k}\times\vec{E}=\omega\vec{B}\). The tangential (\(x\)) magnetic component is set by the normal wavevector component \(k_z=\pm k\cos\theta\). B

\[ H_{x}^{(1)}=H_{x}^{(2)}\ \Rightarrow\ \frac{k_1\cos\theta_i}{\omega\mu_0}\big(E_{0i}-E_{0r}\big)=\frac{k_2\cos\theta_t}{\omega\mu_0}E_{0t} \]

With no free surface current, tangential \(\vec{H}=\vec{B}/\mu_0\) is continuous. The incident and reflected waves carry \(k_z\) of opposite sign, so their \(B_x\) contributions subtract; equating to the transmitted term gives this relation. B

\[ n_1\cos\theta_i\,(E_{0i}-E_{0r})=n_2\cos\theta_t\,E_{0t} \]

Insert \(k_j=n_j\omega/c\) and cancel the common factor \(\omega/(c\mu_0)\). This is the tangential-\(\vec{H}\) condition in terms of refractive indices. A

\[ n_1\cos\theta_i(E_{0i}-E_{0r})=n_2\cos\theta_t(E_{0i}+E_{0r}) \]

Substitute the tangential-\(\vec{E}\) condition \(E_{0t}=E_{0i}+E_{0r}\) from Step 2. We now have one equation in the two amplitudes. A

\[ E_{0i}\big(n_1\cos\theta_i-n_2\cos\theta_t\big)=E_{0r}\big(n_1\cos\theta_i+n_2\cos\theta_t\big) \]

Collect the \(E_{0i}\) and \(E_{0r}\) terms on opposite sides. Pure algebraic rearrangement. A

\[ r_s\equiv\frac{E_{0r}}{E_{0i}}=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t},\qquad t_s\equiv\frac{E_{0t}}{E_{0i}}=1+r_s=\frac{2n_1\cos\theta_i}{n_1\cos\theta_i+n_2\cos\theta_t} \]

Divide by \(E_{0i}\) for the reflection amplitude; the transmission amplitude follows from Step 2, \(t_s=1+r_s\). A

\[ E_{0i}\cos\theta_i+E_{0r}\cos\theta_i=E_{0t}\cos\theta_t,\qquad n_1(E_{0i}-E_{0r})=n_2E_{0t} \]

Repeat Steps 2 and 4 for \(p\)-polarization, where \(\vec{B}\parallel\hat{y}\) and \(\vec{E}\) lies in the plane of incidence. Now it is the tangential \(E_x=E_0\cos\theta\) that is continuous, while \(\vec{H}=\vec{B}/\mu_0\) is purely tangential and gives the second relation. C

\[ r_p=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t},\qquad t_p=\frac{2n_1\cos\theta_i}{n_2\cos\theta_i+n_1\cos\theta_t} \]

Eliminate \(E_{0t}\) between the two Step-9 relations and solve for \(E_{0r}/E_{0i}\) exactly as in Steps 6–8. (Sign of \(r_p\) follows the convention that positive \(E_{0r}\) points as in the incident wave at grazing incidence.) B

\[ r_p=0\ \Rightarrow\ n_2\cos\theta_i=n_1\cos\theta_t \]

The reflected \(p\)-wave vanishes when the numerator of \(r_p\) is zero. This defines Brewster's angle \(\theta_i=\theta_B\). A

\[ n_2\cos\theta_B=n_1\cos\theta_t\ \text{and}\ n_1\sin\theta_B=n_2\sin\theta_t\ \Rightarrow\ \sin\theta_B\cos\theta_B=\sin\theta_t\cos\theta_t \]

Eliminate \(n_2\) using Snell's law: from Snell \(n_2=n_1\sin\theta_B/\sin\theta_t\); substitute into the Brewster condition. B

\[ \sin 2\theta_B=\sin 2\theta_t\ \Rightarrow\ \theta_B+\theta_t=\frac{\pi}{2} \]

Use \(2\sin\theta\cos\theta=\sin 2\theta\). The non-trivial solution is \(2\theta_B=\pi-2\theta_t\): the reflected and refracted rays are mutually perpendicular at Brewster incidence. C

\[ n_1\sin\theta_B=n_2\sin\theta_t=n_2\sin\!\left(\tfrac{\pi}{2}-\theta_B\right)=n_2\cos\theta_B\ \Rightarrow\ \boxed{\tan\theta_B=\frac{n_2}{n_1}} \]

Insert \(\theta_t=\pi/2-\theta_B\) into Snell's law and divide by \(\cos\theta_B\). This is Brewster's angle in closed form. A

Result

\[ r_s=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t},\quad r_p=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t},\quad \tan\theta_B=\frac{n_2}{n_1} \]

Reading. The reflection amplitude for each polarization is the difference of the two media's "effective indices" \(n\cos\theta\) (weighted oppositely for \(s\) and \(p\)) divided by their sum — a wave-impedance mismatch. When the mismatch vanishes the reflection vanishes; for \(p\)-polarization this happens at a real angle, Brewster's angle, where \(\tan\theta_B=n_2/n_1\), and the reflected beam is then perfectly \(s\)-polarized. The transmission amplitudes exceed unity near grazing incidence because amplitude, not power, is being compared.

Units check. Each \(n\cos\theta\) is dimensionless (index times a pure cosine), so \(r_s,r_p,t_s,t_p\) are dimensionless ratios of field amplitudes, as required for coefficients that map one \(\mathrm{V\,m^{-1}}\) onto another. \(\tan\theta_B=n_2/n_1\) is a ratio of indices, dimensionless, and \(\theta_B\) is a genuine angle.

Limiting cases

Normal incidence (\(\theta_i=\theta_t=0\)): both reduce to \(r=\dfrac{n_1-n_2}{n_1+n_2}\), \(t=\dfrac{2n_1}{n_1+n_2}\); the \(s\)/\(p\) distinction disappears.
Grazing incidence (\(\theta_i\to\pi/2\)): \(\cos\theta_i\to0\) gives \(r_s\to-1\) and \(r_p\to-1\) (magnitude 1) — every surface becomes a perfect mirror at glancing angle.
Index matched (\(n_1=n_2\)): \(r_s=r_p=0\), \(t=1\); there is no interface optically.
Brewster incidence (\(\theta_i=\theta_B\)): \(r_p=0\) exactly, \(r_s\neq0\); reflected light is fully \(s\)-polarized.
Near-normal small contrast (\(n_2=n_1+\delta n\)): \(R\approx(\delta n/2n_1)^2\), the basis of low-reflection index matching.

Breaks when

Total internal reflection: for \(n_1>n_2\) and \(\theta_i\) beyond the critical angle \(\theta_c=\arcsin(n_2/n_1)\), Snell's law gives no real \(\theta_t\). \(\cos\theta_t\) turns imaginary, \(|r|=1\), and the transmitted field is evanescent — the real-angle formulas above no longer apply and must be continued to complex \(\theta_t\).
Absorbing or conducting media: a complex index \(n\to n+i\kappa\) makes the coefficients complex; reflection carries a phase shift, Brewster's zero softens to a non-zero minimum (pseudo-Brewster), and energy is dissipated at the interface.
Anisotropic (birefringent) media: with a dielectric tensor the single \(\theta_t\) splits into ordinary and extraordinary rays; a scalar \(r_p\) no longer describes the response and mode conversion occurs.
Rough or structured interfaces: when surface height variations approach \(\lambda\), specular Fresnel reflection is supplemented by diffuse scattering and the boundary conditions cannot be applied on a single flat plane.

Failure modes

Confusing amplitude with power. \(t_s\) can exceed 1 near grazing incidence; students wrongly conclude energy is created. Power transmittance is \(T=\dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}\,t^2\), and \(R+T=1\) always holds.
Writing \(R=t^2\) directly. Reflectance is \(R=r^2\), but transmittance is not \(t^2\); the geometric and impedance factor \(n_2\cos\theta_t/(n_1\cos\theta_i)\) is essential.
Sign-convention panic. \(r_p\) changes overall sign between textbooks (Hecht vs. Griffiths conventions). The physics — magnitude, Brewster zero — is convention-independent; only the reference direction of \(\vec{E}_r\) differs.
Using \(\tan\theta_B=n_1/n_2\). The index in the numerator is that of the second (transmitting) medium: \(\tan\theta_B=n_2/n_1\). Air-to-glass gives \(\theta_B\approx56^\circ\), not \(34^\circ\).
Expecting a Brewster angle for \(s\)-polarization. \(r_s\) has no real zero for ordinary dielectrics; only \(p\)-light can be extinguished on reflection.
Forgetting Snell's law couples \(\theta_t\) to \(\theta_i\). \(\cos\theta_t\) must be computed from \(\cos\theta_t=\sqrt{1-(n_1/n_2)^2\sin^2\theta_i}\), not treated as independent.

Discussion

Physically, the Fresnel coefficients are impedance-matching ratios. Each medium presents a wave impedance to the tangential fields, and the combination \(n\cos\theta\) is precisely the tangential admittance seen at the interface for that polarization. Reflection is the mismatch of these admittances, exactly as a voltage wave reflects from a change in transmission-line impedance. This analogy is not loose: the boundary conditions on tangential \(\vec{E}\) and \(\vec{H}\) are the optical equivalents of continuity of voltage and current, and thin-film optics is literally impedance-transformer design.

Brewster's angle has an elegant microscopic picture. The transmitted wave drives dipoles in medium 2 oscillating along \(\vec{E}_t\); these dipoles radiate the reflected wave. A dipole cannot radiate along its own axis. At Brewster incidence the would-be reflected direction lies exactly along the \(p\)-polarized dipole axis (because reflected and refracted rays are perpendicular), so no \(p\)-reflection is possible. The \(s\)-dipoles point out of the plane of incidence and radiate freely, which is why the reflection is purely \(s\)-polarized.

The equations also encode phase information that pure reflectance hides. For external reflection (\(n_1<n_2\)) below Brewster, \(r_p>0\) while \(r_s<0\), a relative \(\pi\) phase difference; above Brewster \(r_p\) flips sign. Ellipsometry exploits exactly this: measuring the complex ratio \(r_p/r_s\) determines film thickness and index to sub-nanometre precision, because the phase is exquisitely sensitive to the interface structure.

At the deepest level the coefficients are the classical limit of a photon scattering amplitude. Continuing \(\theta_t\) to complex values in the total-internal-reflection regime produces the Goos–Hänchen lateral beam shift and the evanescent tunnelling that frustrated total reflection and near-field microscopy rely on; the same analytic structure, continued to complex frequency, yields the surface-plasmon poles of a metal interface. The Fresnel equations are thus the entry point to the entire analytic theory of interface electromagnetics.

Common misconceptions. Brewster's angle is not the critical angle — the critical angle exists only for \(n_1>n_2\) and concerns transmission cutoff, whereas Brewster exists for any index step and concerns polarization. And "no reflection at Brewster" applies only to \(p\)-polarization; unpolarized light still reflects, just fully \(s\)-polarized.

Worked examples

Example 1 — 4% loss at a glass surface. A laser hits an air–glass interface (\(n_1=1.00\), \(n_2=1.50\)) at normal incidence. Find the reflectance and transmittance.

\[ r=\frac{n_1-n_2}{n_1+n_2}=\frac{1.00-1.50}{1.00+1.50}=\frac{-0.50}{2.50}=-0.200 \]

Normal-incidence limit of both Fresnel coefficients (\(\theta=0\)). A

\[ R=r^2=(-0.200)^2=0.0400 \]

Reflectance is the squared amplitude; 4.0% of the power reflects. A

\[ t=1+r=0.800,\qquad T=\frac{n_2}{n_1}\,t^2=1.50\times0.640=0.960 \]

Transmission amplitude from \(t=1+r\); power transmittance includes the index factor (\(\cos\theta=1\) at normal incidence). B

\[ R=4.0\%,\qquad T=96.0\%,\qquad R+T=1 \]

Reading. Each uncoated glass surface loses 4% — the reason multi-element lenses need anti-reflection coatings. Units check. \(R,T\) are dimensionless power fractions summing to unity, as energy conservation demands.

Example 2 — Brewster's angle for glass and the residual \(s\)-reflection. For the same air–glass interface, find \(\theta_B\), the refraction angle there, and the reflectance of \(s\)-polarized light at Brewster incidence.

\[ \tan\theta_B=\frac{n_2}{n_1}=\frac{1.50}{1.00}=1.50\ \Rightarrow\ \theta_B=\arctan 1.50=56.3^\circ \]

Brewster's angle in closed form. A

\[ \theta_t=90^\circ-\theta_B=33.7^\circ\quad(\text{check: }\tfrac{\sin 56.3^\circ}{\sin 33.7^\circ}=\tfrac{0.832}{0.555}=1.50=n_2/n_1) \]

Reflected and refracted rays are perpendicular at Brewster incidence; Snell's law confirms consistency. A

\[ r_s=\frac{n_1\cos\theta_B-n_2\cos\theta_t}{n_1\cos\theta_B+n_2\cos\theta_t}=\frac{0.555-1.50(0.832)}{0.555+1.50(0.832)}=\frac{-0.693}{1.803}=-0.384 \]

Insert numbers into the \(s\)-coefficient with \(\cos 56.3^\circ=0.555\), \(\cos 33.7^\circ=0.832\). B

\[ R_s=r_s^2=(-0.384)^2=0.148,\qquad R_p=0 \]

Square for reflectance; \(r_p=0\) by construction at Brewster's angle. A

\[ \theta_B=56.3^\circ,\qquad R_p=0,\qquad R_s=14.8\% \]

Reading. At \(56.3^\circ\) the reflected beam contains no \(p\)-component, so it is 100% \(s\)-polarized — the working principle of pile-of-plates polarizers and Brewster windows. Units check. \(\theta_B\) is an angle; \(R_s\) is a dimensionless power fraction between 0 and 1.

Problems

(A) Light strikes an air–water interface (\(n_1=1.00\), \(n_2=1.33\)) at normal incidence. Find the reflectance.
Solution
\(r=\dfrac{1.00-1.33}{1.00+1.33}=\dfrac{-0.33}{2.33}=-0.1416\). Then \(R=r^2=0.0201\approx 2.0\%\). About two percent of light reflects off calm water at normal viewing.
(A) Find Brewster's angle for the air–water interface of Problem 1.
Solution
\(\tan\theta_B=n_2/n_1=1.33\Rightarrow\theta_B=\arctan 1.33=53.1^\circ\). Reflected glare off water is fully \(s\)-polarized (horizontal) at this incidence, which is why polarized sunglasses with a vertical transmission axis cut it.
(B) A ray travels inside glass (\(n_1=1.50\)) toward a glass–air interface (\(n_2=1.00\)). Find both the internal Brewster angle and the critical angle, and explain their different roles.
Solution
Brewster: \(\tan\theta_B=n_2/n_1=1.00/1.50=0.667\Rightarrow\theta_B=33.7^\circ\), where \(r_p=0\). Critical: \(\sin\theta_c=n_2/n_1=0.667\Rightarrow\theta_c=41.8^\circ\), beyond which total internal reflection occurs and no light transmits. Since \(\theta_B<\theta_c\), the Brewster zero lies inside the transmitting range; above \(\theta_c\) the real-angle Fresnel formulas break down and \(|r|=1\).
(B) Unpolarized light in air hits glass (\(n_2=1.50\)) at \(\theta_i=30.0^\circ\). Find the transmission amplitude \(t_s\) for the \(s\)-component.
Solution
Snell: \(\sin\theta_t=\dfrac{n_1\sin\theta_i}{n_2}=\dfrac{\sin 30^\circ}{1.50}=0.3333\Rightarrow\theta_t=19.5^\circ\), \(\cos\theta_t=0.9428\). Then \(t_s=\dfrac{2n_1\cos\theta_i}{n_1\cos\theta_i+n_2\cos\theta_t}=\dfrac{2(1)(0.8660)}{(1)(0.8660)+1.50(0.9428)}=\dfrac{1.732}{0.866+1.414}=\dfrac{1.732}{2.280}=0.760\).
(C) For the air–glass interface at normal incidence (\(n_1=1.00\), \(n_2=1.50\)), verify explicitly that \(R+T=1\) using the amplitude coefficients and the correct power factors.
Solution
\(r=-0.200\Rightarrow R=0.0400\). \(t=1+r=0.800\). Power transmittance at normal incidence: \(T=\dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}t^2=\dfrac{1.50}{1.00}(0.800)^2=1.50\times0.640=0.960\). Sum: \(R+T=0.0400+0.960=1.000\). Note that using \(T=t^2=0.64\) would wrongly give \(R+T=0.68\); the index factor \(n_2/n_1\), from the change in the Poynting flux across the interface, restores energy conservation.