Fresnel Equations from Electromagnetic Boundary Conditions

Derivation

Fresnel Equations from Electromagnetic Boundary Conditions

D-201 Home PU-206 Threads light · fields · energy Depends on plane-wave-solutions, maxwell-boundary-conditions, Laws of Reflection and Refraction from Fermat

Statement

For a monochromatic plane wave incident from a linear, isotropic, non-magnetic medium of refractive index \(n_1\) onto a planar interface (at \(z=0\)) with a second such medium of index \(n_2\), we derive the amplitude reflection and transmission coefficients \(r_s, t_s\) (electric field perpendicular to the plane of incidence) and \(r_p, t_p\) (electric field in the plane of incidence) by imposing continuity of the tangential components of \(\vec{E}\) and \(\vec{H}\) at the interface, and we obtain the energy relations \(R=|r|^2\), \(T=\dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2\) with \(R+T=1\).

Why it matters

The Fresnel equations are the quantitative backbone of every reflection-and-transmission phenomenon in classical optics: anti-reflection coatings, fibre coupling, polarizing beam splitters, the glare that polarized sunglasses reject, and the Brewster windows in laser cavities all follow from them. They convert the abstract statement "Maxwell's equations plus boundary conditions" into concrete numbers for how much light bounces and how much passes.

Conceptually they show that the direction of propagation (Snell's law) and the partitioning of energy are governed by two logically distinct pieces of information: phase matching along the interface fixes the angles, while the field-amplitude matching derived here fixes the amplitudes. The same boundary-condition method scaffolds thin-film optics, total internal reflection with evanescent fields, and the optics of metals and metamaterials.

Assumptions

Both media are linear, homogeneous, isotropic and lossless.If dropped, \(n\) becomes complex or tensorial; \(r,t\) acquire phase shifts and the simple real forms below fail (metals, birefringent crystals).

Both media are non-magnetic, \(\mu_1=\mu_2=\mu_0\).If dropped, the wave impedance is \(\eta=\sqrt{\mu/\varepsilon}\) rather than \(\mu_0 c/n\), and every "\(n\)" in the coefficients must be replaced by \(n/\mu_r\); Brewster's condition changes.

The interface is planar, sharp on the scale of a wavelength, and infinite in extent.If dropped (graded or rough interface), one must integrate through a transition layer or add scattering; a single amplitude coefficient no longer suffices.

The incident field is a single-frequency plane wave; the geometry is time-harmonic \(e^{-i\omega t}\).If dropped, one Fourier-decomposes the beam; each angular/frequency component obeys the Fresnel result but a finite beam also shows the Goos–Hänchen shift near total internal reflection.

There are no free surface charges or currents at the interface.If dropped, the tangential-\(H\) condition gains a surface-current term \(\vec{K}\), valid only for idealized perfect conductors, not dielectrics.

Derivation

\[ \vec{k}_i=\frac{n_1\omega}{c}(\sin\theta_i,\,0,\,\cos\theta_i),\quad \vec{k}_r=\frac{n_1\omega}{c}(\sin\theta_i,\,0,\,-\cos\theta_i),\quad \vec{k}_t=\frac{n_2\omega}{c}(\sin\theta_t,\,0,\,\cos\theta_t) \]

Choose the interface at \(z=0\) and the plane of incidence as the \(xz\)-plane; each wave is a plane-wave solution \(\vec{E}e^{i(\vec{k}\cdot\vec{r}-\omega t)}\) with \(|\vec k|=n\omega/c\). A

\[ (\vec{k}_i)_x=(\vec{k}_r)_x=(\vec{k}_t)_x \;\Longrightarrow\; n_1\sin\theta_i=n_1\sin\theta_r=n_2\sin\theta_t \]

The boundary conditions must hold for all \(x,y,t\) on \(z=0\); this forces the tangential phase factors to be identical, giving \(\theta_r=\theta_i\) and Snell's law. This is the prior result we import; below we match amplitudes. A

\[ \text{s-polarization: } \vec{E}=E\,\hat{y}, \qquad \vec{H}=\frac{1}{\mu_0\omega}\,\vec{k}\times\vec{E}=\frac{n}{\mu_0 c}\,(\hat{k}\times\hat{y})\,E \]

For the transverse-electric case the electric field is perpendicular to the plane of incidence (along \(\hat y\)); Faraday's law for a plane wave gives \(\vec H\) from \(\vec E\), with \(|\vec k|=n\omega/c\) and \(\mu=\mu_0\). B

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

Tangential \(\vec E\) is continuous (from \(\nabla\times\vec E=-\partial_t\vec B\) integrated over a pillbox loop). Here \(E\) lies along \(\hat y\), which is tangential, so the field amplitudes add directly at \(z=0\). A

\[ (\hat{k}_i\times\hat{y})_x=-\cos\theta_i,\quad (\hat{k}_r\times\hat{y})_x=+\cos\theta_i,\quad (\hat{k}_t\times\hat{y})_x=-\cos\theta_t \]

Evaluate the tangential (\(x\)) component of \(\hat k\times\hat y\) for each wave using the wavevectors of Step 1; only the \(x\)-component of \(\vec H\) is tangential to the interface. B

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

Tangential \(\vec H\) is continuous (from \(\nabla\times\vec H=\partial_t\vec D\) with no surface current). Insert the amplitudes \(H_x=(n/\mu_0 c)(\hat k\times\hat y)_x E\) from Step 5; the common factor \(1/\mu_0 c\) cancels. B

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

Eliminate \(E_{0t}\) using Step 4. Two continuity conditions in two unknown ratios now close the system algebraically. A

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

Collect \(E_{0i}\) and \(E_{0r}\) in Step 7 and solve; \(t_s=1+r_s\) follows directly from Step 4. These are the s-polarization Fresnel coefficients. B

\[ \text{p-polarization: } \vec{H}=H\,\hat{y}, \qquad \vec{E}=\eta\,(\hat{H}\times\hat{k})\,H,\quad \eta=\frac{\mu_0 c}{n} \]

For the transverse-magnetic case \(\vec H\) is along \(\hat y\) and \(\vec E\) lies in the plane of incidence; the wave impedance \(\eta=\sqrt{\mu_0/\varepsilon}=\mu_0 c/n\) relates the field magnitudes. B

\[ H_{0i}+H_{0r}=H_{0t},\qquad \eta_1\cos\theta_i\,(H_{0i}-H_{0r})=\eta_2\cos\theta_t\,H_{0t} \]

Now \(\vec H\parallel\hat y\) is tangential (first equation), and the tangential \(x\)-component of \(\vec E\) is \(\eta H(\hat y\times\hat k)_x=\pm\eta\cos\theta\,H\) (second equation). Same two boundary conditions, roles of \(E\) and \(H\) swapped. B

\[ \frac{H_{0r}}{H_{0i}}=\frac{\cos\theta_i/n_1-\cos\theta_t/n_2}{\cos\theta_i/n_1+\cos\theta_t/n_2}=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t} \]

Substitute \(\eta=\mu_0 c/n\) into Step 10, eliminate \(H_{0t}\), and clear denominators by multiplying through by \(n_1 n_2\). B

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

Reflected and incident \(E\) share the same \(\eta_1\), so \(E_{0r}/E_{0i}=H_{0r}/H_{0i}\); the transmitted case carries a factor \(\eta_2/\eta_1=n_1/n_2\), giving \(t_p\). These are the p-polarization coefficients (Verdet sign convention). B

\[ \langle\vec{S}\rangle\cdot\hat{z}=\tfrac{1}{2}\,\frac{n}{\mu_0 c}\,|E|^2\cos\theta,\qquad R=\frac{|E_{0r}|^2}{|E_{0i}|^2}=|r|^2 \]

The normal energy flux is the \(z\)-projection of the time-averaged Poynting vector \(\langle\vec S\rangle=\tfrac12\mathrm{Re}(\vec E\times\vec H^*)\). Reflected and incident waves are in the same medium at the same angle, so their prefactors cancel. C

\[ T=\frac{\langle S_t\rangle\cdot\hat z}{\langle S_i\rangle\cdot\hat z}=\frac{n_2\cos\theta_t}{n_1\cos\theta_i}\,|t|^2,\qquad R+T=1 \]

The transmitted flux carries the factor \(n_2\cos\theta_t\) versus \(n_1\cos\theta_i\); substituting the s- (or p-) coefficients and simplifying with \(a=n_1\cos\theta_i,\ b=n_2\cos\theta_t\) gives \(\frac{(a-b)^2}{(a+b)^2}+\frac{4ab}{(a+b)^2}=1\), i.e. energy conservation. C

Result

\[ r_s=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t},\quad t_s=\frac{2n_1\cos\theta_i}{n_1\cos\theta_i+n_2\cos\theta_t} \] \[ r_p=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t},\quad t_p=\frac{2n_1\cos\theta_i}{n_2\cos\theta_i+n_1\cos\theta_t} \] \[ R=|r|^2,\qquad T=\frac{n_2\cos\theta_t}{n_1\cos\theta_i}\,|t|^2,\qquad R+T=1 \]

Reading. Reflection is set by the mismatch of the quantity \(n\cos\theta\) across the interface, with the two indices entering in opposite order for s- and p-polarization. When the two "effective impedances" \(n_1\cos\theta_i\) and \(n_2\cos\theta_t\) (s) — or \(n_2\cos\theta_i\) and \(n_1\cos\theta_t\) (p) — are equal, the numerator vanishes and there is no reflected wave. The transmission amplitude \(t\) can exceed 1 (the transmitted \(E\)-field can be larger than the incident one) yet energy is still conserved because \(T\) carries the geometric/impedance factor \(n_2\cos\theta_t/(n_1\cos\theta_i)\).

Units check. Every coefficient is a ratio of terms of the form \(n\cos\theta\), which is dimensionless (index and cosine are both pure numbers); hence \(r,t,R,T\) are dimensionless, as required for field ratios and energy fractions. In \(T\) the prefactor \(n_2\cos\theta_t/(n_1\cos\theta_i)\) is also dimensionless, so \(R+T\) is a pure number equal to 1.

Limiting cases

Normal incidence \(\theta_i=0\): both polarizations give \(r=\dfrac{n_1-n_2}{n_1+n_2}\), \(t=\dfrac{2n_1}{n_1+n_2}\); for air–glass (\(1\to1.5\)), \(r=-0.2\), \(R=4\%\).
Brewster angle (p-pol): \(r_p=0\) when \(n_2\cos\theta_i=n_1\cos\theta_t\), equivalently \(\tan\theta_B=n_2/n_1\), giving \(\theta_i+\theta_t=90^\circ\); reflected light is then purely s-polarized.
Grazing incidence \(\theta_i\to90^\circ\): \(\cos\theta_i\to0\) so \(r_s\to-1\) and \(r_p\to-1\), \(R\to1\) — every interface becomes a mirror at glancing angles.
Index matching \(n_2\to n_1\): \(r_s,r_p\to0\), \(t\to1\) — no interface optically exists.
Total internal reflection (\(n_1>n_2\), \(\theta_i>\theta_c\)): \(\cos\theta_t\) becomes imaginary, \(|r|=1\), \(R=1\), and \(r\) acquires a polarization-dependent phase.

Breaks when

Absorbing or conducting media. With complex \(\tilde n=n+i\kappa\), \(\cos\theta_t\) and the coefficients become complex; \(R\ne|r_{\text{real}}|^2\) and one must use the complex Fresnel forms, so the real algebra above is invalid (metals, doped semiconductors).
Beyond the critical angle in TIR. \(\sin\theta_t=(n_1/n_2)\sin\theta_i>1\) has no real \(\theta_t\); the "transmitted" wave is evanescent, \(T=0\), and the naive \(T=\tfrac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2\) with real \(\cos\theta_t\) gives nonsense.
Magnetic media. If \(\mu_r\ne1\), impedance is \(\eta=\mu_0\mu_r c/n\); replacing \(n\) by \(n\) alone is wrong, and Brewster's angle shifts (or disappears for one polarization).
Ultrathin films / near-field / rough interfaces. When the interface transition or a coating is comparable to \(\lambda\), multiple reflections interfere and a single boundary coefficient no longer describes the response.

Failure modes

Writing \(T=|t|^2\) (forgetting the \(n_2\cos\theta_t/(n_1\cos\theta_i)\) factor), then "discovering" that \(R+T\ne1\).
Swapping the index order between \(r_s\) and \(r_p\): the correct forms have \(n_1\cos\theta_i\pm n_2\cos\theta_t\) for s and \(n_2\cos\theta_i\pm n_1\cos\theta_t\) for p.
Using \(\cos\theta_i\) where \(\cos\theta_t\) belongs (or vice versa) after applying Snell's law; the transmitted term always carries \(\theta_t\).
Sign-convention confusion for \(r_p\): different textbooks flip the sign, so quoting a numerical \(r_p\) without stating the convention leads to a wrong reflected-field direction (though \(R=|r_p|^2\) is convention-independent).
Applying the real formulas past the critical angle instead of continuing \(\cos\theta_t\) to imaginary values.
Assuming \(t>1\) violates energy conservation; it does not, because intensity also depends on \(n\) and beam projection.

Discussion

The derivation makes explicit that reflection is an impedance-mismatch phenomenon. Rewriting \(r_s=\dfrac{Z_2^{-1}-Z_1^{-1}}{Z_2^{-1}+Z_1^{-1}}\) with the "transverse impedance" \(Z=\mu_0 c/(n\cos\theta)\) (for s) shows the Fresnel result is formally identical to a wave meeting a change of characteristic impedance on a transmission line. The plane of incidence, the polarization, and the angle all enter only through how they set that transverse impedance, which is why the same algebra transfers to acoustics, quantum-mechanical step potentials, and microwave engineering.

The Brewster angle is the sharpest physical fingerprint of the p-polarization formula. Its microscopic origin is that at \(\theta_i+\theta_t=90^\circ\) the transmitted wave's oscillating dipoles point exactly along the would-be reflected ray; a dipole does not radiate along its own axis, so no p-polarized reflection can be launched. This dipole picture, absent from the boundary-condition algebra, is what connects Fresnel's macroscopic result to the microscopic radiation of the medium's electrons.

The energy relation \(R+T=1\) is not an extra assumption but a theorem: it drops out of the same boundary conditions once the Poynting flux is computed, confirming that no energy is stored or lost at a lossless interface. The asymmetry between \(r\) (a simple field ratio) and \(T\) (which needs the \(n\cos\theta\) projection) is the recurring subtlety students meet — amplitudes and energies scale differently across a change of medium.

At oblique incidence the s- and p-coefficients differ, so a general polarization state is transformed on reflection: linearly polarized light becomes elliptically polarized once \(r_s\) and \(r_p\) carry different phases, which happens automatically beyond the critical angle (the Fresnel rhomb exploits exactly this to make a quarter-wave retarder from total internal reflection). Continuing \(\cos\theta_t=\sqrt{1-(n_1/n_2)^2\sin^2\theta_i}\) analytically into the imaginary axis turns the transmitted plane wave into a surface-bound evanescent field with \(|r|=1\) and a real, polarization-dependent reflection phase — the seed of frustrated TIR, attenuated-total-reflection spectroscopy, and waveguide coupling.

Common misconceptions. "Transmission amplitude greater than one breaks conservation" — false; only \(T\), not \(t\), is bounded by 1. "Unpolarized light has a Brewster angle where it fully transmits" — only the p-component vanishes in reflection; the s-component still reflects, so reflected light is merely fully polarized, not extinguished. "\(R\) depends on which way you traverse the interface" — reflectance is the same for \(1\to2\) and \(2\to1\) at Snell-conjugate angles (Stokes relations), even though the sign of \(r\) flips.

Worked examples

\[ \text{s-polarized light, air}\to\text{glass},\ n_1=1,\ n_2=1.5,\ \theta_i=30^\circ.\ \text{Find } r_s,R,t_s,T. \]

Symbolic form first: \(r_s=\dfrac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\), with \(\theta_t\) from Snell's law. A

\[ \sin\theta_t=\frac{n_1}{n_2}\sin\theta_i=\frac{1}{1.5}(0.5)=0.3333\ \Rightarrow\ \theta_t=19.47^\circ,\ \cos\theta_t=0.9428 \]

Apply Snell's law, then take the cosine of the refraction angle. A

\[ a=n_1\cos\theta_i=1(0.8660)=0.8660,\qquad b=n_2\cos\theta_t=1.5(0.9428)=1.4142 \]

Form the two effective-impedance terms; \(\cos30^\circ=0.8660\). A

\[ r_s=\frac{a-b}{a+b}=\frac{0.8660-1.4142}{0.8660+1.4142}=-0.2404,\quad t_s=\frac{2a}{a+b}=\frac{1.7320}{2.2802}=0.7596 \]

Substitute numbers into the Step-8 formulas. The negative \(r_s\) is the \(180^\circ\) phase flip on reflecting off the denser medium. B

\[ R=r_s^2=0.0578,\qquad T=\frac{b}{a}\,t_s^2=\frac{1.4142}{0.8660}(0.7596)^2=0.9422 \]

Energy fractions; check \(R+T=0.0578+0.9422=1.000\). B

\[ r_s=-0.240,\quad t_s=0.760,\quad R=5.8\%,\quad T=94.2\% \]

Reading. At \(30^\circ\), s-polarized light reflects about 6% of its power from a glass surface, slightly more than the 4% at normal incidence, and \(R+T=1\) confirms the algebra. Units check. All quantities dimensionless; \(R,T\) are power fractions summing to unity.

\[ \text{p-polarized light, air}\to\text{glass},\ n_1=1,\ n_2=1.5.\ \text{Find the Brewster angle and verify } r_p=0. \]

Brewster condition from the \(r_p\) numerator: \(n_2\cos\theta_i=n_1\cos\theta_t\). A

\[ \tan\theta_B=\frac{n_2}{n_1}=\frac{1.5}{1}=1.5\ \Rightarrow\ \theta_B=56.31^\circ,\quad \cos\theta_B=0.5547,\ \sin\theta_B=0.8321 \]

The numerator condition combined with Snell's law reduces to \(\tan\theta_B=n_2/n_1\). B

\[ \sin\theta_t=\frac{1}{1.5}(0.8321)=0.5547\ \Rightarrow\ \theta_t=33.69^\circ,\ \cos\theta_t=0.8321,\quad \theta_B+\theta_t=90.0^\circ \]

Refraction angle from Snell's law; note incidence and refraction rays are perpendicular. B

\[ n_2\cos\theta_i=1.5(0.5547)=0.8321,\qquad n_1\cos\theta_t=1(0.8321)=0.8321 \]

Evaluate the two terms of the \(r_p\) numerator; they are equal. B

\[ r_p=\frac{0.8321-0.8321}{0.8321+0.8321}=0,\qquad R_p=0,\qquad T_p=1 \]

Substitute into the p-reflection formula. A

\[ \theta_B=56.3^\circ,\qquad r_p=0,\qquad R_p=0,\qquad T_p=100\% \]

Reading. At \(56.3^\circ\) the p-polarized reflection is fully suppressed, so glare reflected from a horizontal glass surface is entirely s-polarized — the principle behind polarizing sunglasses and Brewster laser windows. Units check. \(\tan\theta_B\) equals the pure-number index ratio; all coefficients dimensionless.

Problems

Compute the normal-incidence reflectance for a water–air interface (\(n_1=1.33\), \(n_2=1.00\)) and state whether the reflected wave suffers a \(180^\circ\) phase shift.
Solution
\(r=\dfrac{n_1-n_2}{n_1+n_2}=\dfrac{1.33-1.00}{1.33+1.00}=\dfrac{0.33}{2.33}=0.1416\). Then \(R=r^2=0.0201\approx2.0\%\). Because \(n_1>n_2\) the numerator is positive, so \(r>0\): there is no \(180^\circ\) phase flip (reflecting from the less-dense medium, the field keeps its sign). \(T=1-R=98.0\%\).
For s-polarized light incident from air (\(n_1=1\)) onto glass (\(n_2=1.52\)) at \(\theta_i=60^\circ\), find \(r_s\) and \(R_s\).
Solution
Snell: \(\sin\theta_t=\dfrac{1}{1.52}\sin60^\circ=\dfrac{0.8660}{1.52}=0.5698\Rightarrow\theta_t=34.74^\circ,\ \cos\theta_t=0.8213\). Terms: \(a=n_1\cos60^\circ=0.5000\), \(b=n_2\cos\theta_t=1.52(0.8213)=1.2484\). \(r_s=\dfrac{a-b}{a+b}=\dfrac{0.5000-1.2484}{0.5000+1.2484}=\dfrac{-0.7484}{1.7484}=-0.4281\). \(R_s=r_s^2=0.183=18.3\%\).
Using the same air–glass interface and angle as Problem 2, find \(r_p\), \(R_p\), and explain why \(R_p<R_s\).
Solution
Terms for p: \(n_2\cos\theta_i=1.52(0.5000)=0.7600\), \(n_1\cos\theta_t=1(0.8213)=0.8213\). \(r_p=\dfrac{0.7600-0.8213}{0.7600+0.8213}=\dfrac{-0.0613}{1.5813}=-0.0388\). \(R_p=r_p^2=0.00150=0.15\%\). \(R_p<R_s\) because \(60^\circ\) is close to the Brewster angle \(\theta_B=\arctan(1.52)=56.7^\circ\), where the p-reflection is driven to zero; the s-reflection has no such minimum and grows steadily toward grazing.
Light travels inside glass (\(n_1=1.50\)) toward a glass–air interface (\(n_2=1.00\)). Find the critical angle, and state \(R\) for \(\theta_i=50^\circ\).
Solution
Critical angle: \(\sin\theta_c=\dfrac{n_2}{n_1}=\dfrac{1}{1.50}=0.6667\Rightarrow\theta_c=41.8^\circ\). Since \(50^\circ>\theta_c\), the incidence is beyond the critical angle: \(\sin\theta_t=1.5\sin50^\circ=1.149>1\) has no real solution, the transmitted wave is evanescent, \(|r|=1\) and \(R=1\) (100%). All incident power is totally internally reflected; \(T=0\).
Verify \(R+T=1\) explicitly for p-polarization at the interface of Problem 3 (\(n_1=1\), \(n_2=1.52\), \(\theta_i=60^\circ\)) by computing \(t_p\) and \(T_p\).
Solution
From Problem 3, \(n_2\cos\theta_i=0.7600\), \(n_1\cos\theta_t=0.8213\), so denominator \(=1.5813\). \(t_p=\dfrac{2n_1\cos\theta_i}{n_2\cos\theta_i+n_1\cos\theta_t}=\dfrac{2(0.5000)}{1.5813}=0.6324\). Transmittance factor: \(\dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}=\dfrac{1.52(0.8213)}{1(0.5000)}=\dfrac{1.2484}{0.5000}=2.4968\). \(T_p=2.4968(0.6324)^2=2.4968(0.3999)=0.9985=99.85\%\). With \(R_p=0.15\%\), \(R_p+T_p=0.0015+0.9985=1.000\). Energy conserved, and note \(t_p=0.632<1\) here while \(T_p\) is still near unity because of the impedance/projection factor.