Brewster's Angle, Total Internal Reflection, and Evanescent Waves

Derivation

Brewster's Angle, Total Internal Reflection, and Evanescent Waves

D-202 Home PU-206 Threads light · fields Depends on Fresnel Equations from Electromagnetic Boundary Conditions

Statement

Starting from the Fresnel amplitude reflection coefficients for a plane monochromatic wave crossing a planar interface between two lossless, non-magnetic dielectrics (indices \(n_1\), \(n_2\)), we derive three consequences: (i) the Brewster angle \(\tan\theta_B = n_2/n_1\) at which the \(p\)-polarised reflection coefficient vanishes; (ii) the critical angle \(\sin\theta_c = n_2/n_1\) (requiring \(n_1>n_2\)) beyond which no real refraction angle exists and all power is reflected; and (iii) the evanescent field in medium 2 for \(\theta_i>\theta_c\), which propagates along the interface while decaying as \(e^{-\kappa z}\) with \(\kappa = k_0\sqrt{n_1^2\sin^2\theta_i - n_2^2}\).

Why it matters

These three results are the entire optical behaviour of a dielectric interface read off from one pair of formulae. Brewster's angle underlies polarising windows, laser cavities and glare-reducing coatings; the critical angle governs optical fibres, prisms and refractometers; and the evanescent wave is the working principle of total-internal-reflection microscopy (TIRF), frustrated-TIR beam splitters, and the near-field coupling that lets light "tunnel" across sub-wavelength gaps.

Physically they show that the innocuous-looking Fresnel ratios encode a polarisation-selective zero, a real-to-imaginary transition of the transmitted wavevector, and a bound surface field — three qualitatively different phenomena from a single boundary-matching calculation.

Assumptions

Both media are linear, isotropic, homogeneous and non-magnetic (\(\mu_1=\mu_2=\mu_0\)).If magnetic (\(\mu\neq\mu_0\)) the Fresnel coefficients carry impedance factors and the Brewster condition becomes \(\tan^2\theta_B\) dependent on both \(\varepsilon\) and \(\mu\); a magnetic-Brewster angle can then exist for \(s\)-polarisation too.

Both indices are real (no absorption): \(n_1,n_2\in\mathbb{R}\).If \(n\) is complex, \(r_p\) never reaches exactly zero — Brewster is replaced by a "pseudo-Brewster" reflectance minimum, and \(\theta_c\) is smeared because \(\cos\theta_t\) is complex even below the nominal critical angle.

The interface is planar, sharp on the scale of \(\lambda\), and infinite in extent.A graded or rough interface mixes wavevectors, scatters into other directions, and destroys the clean Snell relation that all three results rest on.

A single plane wave of definite frequency \(\omega\); the incident field is decomposed into pure \(s\) and \(p\) linear polarisations.For a beam of finite width (a wavepacket) the results hold per Fourier component; the finite angular spread produces the lateral Goos–Hänchen shift on total reflection, invisible to the ideal-plane-wave treatment.

Derivation

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

Fresnel amplitude reflection coefficients (assumed prior result), with \(\theta_i\) the angle of incidence in medium 1 and \(\theta_t\) the transmission angle in medium 2. A

\[ n_2\cos\theta_i - n_1\cos\theta_t = 0 \]

Impose \(r_p=0\): a fraction vanishes when its numerator does (denominator stays finite and positive). This is the Brewster condition. A

\[ n_1\sin\theta_i = n_2\sin\theta_t \;\Rightarrow\; n_1 = n_2\,\frac{\sin\theta_t}{\sin\theta_i} \]

Snell's law from continuity of the tangential wavevector; solve it for \(n_1\) so we can eliminate the indices from Step 2. A

\[ n_2\cos\theta_i = n_2\,\frac{\sin\theta_t}{\sin\theta_i}\cos\theta_t \;\Rightarrow\; \sin\theta_i\cos\theta_i = \sin\theta_t\cos\theta_t \]

Substitute Step 3 into Step 2 and cancel \(n_2\), then multiply through by \(\sin\theta_i\). Indices have now dropped out entirely. B

\[ \tfrac{1}{2}\sin 2\theta_i = \tfrac{1}{2}\sin 2\theta_t \;\Rightarrow\; 2\theta_i = \pi - 2\theta_t \;\Rightarrow\; \theta_i+\theta_t = \frac{\pi}{2} \]

Apply the double-angle identity \(2\sin\theta\cos\theta=\sin2\theta\). The root \(2\theta_i=2\theta_t\) is rejected (it needs \(n_1=n_2\), no interface); the supplement root gives the physical Brewster geometry: reflected and refracted rays perpendicular. B

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

Put \(\theta_t=\tfrac{\pi}{2}-\theta_B\) into Snell and divide by \(\cos\theta_B\). Brewster angle obtained. A

\[ \sin\theta_t = \frac{n_1}{n_2}\sin\theta_i \]

Return to Snell (Step 3) and solve for \(\sin\theta_t\); this is the master relation for the reflection side. Now take \(n_1>n_2\) (dense-to-rare). A

\[ \sin\theta_t = 1 \;\Rightarrow\; \sin\theta_c = \frac{n_2}{n_1} \quad(n_1>n_2) \]

The largest real \(\theta_t\) is \(\pi/2\) (refracted ray grazing the surface). Setting \(\sin\theta_t=1\) defines the incidence angle \(\theta_c\) at which refraction ceases to have a real solution. A

\[ \theta_i>\theta_c:\quad \sin\theta_t=\frac{n_1}{n_2}\sin\theta_i>1,\qquad \cos\theta_t=\pm\sqrt{1-\sin^2\theta_t}=\pm\, i\sqrt{\sin^2\theta_t-1} \]

Beyond the critical angle \(\sin\theta_t>1\), so \(\theta_t\) is no longer a real angle. Keep Snell's law analytically continued: \(\cos\theta_t\) becomes pure imaginary. C

\[ \mathbf{E}_t \propto \exp\!\big[i(k_{tx}x + k_{tz}z - \omega t)\big],\quad k_{tx}=k_2\sin\theta_t=k_0 n_1\sin\theta_i,\quad k_{tz}=k_2\cos\theta_t \]

Write the transmitted plane wave in medium 2 (interface at \(z=0\), medium 2 in \(z>0\)). Tangential wavevector \(k_{tx}\) is conserved across the boundary and stays real; only \(k_{tz}\) inherits the imaginary \(\cos\theta_t\). Here \(k_0=\omega/c\), \(k_2=n_2k_0\). C

\[ k_{tz}=k_2\cdot i\sqrt{\sin^2\theta_t-1}=i\kappa,\qquad \kappa \equiv k_2\sqrt{\Big(\tfrac{n_1}{n_2}\Big)^2\sin^2\theta_i-1} \]

Insert \(\cos\theta_t=+i\sqrt{\sin^2\theta_t-1}\) and \(\sin\theta_t=(n_1/n_2)\sin\theta_i\). The \(+\) sign is chosen so the field decays (not grows) into medium 2; the \(-\) root is unphysical for a semi-infinite medium as it diverges as \(z\to\infty\). C

\[ \mathbf{E}_t \propto e^{-\kappa z}\,\exp\!\big[i(k_{tx}x-\omega t)\big],\qquad \kappa = k_2\sqrt{\tfrac{n_1^2}{n_2^2}\sin^2\theta_i-1}=k_0\sqrt{n_1^2\sin^2\theta_i-n_2^2} \]

Substitute \(k_{tz}=i\kappa\): the \(z\)-dependence becomes \(e^{i(i\kappa)z}=e^{-\kappa z}\). Multiply the surd by \(k_2=n_2k_0\) and take \(n_2\) inside to reach the index-symmetric form. The wave still travels along \(x\) but is bound to the surface. B

Result

\[ \tan\theta_B=\frac{n_2}{n_1},\qquad \sin\theta_c=\frac{n_2}{n_1}\ (n_1>n_2),\qquad \kappa = k_0\sqrt{n_1^2\sin^2\theta_i-n_2^2},\quad d=\frac{1}{\kappa} \]

Reading. At \(\theta_B\) the reflected beam is purely \(s\)-polarised because the \(p\)-component reflects with zero amplitude — the microscopic dipoles in medium 2 oscillate along the would-be reflected \(p\)-ray and cannot radiate along their own axis. At \(\theta_c\) the refracted ray lies flat in the surface; for any \(\theta_i>\theta_c\) refraction is forbidden, \(|r|=1\) for both polarisations, and 100% of the time-averaged power is reflected. Yet the fields do not stop at the boundary: an evanescent wave leaks a distance \(d=1/\kappa\) (the penetration depth) into the rarer medium, carrying no net energy across but able to couple to a third body placed within \(d\).

Units check. \(\tan\theta_B\) and \(\sin\theta_c\) are ratios of refractive indices — dimensionless, as an angle's trig function must be. \(k_0=\omega/c=2\pi/\lambda_0\) has dimension \(\text{length}^{-1}\); the surd \(\sqrt{n_1^2\sin^2\theta_i-n_2^2}\) is dimensionless, so \(\kappa\) is \(\text{length}^{-1}\) and \(d=1/\kappa\) is a length, as required for a decay distance.

Limiting cases

\(n_1=n_2\): \(\tan\theta_B=1\Rightarrow\theta_B=45^\circ\) but there is no real interface, so the "zero" is vacuous; \(\theta_c\to90^\circ\) (no TIR); \(\kappa\to k_0\sqrt{n_1^2\sin^2\theta_i-n_1^2}\) which is imaginary below grazing — i.e. no evanescent regime, consistent with no reflection.
\(n_2\gg n_1\) (into a very dense medium): \(\theta_B\to90^\circ\); Brewster requires near-grazing incidence, and there is no critical angle (\(\sin\theta_c=n_2/n_1>1\) has no solution) — TIR is impossible going into the denser medium.
\(\theta_i\to\theta_c^{+}\): \(n_1^2\sin^2\theta_i\to n_2^2\), so \(\kappa\to0\) and \(d\to\infty\) — the evanescent tail spreads arbitrarily far; the field is barely bound at onset.
\(\theta_i\to90^\circ\) (grazing) with \(n_1>n_2\): \(\kappa\to k_0\sqrt{n_1^2-n_2^2}\), its maximum; penetration depth is shortest, of order \(\lambda_0/\big(2\pi\sqrt{n_1^2-n_2^2}\big)\), a fraction of a wavelength.
Grazing incidence, \(\theta_i\to90^\circ\), any interface: \(r_s,r_p\to-1\); everything reflects regardless of polarisation, so both Brewster and the evanescent structure are washed out.

Breaks when

Absorbing or metallic media (complex \(n\)). With \(n=n'+in''\) the numerator of \(r_p\) cannot be driven to exactly zero, so true Brewster extinction disappears — one sees only a shallow "pseudo-Brewster" minimum. Likewise \(\cos\theta_t\) is complex for all angles, so reflectance approaches but never reaches unity and the sharp critical angle blurs into a gradual rise.
A second interface within the penetration depth (frustrated TIR). The derivation assumes medium 2 is semi-infinite. If a third medium sits a gap \(g\lesssim d\) away, the evanescent tail is non-zero there and re-radiates a propagating wave: energy tunnels across the "forbidden" gap, \(|r|<1\), and total reflection fails — the basis of variable beam splitters and near-field microscopy.
Anisotropic or optically active media. Snell's law with a single index no longer holds (birefringence splits the transmitted ray), so both \(\tan\theta_B=n_2/n_1\) and \(\sin\theta_c=n_2/n_1\) are replaced by direction-dependent conditions.
Interfaces or beams not much larger than \(\lambda\). Sub-wavelength apertures, tightly focused beams and rough boundaries inject a spread of transverse wavevectors, so a single \(\theta_i\) and the plane-wave Snell relation no longer describe the field.

Failure modes

Inverting the Brewster ratio. Writing \(\tan\theta_B=n_1/n_2\) instead of \(n_2/n_1\). Sanity check: air-to-glass Brewster is \(\approx56^\circ\) (large), so \(\tan\theta_B>1\), forcing \(n_2>n_1\) in the numerator.
Believing TIR happens going into the denser medium. The critical angle only exists for \(n_1>n_2\); students routinely quote a critical angle for air→glass, where \(\sin\theta_c=n_2/n_1>1\) has no solution.
Confusing Brewster's angle with the critical angle. They coincide numerically only by accident; Brewster kills \(r_p\) (a polarisation effect), the critical angle kills real refraction (a total-reflection effect). One can lie either side of the other.
Thinking the reflected beam at \(\theta_B\) is unpolarised or zero. Only the \(p\)-component vanishes; the \(s\)-component still reflects strongly, so the reflected beam is fully \(s\)-polarised, not extinguished.
Claiming the evanescent wave carries energy into medium 2. The time-averaged normal Poynting flux \(\langle S_z\rangle=0\) under pure TIR; energy oscillates in and back out. Net transmission appears only if the semi-infinite assumption is broken (frustrated TIR).
Dropping the tangential-wavevector conservation. Forgetting that \(k_{tx}=k_0n_1\sin\theta_i\) stays real — treating the whole \(\mathbf{k}_t\) as imaginary — gives a field that decays in \(x\) as well, which is wrong: it must propagate along the surface.

Discussion

The three results are unified by watching the transmitted wavevector's normal component \(k_{tz}=k_0\sqrt{n_2^2-n_1^2\sin^2\theta_i}\). Below the critical angle the argument is positive and \(k_{tz}\) is real — an ordinary refracted wave. At \(\theta_c\) it passes through zero (grazing ray). Above \(\theta_c\) the argument goes negative and \(k_{tz}\) becomes pure imaginary \(=i\kappa\): the same square root that gives the refracted ray continuously turns into the evanescent decay constant. Brewster, by contrast, is not about \(k_{tz}\) at all but about the amplitude \(r_p\) vanishing while the wave remains fully propagating — which is why a Brewster angle exists for both \(n_1<n_2\) and \(n_1>n_2\), whereas a critical angle exists only for \(n_1>n_2\).

The Brewster geometry \(\theta_i+\theta_t=90^\circ\) has a vivid microscopic reading: the field transmitted into medium 2 drives its bound electrons to oscillate along the direction the reflected \(p\)-ray would travel. A dipole does not radiate along its own axis, so no \(p\)-polarised light can be launched back into medium 1 — the reflection is nulled by a radiation-pattern zero, not by interference of many layers. This is why Brewster windows in gas lasers transmit \(p\)-light with essentially no reflective loss.

The evanescent wave is a genuinely bound electromagnetic field: it satisfies the wave equation with an imaginary normal wavenumber, propagates phase along the surface faster in wavelength but slower in phase velocity than a bulk wave in medium 2, and stores rather than transports energy. Placing a detector, a second prism, or a fluorophore within \(d=1/\kappa\) lets it couple out — the mechanism behind total-internal-reflection fluorescence, attenuated total reflectance (ATR) spectroscopy, and optical tunnelling. The penetration depth, of order a fraction of \(\lambda_0\) except near \(\theta_c\), is precisely what gives TIRF its prized surface selectivity.

A deeper view treats the interface as a scattering problem for the four Fresnel coefficients as analytic functions of \(\sin\theta_i\). Brewster is a zero of \(r_p(\theta)\) on the real axis; the critical angle is a branch point of \(\cos\theta_t(\theta)\) where the square root's argument changes sign. Above \(\theta_c\), \(r_s\) and \(r_p\) become unimodular complex numbers \(e^{i\phi_s},e^{i\phi_p}\); their differential phase \(\phi_s-\phi_p\) is non-zero and angle-dependent, which is exactly what a Fresnel rhomb exploits to convert linear to circular polarisation using only two total reflections — a purely geometric-phase device with no birefringent material at all.

Common misconceptions. "Total internal reflection means the field is exactly zero beyond the boundary" — false; the field penetrates a distance \(\sim d\) as an evanescent wave, it simply carries no net normal power. "At Brewster the surface is a perfect mirror for one polarisation" — backwards: at Brewster the surface is perfectly transparent (zero reflection) for the \(p\)-component, and a partial mirror only for \(s\).

Worked examples

Brewster angle for an air–glass window and its refracted ray. Light in air (\(n_1=1.00\)) meets crown glass (\(n_2=1.50\)). Find \(\theta_B\) and the refraction angle \(\theta_t\), and verify the perpendicular-ray property. A

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

Refraction angle from Snell: \(\sin\theta_t=(n_1/n_2)\sin\theta_B\).

\[ \sin\theta_t=\frac{1.00}{1.50}\sin 56.3^\circ=\frac{0.832}{1.50}=0.555 \;\Rightarrow\; \theta_t=33.7^\circ \]

Check: \(\theta_B+\theta_t=56.3^\circ+33.7^\circ=90.0^\circ\), confirming reflected and refracted rays are perpendicular.

\[ \theta_B=56.3^\circ,\qquad \theta_t=33.7^\circ,\qquad \theta_B+\theta_t=90^\circ \]

Reading. A window tilted at \(56.3^\circ\) reflects only \(s\)-polarised glare; a \(p\)-polarised laser passes without reflective loss. Units check. Angles in degrees; \(\tan\theta_B=1.50\) dimensionless.

Critical angle and evanescent penetration depth for glass–air TIR. A wave in glass (\(n_1=1.50\)) hits a glass–air surface (\(n_2=1.00\)) at \(\theta_i=50^\circ\) with vacuum wavelength \(\lambda_0=500\ \text{nm}\). Find \(\theta_c\), confirm TIR, and compute \(\kappa\) and \(d\). B

\[ \sin\theta_c=\frac{n_2}{n_1}=\frac{1.00}{1.50}=0.667 \;\Rightarrow\; \theta_c=41.8^\circ \]

Since \(\theta_i=50^\circ>41.8^\circ=\theta_c\), the wave is totally internally reflected. Compute the surd (symbols first).

\[ \kappa=k_0\sqrt{n_1^2\sin^2\theta_i-n_2^2},\qquad k_0=\frac{2\pi}{\lambda_0}=\frac{2\pi}{500\ \text{nm}}=0.01257\ \text{nm}^{-1} \]

\[ n_1^2\sin^2\theta_i-n_2^2=(1.50)^2(0.766)^2-(1.00)^2=1.320-1.000=0.320 \]

\[ \kappa=0.01257\times\sqrt{0.320}=0.01257\times0.566=7.11\times10^{-3}\ \text{nm}^{-1},\qquad d=\frac{1}{\kappa}=141\ \text{nm} \]

\[ \theta_c=41.8^\circ,\qquad \kappa=7.1\times10^{-3}\ \text{nm}^{-1},\qquad d\approx141\ \text{nm} \]

Reading. The evanescent field falls to \(1/e\) of its surface value within \(\approx141\ \text{nm}\) — about \(0.28\,\lambda_0\), safely sub-wavelength, giving TIRF its thin optical section. Units check. \(k_0\) in \(\text{nm}^{-1}\), surd dimensionless, so \(\kappa\) in \(\text{nm}^{-1}\) and \(d\) in \(\text{nm}\).

Problems

Light in air strikes the flat surface of water (\(n=1.33\)) at Brewster's angle. Find \(\theta_B\) and the refraction angle.
Solution
\(\tan\theta_B=n_2/n_1=1.33/1.00=1.33\Rightarrow\theta_B=\arctan(1.33)=53.1^\circ\). Refraction: \(\sin\theta_t=(1/1.33)\sin 53.1^\circ=0.799/1.33=0.601\Rightarrow\theta_t=36.9^\circ\). Check \(\theta_B+\theta_t=53.1^\circ+36.9^\circ=90.0^\circ\).
A beam inside glass (\(n_1=1.50\)) meets a glass–air surface (\(n_2=1.00\)). Find the internal Brewster angle and compare it with the external (air→glass) Brewster angle of \(56.3^\circ\).
Solution
\(\tan\theta_B=n_2/n_1=1.00/1.50=0.667\Rightarrow\theta_B=33.7^\circ\). Note \(33.7^\circ+56.3^\circ=90^\circ\): the internal and external Brewster angles are complementary, because they describe the same perpendicular reflected/refracted geometry traversed in opposite directions. (Also note \(33.7^\circ<\theta_c=41.8^\circ\), so internal Brewster occurs below the critical angle and is physically accessible.)
Find the critical angle for a water–air interface (\(n_1=1.33\), \(n_2=1.00\)), and explain why there is no critical angle for the reverse (air→water) direction.
Solution
\(\sin\theta_c=n_2/n_1=1.00/1.33=0.752\Rightarrow\theta_c=48.8^\circ\). For air→water, \(\sin\theta_c=n_2/n_1=1.33/1.00=1.33>1\), which has no real solution — TIR requires going from the denser to the rarer medium (\(n_1>n_2\)).
A wave in diamond (\(n_1=2.42\)) undergoes TIR at a diamond–air surface (\(n_2=1.00\)) at \(\theta_i=30^\circ\) with \(\lambda_0=633\ \text{nm}\). Find the critical angle, confirm TIR, and compute the penetration depth \(d\).
Solution
\(\sin\theta_c=1/2.42=0.413\Rightarrow\theta_c=24.4^\circ\); since \(30^\circ>24.4^\circ\), TIR holds. \(k_0=2\pi/633\ \text{nm}=9.93\times10^{-3}\ \text{nm}^{-1}\). Surd: \(n_1^2\sin^2\theta_i-n_2^2=(2.42)^2(0.500)^2-1=5.856\times0.250-1=1.464-1=0.464\); \(\sqrt{0.464}=0.681\). \(\kappa=9.93\times10^{-3}\times0.681=6.76\times10^{-3}\ \text{nm}^{-1}\), \(d=1/\kappa=148\ \text{nm}\) (\(\approx0.23\,\lambda_0\)).
For a glass–air interface (\(n_1=1.50\), \(n_2=1.00\), \(\lambda_0=500\ \text{nm}\)), compute the penetration depth at \(\theta_i=42^\circ\) (just above \(\theta_c=41.8^\circ\)) and at \(\theta_i=80^\circ\). Comment on the trend as \(\theta_i\to\theta_c^{+}\).
Solution
\(k_0=0.01257\ \text{nm}^{-1}\). At \(42^\circ\): \(n_1^2\sin^2\theta_i-n_2^2=2.25\times(0.669)^2-1=2.25\times0.4477-1=1.007-1=0.00742\); \(\sqrt{}=0.0861\); \(\kappa=1.083\times10^{-3}\ \text{nm}^{-1}\), \(d=924\ \text{nm}\ (\approx1.85\,\lambda_0)\). At \(80^\circ\): \(2.25\times(0.985)^2-1=2.25\times0.970-1=2.183-1=1.183\); \(\sqrt{}=1.088\); \(\kappa=0.01367\ \text{nm}^{-1}\), \(d=73.2\ \text{nm}\ (\approx0.15\,\lambda_0)\). Trend: as \(\theta_i\to\theta_c^{+}\) the surd \(\to0\), so \(\kappa\to0\) and \(d\to\infty\) — the evanescent tail delocalises at the onset of TIR, then tightens rapidly toward grazing incidence.