advanced30 min studyLesson 220 in Physics

Evanescent waves and frustrated reflection

Derive imaginary normal wavenumber from interface phase matching, distinguish amplitude and intensity decay, and explain coupling across a finite gap.

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What you’ll understand

  • Choose the decaying evanescent branch using a physical boundary condition.
  • Compute critical angle and field-amplitude penetration length.
  • Explain why a second boundary changes the energy-flow conclusion of single-interface total reflection.

Total internal reflection returns all incident normal power at an ideal single lossless interface, but the lower-index side still contains a nonzero electromagnetic field. Its amplitude decreases exponentially away from the boundary while its phase travels parallel to that boundary. This evanescent field is required by electromagnetic boundary conditions. Bringing a second high-index object close enough changes those conditions and can convert some of the field into a transmitted propagating wave.

Tangential phase matching forces an imaginary component

Place the planar interface at z=0z=0, with a transparent nonmagnetic medium of index n1n_1 in z<0z<0 and a lower index n2n_2 in z>0z>0. Let incidence lie in the xz plane. Continuity of tangential phase requires the same real kx=k0n1sin⁡θk_x=k_0n_1\sin\theta on both sides. The lower-index dispersion relation then determines the squared normal wavenumber; it is not free to adjust independently to preserve a geometrical ray.

Above the critical angle, the required tangential component exceeds the entire propagating wavenumber magnitude in medium two. Its squared normal component is negative. With the e−iωte^{-i\omega t} convention and a factor eikzze^{ik_z z}, select kz=iκk_z=i\kappa with positive κ\kappa in the semi-infinite upper medium. The alternative branch grows without bound as z increases and violates the radiation or boundedness condition for this source-free half-space.

sin⁡θc=n2n1,kx=k0n1sin⁡θ,kz2=iκ,κ=k0n12sin⁡2θ−n22,k0=2πλ0\sin\theta_c=\frac{n_2}{n_1},\qquad k_x=k_0n_1\sin\theta,\qquad k_{z2}=i\kappa,\quad \kappa=k_0\sqrt{n_1^2\sin^2\theta-n_2^2},\quad k_0=\frac{2\pi}{\lambda_0}
n1>n2 are real indices; theta is the incidence angle from the normal and theta>theta_c. Vacuum wavelength lambda0 is in metres; kx and kappa are in inverse metres. This evanescent branch describes the bounded field in a semi-infinite lossless upper medium.↔ Scroll wide equations horizontally.

A nonzero field need not carry normal power

The electric and magnetic fields must obey continuity of their tangential components when there is no free surface current. Normal displacement and magnetic flux satisfy their respective jump conditions. In particular, normal electric field need not be continuous across different permittivities. The Fresnel coefficients derived from these conditions have unit reflected magnitude above critical incidence, but generally nontrivial polarization-dependent phase. That phase can alter a reflected beam's polarization without implying energy absorption.

The evanescent field contains an exponential factor e−κze^{-\kappa z}. Its amplitude penetration length is 1/κ1/\kappa, whereas a quadratic local field magnitude falls by a factor e over distance 1/(2κ)1/(2\kappa). Calling both lengths simply penetration depth invites a factor-of-two mistake. For one lossless interface and a single decaying evanescent wave, the cycle-averaged Poynting vector has no normal component but can have a tangential component. A nonzero stored field and zero outgoing normal flux are therefore compatible.

U(x,z)=U(0)eikxx−κz,ℓE=1κ,∣U(z)∣2∣U(0)∣2=e−2κz,ℓ∣U∣2=12κU(x,z)=U(0)e^{ik_xx-\kappa z},\qquad \ell_E=\frac{1}{\kappa},\qquad \frac{|U(z)|^2}{|U(0)|^2}=e^{-2\kappa z},\quad \ell_{|U|^2}=\frac{1}{2\kappa}
U denotes one field component, ellE is its amplitude e-folding length in metres, and the squared-magnitude length is half as large. Squared local field magnitude is not a transmitted normal irradiance for a single lossless evanescent wave.↔ Scroll wide equations horizontally.

A finite gap is a new boundary problem

Put another high-index transparent medium beyond a gap of thickness gg. If that medium supports the same tangential wavenumber as a propagating wave, the two interfaces can couple light across the gap. The gap field now generally includes both exponential branches because it occupies a finite interval; boundedness at infinity no longer rules out one branch inside that interval. Their relative amplitudes are fixed by matching both boundaries and can produce nonzero normal energy flux.

This frustrated total internal reflection is a classical Maxwell boundary phenomenon with a useful tunneling analogy. The factor e−2κge^{-2\kappa g} suggests the large-gap attenuation scale but is not the exact transmitted-power coefficient: polarization, impedance, angle, and multiple reflections supply additional factors. For an ideal symmetric high-index pair, transmission approaches the direct-contact result as the gap vanishes. Large gaps recover isolated-interface total reflection, while absorption or a different receiving medium changes the limiting power balance.

A simple evanescent decay length

A vacuum wavelength of 600 nm600\,\mathrm{nm} enters an interface from index n1=1.5n_1=1.5 toward air n2=1n_2=1. Choose an incidence angle with sin⁡θ=0.80\sin\theta=0.80. Find the critical angle, amplitude penetration length, and squared-field ratio at z=100 nmz=100\,\mathrm{nm} in the air half-space.

  1. The critical sine is 1/1.5=2/31/1.5=2/3, giving θc≃41.81∘\theta_c\simeq41.81^\circ. The incident angle is about 53.13∘53.13^\circ, so the specified geometry is above critical incidence.

  2. The dimensionless square root is (1.5)2(0.80)2−1=0.44≃0.6633\sqrt{(1.5)^2(0.80)^2-1}=\sqrt{0.44}\simeq0.6633. Hence κ=(2π/600 nm)(0.6633)≃0.006946 nm−1\kappa=(2\pi/600\,\mathrm{nm})(0.6633)\simeq0.006946\,\mathrm{nm^{-1}}, and ℓE≃144 nm\ell_E\simeq144\,\mathrm{nm}.

  3. At 100 nm, the amplitude ratio is e−0.6946≃0.499e^{-0.6946}\simeq0.499. Squaring it gives e−1.389≃0.249e^{-1.389}\simeq0.249. This is a local field-squared ratio, not a quarter of the incident power transmitted normally into air.

The result

The critical angle is 41.81°, the amplitude e-folding length is about 144 nm, and the local squared field at 100 nm is about 0.249 of its boundary value.

Explore the model

Interactive lab

Interactive lab · function

The square-root threshold beneath decay

For each independent run, Reset and set BOTH Coefficient a=1 and Position x, comparing x=1 and x=1.50. Interpret x as n1 sinθ/n2; predict y=1 and 2.25. Above the threshold y=1, κ/(k0 n2)=sqrt(y−1). The lab plots y=ax² and its tangent and signed area, not an electromagnetic interface.

The square-root threshold beneath decayThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe quadratic reveals the propagation threshold algebra only. It has no refractive-index controls, evanescent field, Fresnel coefficients, or gap-transmission readout.

Check your understanding

For one lossless total-internal-reflection interface, which statement about the lower-index half-space is correct?

Try it yourself

An evanescent field has κ=5.0 μm−1\kappa=5.0\,\mathrm{\mu m^{-1}}. At what distance is its amplitude one half of the boundary value? Find its squared-field ratio there. Can that ratio alone determine transmission through a gap of the same thickness?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Solve the exponential amplitude equation using the natural logarithm. A second interface requires additional boundary matching before field magnitude becomes a power transmission.

Compare with the solution

The amplitude condition is e−κz=1/2e^{-\kappa z}=1/2, so z=ln⁡2/κ=0.6931/5.0=0.1386 μmz=\ln2/\kappa=0.6931/5.0=0.1386\,\mathrm{\mu m}, or about 139 nm. The local squared magnitude is one quarter of its boundary value. This does not determine the gap transmission: the boundary amplitude itself depends on the full two-interface problem, and both gap branches generally contribute. The receiving medium's impedance and polarization also affect the normal Poynting flux. The exponential gives a useful attenuation scale, but an exact frustrated-reflection coefficient requires matching the electromagnetic fields at both surfaces.

Take it with you

  • Tangential phase matching determines whether the normal wavenumber is real or imaginary.
  • Field-amplitude and squared-field penetration lengths differ by two.
  • Frustrated reflection changes the boundary problem and cannot be predicted from a lone decay factor.

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