advanced35 min studyLesson 274 in Physics

Complex permittivity and attenuation

Choose the passive propagation branch and distinguish amplitude attenuation from intensity absorption.

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What you’ll understand

  • Relate complex refractive index to scalar permittivity.
  • Compute amplitude and intensity decay lengths separately.
  • Distinguish bulk absorption from reflection and coherent cancellation.

A complex permittivity changes both the spatial phase and the magnitude of a monochromatic wave. Measuring a darker transmitted beam, however, does not immediately tell us the imaginary permittivity: interfaces can reflect light and coherent amplitudes can interfere. A careful calculation first specifies a traveling bulk solution, selects the decaying branch in a passive half-space, and then distinguishes its field amplitude from the energy flux carried by that field.

Fix propagation and branch conventions

Use real fields obtained from E0eikz−iωtE_0e^{ikz-i\omega t} and take positive frequency. For a homogeneous local isotropic material with relative permeability equal to one, write the complex index as n~=n+iκ\widetilde n=n+i\kappa. Then its imaginary part gives a factor that decreases with positive depth when κ≥0\kappa\ge0. The sign follows jointly from the time and spatial exponent conventions. Copying an index from a source using the opposite convention without converting its sign would turn loss into growth.

Squaring the index gives two real equations connecting its phase and attenuation components to the real and imaginary permittivity. For an ordinary passive nonmagnetic branch with nonnegative real index, both components can be obtained from the magnitude of permittivity. A lossless negative real permittivity produces a purely imaginary index and an evanescent field. More general magnetic, anisotropic or active media require additional branch analysis; the simple positive-root formulas here are not universal prescriptions for those systems.

n~2=ϵr,k=ωc(n+iκ),n2−κ2=ϵr′,2nκ=ϵr′′\widetilde n^2=\epsilon_r,\quad k=\frac{\omega}{c}(n+i\kappa),\quad n^2-\kappa^2=\epsilon_r^{\prime},\quad 2n\kappa=\epsilon_r^{\prime\prime}
n and κ are dimensionless, k is in m⁻¹, and c is the vacuum light speed. These equations assume μr = 1 and a scalar local permittivity.↔ Scroll wide equations horizontally.

Keep the factor of two visible

The field amplitude decreases exponentially with the imaginary wavenumber. Time-averaged intensity is quadratic in the field amplitude, so it decreases twice as fast in its exponential argument. An absorption coefficient quoted for intensity is therefore twice the inverse amplitude decay length. Stating only a penetration depth without identifying the measured quantity leaves an ambiguity that can produce a factor-of-two error in a thickness specification or an inferred extinction coefficient.

For a single forward wave in a homogeneous medium, the proportionality between intensity and squared electric amplitude is independent of position, so the ratio of intensities is simple even when the wave impedance is complex. This statement does not apply unchanged to a superposition of incident and reflected waves forming a standing pattern. A finite film requires boundary matching and can display interference fringes on top of absorption. The attenuation coefficient by itself describes propagation after the interface has selected the transmitted wave.

∣E(z)∣=∣E(0)∣e−ωκz/c,I(z)=I(0)e−αz,α=2ωκc=4πκλ0|E(z)|=|E(0)|e^{-\omega\kappa z/c},\quad I(z)=I(0)e^{-\alpha z},\quad \alpha=\frac{2\omega\kappa}{c}=\frac{4\pi\kappa}{\lambda_0}
α is the intensity absorption coefficient in m⁻¹; λ0 is vacuum wavelength in m. The amplitude decay length is c/(ωκ), while the intensity decay length is 1/α.↔ Scroll wide equations horizontally.

Separate two ways of losing transmission

At a planar interface some energy can be reflected without being converted to heat. In a passive half-space with a propagating transmitted wave, bulk heating removes the remaining forward energy flux with increasing depth. Energy accounting therefore compares incident, reflected, transmitted and absorbed powers using the actual boundary conditions. A small transmission coefficient is not sufficient evidence for a large bulk imaginary permittivity, especially when the material is highly reflective or the sample contains resonant cavities.

Destructive interference presents another diagnostic trap. Two coherent amplitudes can sum to zero in one observation channel even when the medium is perfectly lossless. Their energy is redistributed among channels rather than necessarily dissipated. The wave lab makes this distinction testable because its controls change phase and relative amplitude while introducing no absorptive material. A genuine extraction of complex permittivity needs phase information or an adequate measurement model, not a single dark spot on a screen.

n=∣ϵr∣+ϵr′2,κ=∣ϵr∣−ϵr′2n=\sqrt{\frac{|\epsilon_r|+\epsilon_r^{\prime}}2},\qquad \kappa=\sqrt{\frac{|\epsilon_r|-\epsilon_r^{\prime}}2}
These nonnegative roots select the ordinary passive nonmagnetic branch for εr″ ≥ 0. At κ = 0 the bulk absorption coefficient vanishes, although interface reflection can remain.↔ Scroll wide equations horizontally.

Two different penetration lengths

At vacuum wavelength λ0=600 nm\lambda_0=600\,\mathrm{nm}, a nonmagnetic material has ϵr=3+4i\epsilon_r=3+4i. Find its index, amplitude decay length and intensity decay length, excluding interface reflection.

  1. The permittivity magnitude is 32+42=5\sqrt{3^2+4^2}=5. Hence n=(5+3)/2=2n=\sqrt{(5+3)/2}=2 and κ=(5−3)/2=1\kappa=\sqrt{(5-3)/2}=1, confirming (2+i)2=3+4i(2+i)^2=3+4i.

  2. The imaginary wavenumber is 2πκ/λ02\pi\kappa/\lambda_0. Its inverse is the amplitude decay length λ0/(2π)=95.5 nm\lambda_0/(2\pi)=95.5\,\mathrm{nm}. At this depth the field magnitude is reduced by a factor of e.

  3. The intensity coefficient is twice the imaginary wavenumber. Its inverse is λ0/(4π)=47.7 nm\lambda_0/(4\pi)=47.7\,\mathrm{nm}. At the amplitude decay length the intensity has therefore fallen by e−2e^{-2}, not by e−1e^{-1}.

The result

The passive index is 2 + i. The amplitude and intensity decay lengths are 95.5 nm and 47.7 nm, respectively.

Explore the model

Interactive lab

Interactive lab · waves

Cancellation is not absorption

Reset and set BOTH Phase difference = 0° and Amplitude ratio = 1; predict a sum with twice the component amplitude. Reset and set Phase difference = 180° and Amplitude ratio = 1; predict cancellation at every position and time. The equal-frequency wave sum contains no loss parameter, material boundary or dissipated-power readout. Its dark result illustrates cancellation, not an absorbing medium.

Cancellation is not absorptionTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forA vanished coherent sum alone does not establish conversion of electromagnetic energy to heat.

Check your understanding

At one amplitude decay length in a homogeneous forward wave, what fraction of the initial intensity remains?

Try it yourself

A homogeneous forward wave has intensity absorption coefficient α=200 m−1\alpha=200\,\mathrm{m^{-1}}. Find the thickness that reduces intensity to one quarter, and the field-amplitude ratio at that depth. Ignore interface reflections.

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

Give me a hint

Take the logarithm of I/I0 = exp(−αz); the amplitude ratio is the positive square root of the intensity ratio.

Compare with the solution

Solve e−200z=1/4e^{-200z}=1/4 to obtain z=ln⁡4/200=6.93×10−3 mz=\ln4/200=6.93\times10^{-3}\,\mathrm m, or 6.93 mm6.93\,\mathrm{mm}. The field-amplitude ratio is 1/4=1/2\sqrt{1/4}=1/2. The amplitude attenuation coefficient is only 100 m−1100\,\mathrm{m^{-1}}, which gives the same amplitude reduction at this thickness. A measured total transmission of one quarter could require a different thickness if interface reflections were included, so the bulk assumption is essential to the answer.

Take it with you

  • The passive spatial branch must decay into a passive half-space.
  • Intensity and amplitude decay lengths differ by a factor of two.
  • Reflection and interference can reduce transmission without proving bulk absorption.

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