Questions: Electromagnetic Waves in Dielectric Materials

Question 1

A glass prism spreads white light into a spectrum because blue light bends more than red light. What does this tell you about the dielectric function ε(ω) in glass?

• A. Glass has a higher absorption coefficient for red light than blue light
• B. ε(ω) is larger at blue frequencies than red frequencies, giving blue a higher refractive index
• C. ε(ω) is the same for all visible frequencies but the absorption differs
• D. Blue photons carry more energy so they interact more strongly with the glass boundary

Answer: (B)

The refractive index n(ω) = √ε(ω), so a higher refractive index for blue means ε is larger at blue (higher) frequencies. This is normal dispersion: the dielectric function encodes the frequency-dependent material response, and in the visible range of glass, bound electrons are driven below their resonance frequency, making ε increase with ω. Option D conflates photon energy with the macroscopic refractive index — the key is the resonance structure of ε(ω), not photon energy per se.

Question 2

Below the plasma frequency of a metal, ε(ω) is negative. A student claims this means the metal strongly absorbs the light. What is wrong with this reasoning?

• A. Nothing — negative ε always means absorption
• B. Negative ε makes n(ω) imaginary, so waves decay exponentially and the metal reflects rather than absorbs
• C. Metals have no bound electrons, so ε(ω) does not apply to them
• D. Absorption requires the imaginary part of ε to be nonzero, but a purely negative real ε still means transmission

Answer: (B)

When ε(ω) is negative (and real), n = √ε is purely imaginary. A purely imaginary refractive index means the wave decays exponentially into the material with no propagating oscillation — the wave is reflected, not transmitted or absorbed. Absorption, in contrast, requires n to have a positive imaginary part (which comes from a nonzero imaginary part of ε). This is why metals below the plasma frequency are shiny and reflective: the wave cannot propagate, so it bounces back.

Question 3

The refractive index of a dielectric is a constant that does not depend on the frequency of light.

• T. True
• F. False

Answer: False

The refractive index n(ω) = √ε(ω) inherits its frequency dependence directly from the dielectric function ε(ω). Because the polarization response of bound charges depends on how close the driving frequency is to a resonance, ε — and therefore n — varies with ω. This is dispersion, and it is why prisms separate white light and why optical fiber design must account for pulse spreading. A constant n would mean no dispersion, which is only approximately true in a narrow frequency range far from any resonance.

Question 4

Near a resonance frequency, the imaginary part of ε(ω) becomes large. This means the refractive index becomes complex, and the wave is absorbed.

• T. True
• F. False

Answer: True

A complex dielectric function ε = ε₁ + iε₂ produces a complex refractive index n = n₁ + iκ, where κ (the extinction coefficient) is related to ε₂. The wave amplitude then decays as e^{-κωz/c} — exponential attenuation. This is the physical origin of Beer's law. The imaginary part of ε arises because near a resonance, the bound charges are driven out of phase with the field (energy is dissipated), which corresponds to absorption. Far from resonances, ε is approximately real and n is approximately real, giving low-loss propagation.

Question 5

Why does a dielectric material slow down electromagnetic waves, and how does that slowdown depend on frequency?

Model answer: An EM wave drives bound charges in the dielectric into oscillation. These oscillating charges radiate their own fields that interfere with the incident wave, effectively slowing the combined wave's phase velocity to c/n(ω). The frequency dependence arises because the amplitude and phase of the bound-charge response depends on how close the driving frequency is to the charges' natural resonance frequencies: near a resonance, the response is large and strongly frequency-dependent, producing large changes in n(ω) and strong absorption; far from resonances, n varies slowly with ω.

The key is that the dielectric does not passively slow the wave — the wave and the bound charges are coupled. The charges re-radiate, and the superposition of incident plus re-radiated fields travels at c/n. At ω → 0, charges follow the field easily (large polarization, large ε, slow wave). At ω → ∞, charges cannot follow (ε → 1, wave travels at c). The resonance structure of ε(ω) between these limits encodes the full optical behavior of the material.