Questions: Frequency-Dependent Permittivity and Dispersion
2 questions to test your understanding
Score: 0 / 2
Question 1 Short Answer
At frequencies well above all resonances of a material, what happens to the permittivity, and what does this imply for the refractive index?
Think about your answer, then reveal below.
Model answer: ε(ω) → ε₀ from below as ω → ∞, meaning the real part of ε can drop below ε₀ (and even below zero near plasma frequency for conductors). The refractive index n = √(ε/ε₀) approaches 1. Physically, the charges are too slow to respond, so the material becomes transparent and behaves like vacuum.
This is why high-energy X-rays are transmitted through most materials with little interaction — the photon frequency far exceeds any electronic resonance. The result also explains why the refractive index of glass decreases toward 1 as you go to shorter wavelengths past the UV absorption band.
Question 2 Short Answer
A dielectric has a strong absorption resonance at frequency ω₀. Describe qualitatively how the real part of the refractive index behaves just below and just above ω₀.
Think about your answer, then reveal below.
Model answer: Just below ω₀, the real part of n increases with frequency (normal dispersion — common in transparent materials). Just above ω₀, it decreases sharply (anomalous dispersion). This dip in n just above the resonance is accompanied by high absorption. The group velocity can become very small or even negative in the anomalous dispersion region.
This behavior is a signature of any driven resonator: the response function has a characteristic shape where the real part (dispersion) goes through a steep S-curve centered on the resonance while the imaginary part (absorption) peaks at ω₀. In optics, anomalous dispersion is observed in materials excited near absorption lines and is exploited in slow-light experiments.