A glass prism separates white light into a spectrum. Which color exits the prism at the largest deflection angle, and what property of the glass causes this?
ARed light, because red has the longest wavelength and is therefore deflected most by wave-mechanical effects
BViolet/blue light, because glass exhibits normal dispersion in the visible range — higher frequency means higher refractive index, so blue light refracts more
CRed light, because higher energy photons penetrate deeper into the glass and are deflected by more interactions
DAll colors exit at the same angle; the spectrum is created by internal interference effects, not differential refraction
In glass, all important atomic resonances lie in the ultraviolet — well above visible frequencies. This places the entire visible spectrum in the normal dispersion regime (dn/dω > 0), where higher-frequency (shorter-wavelength, bluer) light experiences a higher refractive index. A higher n means a greater phase velocity reduction (v = c/n), and by Snell's law, a larger refraction angle. So violet/blue light bends more than red at the glass-air interface, and exits at a larger angle relative to the original beam direction. The spectrum with violet on the high-angle side is characteristic of normal dispersion in glass.
Question 2 Multiple Choice
In a medium with normal dispersion (dn/dω > 0), how does the group velocity of a pulse compare to the phase velocity of the individual frequency components?
AGroup velocity equals phase velocity; dispersion affects wavelength but not propagation speed
BGroup velocity is greater than phase velocity, because the pulse envelope moves faster than the wave crests
CGroup velocity is less than phase velocity; the pulse envelope moves more slowly than the wave crests within it
DGroup velocity exceeds the speed of light c, because the interference of dispersed frequencies accelerates the envelope
Group velocity is v_group = dω/dk = c/(n + ω dn/dω). In normal dispersion, dn/dω > 0, so the denominator n + ω(dn/dω) is larger than n alone, meaning v_group = c/(larger number) < c/n = v_phase. The pulse envelope travels more slowly than the individual wave crests it contains — you can imagine the crests 'walking through' the pulse envelope from the back to the front. This has practical consequences for fiber optic communication: a pulse containing multiple frequencies spreads out in time as it travels, because different frequency components move at different phase velocities.
Question 3 True / False
The frequency-dependence of the refractive index — dispersion — arises because the bound electrons in a material cannot respond instantaneously to the oscillating electromagnetic field.
TTrue
FFalse
Answer: True
The physical origin of dispersion is the resonance behavior of bound electrons modeled as harmonic oscillators. When driven at frequency ω, their oscillation amplitude and phase depend on the proximity of ω to the natural resonance ω₀. The response is not instantaneous: electrons have inertia and a restoring force, so their displacement lags behind the driving field by a frequency-dependent phase. This phase lag determines the polarization response of the medium, which is encoded in the complex permittivity ε(ω), and hence in n(ω). If the response were instantaneous (perfect following), n would be frequency-independent and there would be no dispersion.
Question 4 True / False
Near an atomic resonance frequency, a material exhibits anomalous dispersion (dn/dω < 0) and low optical absorption, since the energy is being used to drive the resonance rather than being absorbed.
TTrue
FFalse
Answer: False
Near resonance, anomalous dispersion (dn/dω < 0) and *strong* absorption occur together — they are two manifestations of the same resonance phenomenon. When the driving frequency ω is close to ω₀, the electrons are driven efficiently and absorb energy from the field (high imaginary part of ε). The rapid change in refractive index (anomalous dispersion) and the absorption peak are both consequences of the resonance response. Far from resonance — in the transparent regions between resonances — the material exhibits normal dispersion and low absorption. The misconception arises from thinking dispersion and absorption are independent; the Kramers-Kronig relations formally connect them.
Question 5 Short Answer
Explain why the group velocity of a pulse in a dispersive medium differs from the phase velocity of its individual frequency components, and what determines whether the group velocity is larger or smaller than the phase velocity.
Think about your answer, then reveal below.
Model answer: Phase velocity v_phase = c/n(ω) is the speed at which a single-frequency wave crest propagates. A real pulse contains a range of frequencies, and its energy envelope travels at the group velocity v_group = dω/dk. Since k = nω/c, we get v_group = c/(n + ω dn/dω). In normal dispersion (dn/dω > 0), the extra term ω dn/dω is positive, making the denominator larger than n, so v_group < v_phase. In anomalous dispersion (dn/dω < 0), the denominator is smaller, so v_group > v_phase — the envelope moves faster than individual crests. This is not a violation of relativity: information cannot travel faster than c because anomalous dispersion is always accompanied by absorption that distorts the pulse.
The distinction between phase and group velocity is fundamental to understanding how signals propagate in real media. Phase velocity tells you how fast a pattern of oscillations moves; group velocity tells you how fast the modulation (information) moves. In a non-dispersive medium (dn/dω = 0), they are equal. Dispersion separates them, and this separation has engineering consequences: pulse spreading (group velocity dispersion) limits the bandwidth of optical fiber communications, and designers must engineer dispersion-shifted fibers to minimize pulse broadening over long distances.