A copper cavity is designed with its fundamental resonant mode at 2.4 GHz. An engineer drives it at 3.1 GHz, which corresponds to no cavity mode. What happens?
AThe cavity resonates at reduced amplitude since 3.1 GHz is near the fundamental
BThe cavity does not resonate — only discrete frequencies matching cavity modes are supported
CThe field propagates through the cavity as through an open waveguide at that frequency
DThe cavity heats and thermally expands, shifting its resonant frequency toward 3.1 GHz
Resonant cavities support standing-wave modes only at specific discrete frequencies determined by the cavity geometry. A frequency that doesn't match a cavity mode cannot establish a standing wave — destructive interference prevents energy buildup. This is the fundamental difference between an open waveguide (continuous band above cutoff) and a closed cavity (discrete resonances). Option A is wrong: there is no partial resonance at nearby off-mode frequencies. Option C is wrong: closing both ends of the waveguide is exactly what creates the discrete spectrum rather than a continuous one.
Question 2 Multiple Choice
A cavity operates at 1 GHz with quality factor Q = 10^4. What is its approximate resonance linewidth (half-power bandwidth)?
A1 MHz — computed as 1 GHz divided by 10^3
B100 kHz — since Q = f₀/Δf, so Δf = 10^9 Hz / 10^4 = 10^5 Hz
C100 MHz — Q multiplies the bandwidth
D10 kHz — only superconducting cavities achieve this linewidth
Quality factor is defined as Q = f₀/Δf, where Δf is the half-power bandwidth. Rearranging: Δf = f₀/Q = 10^9 Hz / 10^4 = 10^5 Hz = 100 kHz. A narrower linewidth means the cavity is highly frequency-selective — it stores energy efficiently only within a very narrow band around the resonant frequency. Option A has a unit error: 1 GHz / 10^4 = 100 kHz, not 1 MHz. High-Q cavities are sharply resonant; low-Q cavities respond over a broader band.
Question 3 True / False
Superconducting resonant cavities achieve dramatically higher Q values than copper cavities because their walls have near-zero electrical resistance.
TTrue
FFalse
Answer: True
Q = ω × (stored energy) / (power loss). Power loss in cavity walls comes from resistive heating — currents induced in the walls by the oscillating electromagnetic field dissipate energy as heat. Superconducting walls have essentially zero resistance, reducing wall losses by many orders of magnitude. This raises Q from ~10^4–10^5 for copper to Q > 10^10 for superconducting niobium cavities used in particle accelerators. The consequence is that particle bunches receive energy from the oscillating field extremely efficiently.
Question 4 True / False
A resonant cavity supports most electromagnetic frequencies above its lowest cutoff frequency, just as an open waveguide does.
TTrue
FFalse
Answer: False
This is the key distinction between a waveguide and a cavity. An open waveguide supports all frequencies above the cutoff of each mode — the spectrum is continuous. A resonant cavity, formed by closing both ends, supports only discrete resonant frequencies: those for which an integer number of half-wavelengths fit between the two end walls. Constructive interference occurs only at these specific frequencies; all others suffer destructive interference. This discreteness is what makes cavities useful for frequency-selective applications.
Question 5 Short Answer
Why does closing both ends of a waveguide produce a discrete set of resonant frequencies rather than a continuous band of supported frequencies?
Think about your answer, then reveal below.
Model answer: An open waveguide supports traveling waves; any frequency above cutoff can propagate. When you close both ends with conducting walls, waves reflect back and forth. For a stable standing wave to exist, the round-trip must produce constructive interference — the wave must return to its starting point in phase. This condition is only met when an integer number of half-wavelengths fit exactly between the two end walls: L = p·(λ/2) for integer p. Only these specific wavelengths satisfy the boundary conditions; all other frequencies produce destructive interference and cannot sustain a standing wave.
Mathematically, the boundary conditions (tangential E = 0 at both end walls) quantize the allowed wavenumber in the propagation direction: k_z = pπ/d. Combined with the transverse mode structure from the waveguide, this gives discrete resonant frequencies f_{mnp} = (c/2)√((m/a)² + (n/b)² + (p/d)²). The discreteness is a direct consequence of imposing boundary conditions at both ends — the same physics that quantizes energy levels in a quantum particle-in-a-box.