 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-2/section-a/a3/solution

Assume an air-filled or evacuated lossless electromagnetic waveguide, and a pulse bandwidth narrow enough to use its carrier's local group velocity. The longitudinal dispersion relation is

ω^{2} = c^{2} k_{z}^{2} + ω_{c}^{2} .

It gives

v_{p} = \frac{ω}{k _{z}} = \frac{c}{1 - ( f _{c} / f ) ^{2}}, v_{g} = \frac{d ω}{d k _{z}} = c 1 - (f_{c} / f)^{2} .

Here

f_{c} / f = 20/40 = 1/2

, so

v_{p} = \frac{2 c}{3} = 3.46 \times 1 0^{8} m s^{- 1}, v_{g} = \frac{3 c}{2} = 2.60 \times 1 0^{8} m s^{- 1} .

c

, but the pulse envelope travels at the subluminal group velocity; a moving constant-phase surface does not by itself transport information. Their product is

v_{p} v_{g} = c^{2}

. A broad pulse can distort because this waveguide is dispersive.

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