Uniform Plane Waves Questions and Answers

Answer: C

Correct answer: 188 ohms

\[ \eta = \sqrt{\frac{\mu}{\epsilon}} = \frac{\eta_0}{\sqrt{\epsilon_r}} = \frac{377}{\sqrt{4}} = \frac{377}{2} = 188\ \Omega. \] The square root is what catches people out: dividing by 4 gives the 94 ohms offered. The wavelength and the velocity are shortened by the same factor of two in this medium.

Answer: D

The free space wavelength is three metres and the square root of 2.25 is 1.5, so the wavelength in the medium is two metres. A relative permittivity of 2.25 corresponds to a refractive index of exactly 1.5, which is that of ordinary glass.

Answer: A

Correct answer: 3

\[ \mathrm{SWR} = \frac{1+|\Gamma|}{1-|\Gamma|} = \frac{1.5}{0.5} = 3. \] The ratio runs from 1 for a matched boundary to infinity for total reflection, so it is a convenient measure precisely because it turns the small numerical range of \( |\Gamma| \) into something easier to read on an instrument.

Answer: B

Power is proportional to the square of the field amplitude, so half the amplitude is a quarter of the power. Three quarters of the incident power is therefore transmitted, which is a useful check on any such calculation.

Answer: C

Correct answer: about 5 per cent of its initial value

The amplitude falls as \( e^{-\alpha z} \), so after 3 m at 1 neper per metre it is \[ e^{-3} = 0.0498, \] about 5 per cent. The 37 per cent offered is \( e^{-1} \), the reduction after a single metre - the trap is treating the neper as a fixed fraction rather than as an exponent that accumulates with distance.

Answer: D

Its exact value is \( 120\pi \) ohms in the old system of units, and it fixes the ratio of the electric to the magnetic field in any plane wave in vacuum. It has nothing to do with the characteristic impedance of a particular cable.

Answer: A

Dividing by the intrinsic impedance gives \( 10/377 = 2.65 \times 10^{-2} \) A/m. The great disparity in the numerical values is entirely an artefact of the units chosen, the two fields carrying equal energy densities in a plane wave.

Answer: B

The divergence condition forces the longitudinal component to be constant along the direction of propagation, and for a wave it must therefore vanish. Guided waves in a hollow waveguide do have longitudinal components, but they are not uniform plane waves.

Answer: C

The phase velocity is the speed of a point of constant phase and carries no information, so no principle forbids it exceeding c, and in a waveguide it does. The group velocity, which carries the energy, remains below c.

Answer: D

Unpolarised light is a rapid and random superposition of all polarisations, so the field is present at every instant but its direction changes unpredictably. A single monochromatic plane wave is always polarised in some sense.