Uniform Plane Waves Questions and Answers

Answer: D

Lossless propagation is not practically possible.

Lossless propagation requires a perfect conductor and a perfect dielectric: \(\sigma = \infty\) for the guiding walls or \(\sigma = 0\) for the medium. Every real conductor has finite conductivity, so currents in it dissipate power, and every real dielectric has some loss tangent. Lossless lines and guides are useful idealisations for analysis, but some attenuation always occurs in practice.

Dielectric waveguides such as optical fibre have very low loss but still attenuate, about 0.2 dB/km at 1,550 nm. Conductor propagation has ohmic loss, and cavity resonators have finite Q because of wall losses.

Answer: C

The example is a mobile transponder.

A transponder receives a radio signal and automatically transmits a reply, as in mobile phones, RFID tags, aircraft transponders and satellite transponders. Both its reception and its transmission depend on electromagnetic waves travelling through space between antennas.

A refrigerator works by a vapour-compression cycle driven by an electric motor. An electric fan is driven by a motor. Relays in actuators are electromagnetic switches operated by current in a coil. These devices use electricity and magnetism but not the propagation of electromagnetic waves.

Answer: A

Copper behaves as a conductor or dielectric depending on the frequency.

Whether a material acts as a conductor or a dielectric depends on the ratio of conduction current to displacement current, \(\sigma/\omega\varepsilon\). For copper, \(\sigma \approx 5.8 \times 10^{7}\) S/m, so this ratio is enormous at all ordinary frequencies and copper is a good conductor. At extremely high frequencies, in the far ultraviolet and X-ray range, the ratio falls and copper behaves more like a dielectric, becoming partly transparent.

So copper is not a conductor 'always'. The classification depends on frequency, not on the applied field strength or the current density, since \(\sigma\) and \(\varepsilon\) do not change with these in ordinary conditions.

Answer: C

The wavelength is 12 metres.

For an electromagnetic wave in free space, wavelength is the speed of light divided by frequency:

\[ \begin{aligned} \lambda &= \frac{c}{f} = \frac{3 \times 10^{8}}{25 \times 10^{6}} \\ &= 12\ \text{m} \end{aligned} \]

15 m would correspond to 20 MHz, 4 m to 75 MHz and 32 m to about 9.4 MHz. Only 12 m fits 25 MHz, a frequency in the HF band.

Answer: A

The pair that represents a propagating wave is the Gaussian-shaped pulse in \(\hat{k}\cdot\vec{r} - ct\), with \(\vec{B} = \hat{k}\times\vec{E}/c\).

A travelling wave in vacuum is any function of the single combination \(\hat{k}\cdot\vec{r} - ct\), so it moves along \(\hat{k}\) at speed \(c\) without changing shape, and its magnetic field is \(\vec{B} = \hat{k}\times\vec{E}/c\), perpendicular to both \(\vec{E}\) and \(\hat{k}\). The pulse in the keyed option has exactly this form (it need not be sinusoidal).

The others fail. A wave with \(\vec{B} = 0\) cannot satisfy Faraday's law, since a changing \(\vec{E}\) needs a \(\vec{B}\). The product \(\sin(\hat{k}\cdot\vec{r})\cos(ct)\) is a standing wave, which does not propagate. The last option's \(\vec{E}\) is not a function of \(\hat{k}\cdot\vec{r} - ct\) alone, because of the extra factor \(\hat{k}\cdot\vec{r}\), and its \(\vec{B}\) is not \(\hat{k}\times\vec{E}/c\).

Answer: C

The correct statement is \(\theta_{B} \lt \theta_{C}\).

For light going from medium 1 (index \(n_{1}\)) into the rarer medium 2:

\[ \begin{aligned} \tan\theta_{B} &= n_{2}/n_{1} \\ \sin\theta_{C} &= n_{2}/n_{1} \end{aligned} \]

Both equal the same number less than 1. For any acute angle \(\tan\theta \gt \sin\theta\), so reaching the same value needs a smaller angle for the tangent: \(\theta_{B} \lt \theta_{C}\). This makes sense physically, because Brewster reflection needs a transmitted beam, which exists only below the critical angle.

The formulas \(\tan\theta_{B} = n_{1}/n_{2}\) and \(\sin\theta_{C} = n_{1}/n_{2}\) have the ratio upside down; the second is impossible since \(n_{1}/n_{2} \gt 1\). That also rules out \(\theta_{B} \gt \theta_{C}\).

Answer: B

The single-mode range is 6.6 GHz to 13.2 GHz.

The cutoff frequencies of a rectangular guide with sides \(a = 2.28\) cm and \(b = 1.01\) cm are

\[ \begin{aligned} \nu_{mn} &= \frac{c}{2}\sqrt{\frac{m^{2}}{a^{2}} + \frac{n^{2}}{b^{2}}} \\ \nu_{10} &= \frac{3 \times 10^{10}}{2 \times 2.28} = 6.58\ \text{GHz} \\ \nu_{20} &= 13.2\ \text{GHz}, \quad \nu_{01} = 14.9\ \text{GHz} \end{aligned} \]

and \(\nu_{11} \approx 16.2\) GHz. At 17 GHz the TE10, TE20, TE01 and TE11 modes all propagate. Only TE10 propagates between its own cutoff and the next one, so the single-mode band is about 6.6 GHz to 13.2 GHz.

3.3 to 6.6 GHz lies below the lowest cutoff, where nothing propagates; 1.62 to 6.6 GHz is also below cutoff; and 66 to 132 GHz is ten times too high and full of higher modes.