Uniform Plane Waves Questions and Answers

Answer: C

The reflected wave is polarized perpendicular to the plane of incidence.

tan^-1(n₂/n₁) is Brewster's angle. At this angle the reflection coefficient for the component polarized in the plane of incidence (p-polarization) is zero, because the reflected and refracted rays are perpendicular and the dipoles in medium 2 cannot radiate along the reflected direction. Only the s-component, polarized perpendicular to the plane of incidence, is reflected.

So the reflected light is not unpolarized, and it is not polarized parallel to the plane of incidence, which is exactly the component that vanishes. There is a transmitted wave: both components are transmitted, and total internal reflection cannot occur since the wave goes into the denser medium (n₂ > n₁).

Answer: D

The correct statement is that for matter waves in free space, \(v_{p} \ne v_{g}\).

For a free relativistic particle \(E^{2} = p^{2}c^{2} + m^{2}c^{4}\). The group velocity \(\frac{dE}{dp} = \frac{pc^{2}}{E}\) equals the particle speed \(v\), while the phase velocity \(\frac{E}{p} = \frac{c^{2}}{v}\) exceeds \(c\). They differ for any massive particle.

The other statements fail. The same relations give \(v_{p}v_{g} = c^{2}\) exactly, not greater. Energy in a dispersive medium travels at the group velocity (in normal dispersion), not the phase velocity. And while \(v_{p}\) can exceed \(c\) in a medium, the group velocity in the normal case does not, so 'both can be more than \(c\)' is not the accepted statement.

Answer: B

The magnetic field is \(10^{-6}(\sqrt{3}\hat{x} - \hat{y})\cos(\vec{k}\cdot\vec{r} - \omega t)\) V.m⁻².s (that is, tesla).

For a plane wave in free space \(\vec{B} = \frac{1}{c}\hat{k} \times \vec{E}\). Here \(|\vec{k}| = 10\sqrt{1 + 3} = 20\), so \(\hat{k} = (\hat{x} + \sqrt{3}\hat{y})/2\):

\[ \begin{aligned} \hat{k} \times \hat{z} &= \tfrac{1}{2}(\hat{x} \times \hat{z} + \sqrt{3}\,\hat{y} \times \hat{z}) \\ &= \tfrac{1}{2}(\sqrt{3}\hat{x} - \hat{y}) \\ B_{0} &= \frac{600}{3 \times 10^{8}} = 2 \times 10^{-6} \end{aligned} \]

So the amplitude vector is \(2 \times 10^{-6} \cdot \frac{1}{2}(\sqrt{3}\hat{x} - \hat{y})\), which is \(10^{-6}(\sqrt{3}\hat{x} - \hat{y})\). The first option forgets that \((\sqrt{3}\hat{x} - \hat{y})\) has length 2, the third reverses the direction and the powers of ten in the third and fourth are too large by 10.

Answer: D

The intrinsic impedance is \(z_{0} = \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}\).

The intrinsic impedance of a medium is the ratio of the electric field to the magnetic field in a plane wave, \(\eta = \sqrt{\mu/\varepsilon}\). For free space:

\[ \begin{aligned} \eta_{0} &= \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} \\ &\approx 120\pi \approx 377\ \Omega \end{aligned} \]

The other forms are different quantities. \(\frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}\) is the speed of light \(c\), and \(\sqrt{\mu_{0}\varepsilon_{0}}\) is \(1/c\). \(\sqrt{\varepsilon_{0}/\mu_{0}}\) is the intrinsic admittance, the reciprocal of the impedance.

Answer: D

The propagation constant is \(j\,2.09\ \text{m}^{-1}\).

In lossless free space the attenuation constant is zero, so the propagation constant \(\gamma = \alpha + j\beta\) is purely imaginary, with \(\beta = \omega/c\).

\[ \begin{aligned} \beta &= \frac{2\pi f}{c} = \frac{2\pi \times 10^{8}}{3 \times 10^{8}} \\ &\approx 2.09\ \text{rad/m} \end{aligned} \]

So \(\gamma = j\,2.09\ \text{m}^{-1}\), matching a wavelength of 3 m. \(j\,1.05\) leaves out the factor 2 in \(2\pi\), and \(j\,0.03\ \text{nm}^{-1}\) has the wrong unit. The constant is finite, so infinity is wrong.

Answer: D

Both Statement I and Statement II are true.

Statement I is true: a perfect conductor has infinite conductivity and zero intrinsic impedance, so the reflection coefficient at its surface is −1. A wave striking it normally is entirely reflected, setting up a standing wave in the air in front of it.

Statement II is true: the fields inside a perfect conductor are zero, and with no resistance there is no ohmic loss, so no energy is absorbed. Both statements describe the same ideal case from two sides, so every option calling one of them false is wrong.

Answer: C

The phase changes by \(\pi\).

When a wave is reflected at a rigid barrier, or more generally at a boundary with a denser (higher impedance) medium, the reflected wave is inverted. For a string fixed at one end, the displacement at the fixed point must stay zero, which forces the reflected wave to cancel the incident one there: a phase change of \(\pi\) (half a wavelength). Light reflected from a medium of higher refractive index behaves the same way.

A change of zero occurs only at a free end or a rarer medium, which a barrier is not, and a shift of \(\pm\pi/2\) is not produced by simple reflection at such a boundary.

Answer: B

The phase difference is \( \pi/4 \).

When \( \omega \ll g/|\epsilon| \) the material is a good conductor. The wave number is then complex with equal real and imaginary parts:

\[ k \approx (1 + i)\sqrt{\frac{\mu g \omega}{2}} \]

Since \( B = \frac{k}{\omega} E \), the magnetic field carries the phase of \( 1 + i \), which is \( 45^{\circ} \): \( \bar{B} \) lags \( \bar{E} \) by \( \pi/4 \).

A zero phase difference holds in a perfect insulator, where \( k \) is real. A lag of \( \pi/2 \) would need a purely imaginary \( k \), and \( \pi \) would only reverse the sign of the field.

Answer: C

The energy propagates with the speed \( c \sin\theta \).

Between the plates the wave travels in a zig-zag at speed \( c \), reflecting from each plate with angle of incidence \( \theta \), measured from the normal. Only the component of its velocity parallel to the plates carries energy down the guide, and that component is \( c\sin\theta \), the group velocity.

The wave fronts sweep along the plates faster, at the phase velocity \( \frac{c}{\sin\theta} \), whose product with the group velocity is \( c^{2} \). That value exceeds \( c \) and carries no energy. The cosine forms would follow only if \( \theta \) were measured from the plates.

Answer: C

The most important mode is the one with a phase shift of \( \pi \) between adjacent cavities, the pi-mode.

In the pi-mode neighbouring cavities oscillate in antiphase, so the RF field at successive gaps alternates in direction. Electrons moving round the cathode in step with this pattern give up energy at every gap, forming the spokes of the space charge; this gives the highest efficiency and is the mode in which a magnetron is designed to run.

Other modes, such as \( \pi/2 \) or \( \pi/4 \) per cavity, lie close in frequency and are suppressed by strapping alternate vanes, which is done precisely to separate them from the pi-mode. \( 3\pi/2 \) is not a standard operating mode.