Uniform Plane Waves Questions and Answers

Answer: C

The ratio is about 300.

The skin depth \( \delta = \sqrt{\frac{2}{\omega\mu\sigma}} \) varies as \( 1/\sqrt{f} \), so

\[ \begin{aligned} \frac{\delta_{1\,\text{kHz}}}{\delta_{100\,\text{MHz}}} &= \sqrt{\frac{10^{8}}{10^{3}}} \\ &= \sqrt{10^{5}} \approx 316 \end{aligned} \]

In copper that is about 2 mm at 1 kHz against 6.6 μm at 100 MHz. 30 and 3000 would follow from frequency ratios of \( 10^{3} \) or \( 10^{7} \), not \( 10^{5} \), and 3 is far too small.

Answer: A

The answer is both (A) and (R) are true and (R) is the correct explanation of (A).

A wave is transverse when its oscillations are at right angles to the direction in which it travels. In an electromagnetic wave in free space, the electric field E and the magnetic field B are perpendicular to each other and both perpendicular to the direction of propagation, given by \(E \times B\).

That perpendicularity of the fields to the direction of propagation is exactly what makes the wave transverse, so the Reason explains the Assertion rather than merely being true alongside it.

Answer: C

\(\sqrt{\mu/\varepsilon}\) is the intrinsic impedance of the medium.

For a plane wave, the ratio of the electric to the magnetic field amplitude is \(E/H = \sqrt{\mu/\varepsilon} = \eta\), measured in ohms. In free space it is about 377 Ω.

The refractive index is a different combination, \(n = \sqrt{\mu\varepsilon/\mu_{0}\varepsilon_{0}}\), which involves the product, not the ratio. The square root of n has no special role, and energy density depends on the field strengths, not on the material constants alone.

Answer: A

The phase difference is Zero.

In a perfect (lossless) dielectric, Maxwell's equations give \(\overline{B} = \frac{1}{v}\hat{k} \times \overline{E}\) with a real wave speed \(v = 1/\sqrt{\mu\varepsilon}\). The ratio of the fields is real, so E and B oscillate in phase.

The value of the dielectric constant (here \(\pi\)) only changes the speed and the ratio of amplitudes, not the phase. A phase lag appears only in a conducting medium, where it approaches \(\pi/4\) for a good conductor.

Answer: A

The velocity is \(\sqrt{\frac{3}{2}} \times 10^{8}\) m/s.

The speed of light in a medium is \(v = \frac{c}{\sqrt{\epsilon_{r}\mu_{r}}}\):

\[ \begin{aligned} v &= \frac{3 \times 10^{8}}{\sqrt{3 \times 2}} \\ &= \frac{3}{\sqrt{6}} \times 10^{8} = \sqrt{\frac{9}{6}} \times 10^{8} \\ &= \sqrt{\frac{3}{2}} \times 10^{8}\ \text{m/s} \end{aligned} \]

That is about \(1.22 \times 10^{8}\) m/s, a refractive index of \(\sqrt{6} \approx 2.45\). The option \(\sqrt{2} \times 10^{8}\) uses only one of the two factors, and the forms with \(\sqrt{3}\) in the numerator invert part of the expression.

Answer: C

The key's answer is \(\frac{\pi c}{l}\), the cut-off expressed as an angular frequency.

In a rectangular guide of sides a and b, the cut-off angular frequency of the \(TE_{mn}\) mode is \(\omega_{c} = c\pi\sqrt{(m/a)^{2} + (n/b)^{2}}\). The lowest mode in a square guide of side l is \(TE_{10}\) (or \(TE_{01}\)):

\[ \begin{aligned} \omega_{c} &= \frac{\pi c}{l} \\ f_{c} &= \frac{\omega_{c}}{2\pi} = \frac{c}{2l} \end{aligned} \]

Waves below this cut-off cannot propagate. The cycle frequency c/2l is not among the options, so the paper's 'frequency' is the angular frequency. \(\sqrt{2}\pi c/l\) is the cut-off of the higher \(TE_{11}\) and \(TM_{11}\) modes, and \(c/l\) and \(2c/l\) correspond to no lowest mode.

Answer: C

Circular polarization needs two orthogonal linear components with the same magnitude and a phase difference in odd multiples of 90°.

If \(E_{x} = E_{0}\cos\omega t\) and \(E_{y} = E_{0}\sin\omega t\), equal in amplitude and 90° apart, the tip of the field vector traces a circle of constant radius E₀ as it rotates.

With a phase difference of 0 or 180° (even multiples of 90°) the components stay in step and the result is linear polarization. Unequal magnitudes with a 90° difference give elliptical polarization, and merely having non-zero transverse components says nothing about the shape traced.

Answer: B

The ratio E/H in free space is about 377 Ω.

This ratio is the intrinsic impedance of free space:

\[ \begin{aligned} \eta_0 &= \sqrt{\frac{\mu_0}{\varepsilon_0}} \\ &= 120\pi \approx 377\ \Omega \end{aligned} \]

75 Ω is the characteristic impedance of common coaxial cable used for television, and 35 Ω and 225 Ω have no special meaning for plane waves in free space.

Answer: B

Statements 1 and 2 are correct, so the answer is (1) and (2).

The skin depth of a good conductor is

\[ \delta = \frac{1}{\sqrt{\pi f \mu \sigma}} \]

so it is inversely proportional to the square root of μ and σ (1) and to the square root of the frequency f (2).

It follows that the skin depth decreases as frequency increases, so statement 3 is false, and it is not directly proportional to the square root of μ and σ, so statement 4 is false. Every option containing 3 or 4 is wrong.

Answer: A

The phenomenon is dispersion.

A medium is dispersive when its refractive index, and so the phase velocity of waves in it, depends on frequency. Water is strongly dispersive for electromagnetic waves, so different frequency components travel at different speeds and a pulse spreads out.

Polarization is the orientation of the electric field vector, absorption is the loss of wave energy in the medium, and scattering is the redirection of the wave by particles or irregularities. None of these is the frequency dependence of phase velocity.