Uniform Plane Waves Questions and Answers

Answer: B

The boundary condition forces the tangential electric field to zero there, so the surface is a node for E and, a quarter wavelength away, an antinode. The magnetic field behaves in the opposite sense, having an antinode at the surface.

Answer: C

With a reflection coefficient of unit magnitude the minima of the standing wave fall to zero, so the ratio of maximum to minimum amplitude is unbounded. A perfectly matched boundary gives a ratio of one and no standing wave at all.

Answer: D

No reflected wave means the amplitude is the same everywhere and the maximum equals the minimum. Achieving it is the object of every matching arrangement, from a quarter-wave layer on a lens to a stub on a transmission line.

Answer: A

The quarter-wave section transforms the load impedance to the square of its own impedance divided by that load, so choosing the geometric mean makes the transformed value equal the source impedance. It is the principle of the antireflection coating on a lens.

Answer: B

Beyond the critical angle Snell's law admits no real angle of refraction and the wave is entirely reflected, an evanescent field decaying into the second medium. It is the mechanism that guides light along an optical fibre.

Answer: C

One of the two polarisations suffers no reflection at that angle, so the reflected wave contains only the other. Polarising sunglasses exploit the fact that light reflected from a horizontal surface is largely horizontally polarised.

Answer: D

In a dispersive medium the phase velocity depends on frequency and a modulated signal travels at the group velocity instead, which is the derivative of angular frequency with respect to the wave number. In a non-dispersive medium the two coincide.

Answer: A

Correct answer: remains below it, so no information travels faster than light

The phase velocity in a guide is \( v_p = c/\sqrt{1-(f_c/f)^2} \), which exceeds \( c \), and the group velocity is \( v_g = c\sqrt{1-(f_c/f)^2} \), which falls below it. Their product is \( c^2 \). Only the group velocity carries energy and information, so relativity is not violated - the phase velocity describes the motion of a pattern, which need carry nothing with it.

Answer: B

The phase velocity falls by the square root of the relative permittivity while the frequency is fixed by the source, so the wavelength falls by the same factor of three. This shortening is why a dielectric-loaded antenna can be made smaller.

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.