Uniform Plane Waves Questions and Answers

Answer: A

Correct answer: False for a passive boundary, the reflected power then exceeding the incident

A passive boundary cannot return more power than arrives, and since the reflected power goes as \( |\Gamma|^2 \), a magnitude above one would violate conservation of energy. The coefficient can be negative, as at a perfect conductor where it is exactly \( -1 \), and it can be complex for a lossy second medium - but its magnitude never exceeds unity.

Answer: B

The only way to satisfy the boundary condition is to add a reflected wave whose tangential field cancels the incident one at the surface, and that requires a coefficient of minus one. The condition and the total reflection are two aspects of the same fact.

Answer: C

In a simple waveguide the product of the two velocities is the square of the free space speed, so one exceeding it forces the other below. The reason therefore does account for the assertion, and relativity constrains only the group velocity.

Answer: D

A quarter-wave section presents the square of its own impedance over the load, so with the geometric mean the transformed value equals the source and the boundary is matched. It works only at the frequency for which the layer is a quarter wavelength thick.

Answer: A

The three cases correspond to the three values the ratio of conduction to displacement current can take, and each has its own characteristic impedance and attenuation. Recognising which applies is the first step in any propagation calculation.

Answer: B

The three are points on one continuum: as the phase difference grows from zero to ninety degrees the line opens into an ellipse and then closes into a circle. Unequal amplitudes give an ellipse at every phase difference except zero.

Answer: C

The impedances come first because everything else is computed from them, then the reflection coefficient, then the transmission coefficient which is one plus it, and the standing wave ratio follows from the magnitude of the reflection coefficient last.

Answer: D

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: in a lossless medium the electric and magnetic fields are exactly in phase, and both lie transverse to the direction of travel. Statement 3 is wrong - \( \eta = \sqrt{\mu/\epsilon} \) is purely real when there is no loss, and it is exactly that reality which puts the two fields in phase. A complex impedance is the signature of a lossy medium.

Answer: A

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: matched impedances give no reflection, and a perfect conductor gives \( \Gamma = -1 \). Statement 3 has the sign wrong - the transmission coefficient is \[ \tau = 1 + \Gamma, \] not \( 1-\Gamma \). The relation looks odd until one notices that the tangential electric field must be continuous, so the transmitted amplitude equals the incident plus the reflected.

Answer: B

The frequency is fixed by the source and is the same everywhere along the path, which is why the wavelength must change when the velocity does. The other three all scale with the square root of the relative permittivity, two downward and one downward as well.