Questions: Impedance Matching and Wave Reflection at Boundaries

Question 1

A wave travels from a medium with impedance Z₁ = 4 into a medium with impedance Z₂ = 12. What is the reflection coefficient R?

• A. R = 0.5
• B. R = -0.5
• C. R = 1/3
• D. R = -1/3

Answer: (A)

R = (Z₂ - Z₁)/(Z₂ + Z₁) = (12 - 4)/(12 + 4) = 8/16 = 0.5. Since Z₂ > Z₁, R is positive — no phase inversion. The reflected wave has the same phase as the incident wave. Option C inverts the formula; option B confuses the sign convention.

Question 2

An engineer measures that the amplitude of a transmitted wave is larger than the amplitude of the incident wave at an impedance boundary. What does this tell us about energy conservation?

• A. Energy conservation is violated because the transmitted wave has more amplitude
• B. The transmitted wave carries more power since power scales with amplitude
• C. Energy is conserved because the transmitted medium has lower impedance, so higher amplitude can carry less power
• D. This situation is impossible — transmitted amplitude can never exceed incident amplitude

Answer: (C)

Power is proportional to amplitude² × impedance. A wave entering a low-impedance medium can have larger amplitude while carrying less power, because the lower impedance factor compensates. The amplitude transmission coefficient t = 2Z₂/(Z₁+Z₂) can exceed 1 while the power transmission coefficient T = 1 − R² is always ≤ 1. Energy is fully conserved.

Question 3

A wave reflecting from a boundary where Z₂ < Z₁ undergoes a phase inversion.

• T. True
• F. False

Answer: True

When Z₂ < Z₁, the reflection coefficient R = (Z₂ − Z₁)/(Z₂ + Z₁) is negative. A negative R means the reflected wave is inverted in phase relative to the incident wave. This is why a string tied to a fixed wall (Z₂ → ∞) reflects with inversion, while a string attached to a free end (Z₂ = 0) reflects without inversion — both cases are captured by the sign of R.

Question 4

Perfect impedance matching (Z₁ = Z₂) is very difficult to achieve in practice, so reflection can mainly be minimized but rarely eliminated.

• T. True
• F. False

Answer: False

Impedance matching can be exact. Two cable sections with identical characteristic impedance have zero reflection at their junction. Optical coatings and electrical matching networks can also achieve near-zero reflection. The formula R = (Z₂ − Z₁)/(Z₂ + Z₁) shows R = 0 exactly when Z₁ = Z₂, which is an achievable condition.

Question 5

Why does an acoustic impedance mismatch between air and biological tissue cause nearly total reflection of sound waves at the skin surface, and how does ultrasound gel solve this problem?

Model answer: Air has acoustic impedance ~400 Pa·s/m while tissue has ~1.5 × 10⁶ Pa·s/m — a ratio of ~3,750:1. Plugging into R = (Z₂ − Z₁)/(Z₂ + Z₁) gives R ≈ 1, so nearly all acoustic energy reflects. Ultrasound gel has impedance close to tissue, replacing the air-tissue boundary with a gel-tissue boundary where the impedance difference is small, dramatically reducing reflection and allowing sound to enter the body.

The reflection coefficient is determined by the ratio of impedances, not their absolute values. A large ratio produces near-total reflection regardless of actual impedance values. The gel acts as an impedance-matching intermediate layer — the same principle as a quarter-wave transformer in electronics or an anti-reflection coating on a lens.