10.5 Reflection and transmission at interfaces

When a wave reaches a boundary between two media, part of it crosses and part turns back. How much does each depends entirely on the impedances of the two media. This lesson derives the reflection and transmission coefficients, converts them to power, and shows why a large impedance mismatch is nearly a perfect mirror — and how a matching layer defeats it.

Matching the fields across the boundary

▶ Reflection and transmission coefficients Derivation

At an interface between media of impedance $Z_1$ and $Z_2$ , two physical conditions must hold: the pressure is continuous (no net force on a massless interface), and the normal velocity is continuous (the media stay in contact). Write incident, reflected, and transmitted amplitudes $P_i, P_r, P_t$ . Pressure continuity gives $P_i + P_r = P_t$ . Velocity continuity, using $u = p/Z$ with the reflected wave travelling backwards ( $u_r = -P_r/Z_1$ ), gives

\frac{P_i}{Z_1} - \frac{P_r}{Z_1} = \frac{P_t}{Z_2}.

Solving the two equations for the ratios,

R \equiv \frac{P_r}{P_i} = \frac{Z_2 - Z_1}{Z_2 + Z_1}, \qquad T \equiv \frac{P_t}{P_i} = \frac{2Z_2}{Z_2 + Z_1}.

where

$R$: pressure reflection coefficient —
$T$: pressure transmission coefficient —
$Z_1, Z_2$: impedances of the incident and transmitting media Pa·s/m

The reflection coefficient depends only on the contrast of impedances. When $Z_2 = Z_1$ the media are matched, $R = 0$ , and the wave crosses as if the boundary were not there. When the two differ greatly $R\to\pm1$ and the boundary is a near-perfect mirror.

R (amplitude)0.600

T (amplitude)1.600

R_P (power)36.0%

T_P (power)64.0%

Z₁ = 1.00 Z₂ = 4.00

At an interface between two media with impedance Z₁ and Z₂, the reflection amplitude is R = (Z₂−Z₁)/(Z₂+Z₁). For air-to-water (Z₂/Z₁ ≈ 3500), R ≈ 1.0 and 99.9% of the power reflects — the impedance-matching problem the middle ear is built to solve. When Z₁ = Z₂ (perfect match), R = 0 and all power transmits.

Power and the mismatch problem

The pressure coefficients convert to power coefficients by energy accounting across the interface:

R_P = R^2, \qquad T_P = 1 - R^2,

which sum to one, as energy conservation requires. The consequence of a large mismatch is stark. For an air-to-water interface ( $Z_1 = 410$ , $Z_2 = 1.5\times10^6$ ), $R^2 \approx 0.999$ : about $99.9\%$ of the incident acoustic power reflects, and only a thousandth crosses. Two media whose impedances differ by thousands are, acoustically, nearly opaque to one another.

The remedy is an impedance-matching layer. Insert between the two media a layer of intermediate impedance $Z_m = \sqrt{Z_1 Z_2}$ and thickness a quarter wavelength, and the reflections from its two faces cancel, allowing near-total transmission. The same quarter-wave trick antireflection-coats camera lenses and matches electrical transmission lines; it is the general strategy by which nature and engineering push energy across an impedance step. Standing-wave resonances are the same reflection physics seen in steady state: a wave trapped between two boundaries reinforces itself only at frequencies where a whole number of half- or quarter-wavelengths fits, which is why a pipe closed at one end sounds its fundamental at $f = c/4L$ .