7.1 Reflection at a boundary (normal incidence)

A plane wave in medium 1 of impedance $Z_1 = \rho_1 c_1$ encounters a flat boundary with medium 2 of impedance $Z_2 = \rho_2 c_2$ . At normal incidence (the wave travels along the surface normal), what happens?

The boundary conditions

Two physical requirements at the interface:

Pressure continuity. The pressure on the two sides of an infinitesimally thin interface must match — otherwise an infinite force per unit volume would act on the interface. So $p_1' = p_2'$ at the boundary.
Normal velocity continuity. The fluids on the two sides must move together at the interface — otherwise a vacuum or interpenetration would form. So $v_{1,n}' = v_{2,n}'$ at the boundary.

These two conditions, applied to plane-wave fields, determine the reflection coefficient $R$ (ratio of reflected to incident pressure amplitudes) and transmission coefficient $T$ (ratio of transmitted to incident).

Plane-wave decomposition

In medium 1, the field is incident plus reflected:

p_1'(x, t) \;=\; P_i\, e^{i(\omega t - k_1 x)} \;+\; P_r\, e^{i(\omega t + k_1 x)}.

The velocity is in phase with each separately, with the right sign:

v_1'(x, t) \;=\; \frac{1}{Z_1}\!\left[P_i\, e^{i(\omega t - k_1 x)} - P_r\, e^{i(\omega t + k_1 x)}\right].

(Note the minus on the reflected wave: it travels in $-x$ , so its velocity points in $-x$ .)

In medium 2, only the transmitted wave (no source in medium 2, no reflection from infinity):

p_2'(x, t) \;=\; P_t\, e^{i(\omega t - k_2 x)}, \qquad v_2'(x, t) \;=\; \frac{P_t}{Z_2}\, e^{i(\omega t - k_2 x)}.

Matching at $x = 0$

Pressure continuity gives $P_i + P_r = P_t$ . Velocity continuity gives $(P_i - P_r)/Z_1 = P_t/Z_2$ . Two linear equations, two unknowns ( $P_r/P_i$ and $P_t/P_i$ ). Solving:

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

These are the Fresnel-like coefficients for normal-incidence acoustic reflection. They are the most-used formulas in this chapter.

▶ Working through the algebra

From $P_i + P_r = P_t$ and $(P_i - P_r)/Z_1 = P_t/Z_2$ :

Substituting the first into the second: $(P_i - P_r)/Z_1 = (P_i + P_r)/Z_2$ . Multiply out: $Z_2(P_i - P_r) = Z_1(P_i + P_r)$ , i.e. $(Z_2 - Z_1) P_i = (Z_1 + Z_2) P_r$ . Divide: $R = P_r/P_i = (Z_2 - Z_1)/(Z_2 + Z_1)$ .

Then $T = 1 + R = 2 Z_2 / (Z_1 + Z_2)$ .

Three regimes

Hard wall ( $Z_2 \to \infty$ ): $R \to +1$ , $T \to 2$ . Wait — $T = 2$ ? The pressure transmitted is twice the incident? Yes: at the hard wall the incident and reflected pressures add coherently, doubling the pressure at the wall. No energy is transmitted (the velocity is zero in medium 2). The total reflected power is 100%.
Soft termination ( $Z_2 \to 0$ , e.g. air-to-vacuum): $R \to -1$ , $T \to 0$ . The reflected pressure is the inverse of the incident, so $p$ vanishes at the boundary. Again 100% reflected power.
Impedance matched ( $Z_2 = Z_1$ ): $R = 0$ , $T = 1$ . No reflection, full transmission. The wave continues into medium 2 as if the boundary weren’t there.

Power reflection and transmission

The power reflection and transmission coefficients are

R_P = |R|^2 = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2, \qquad T_P = 1 - R_P = \frac{4 Z_1 Z_2}{(Z_1 + Z_2)^2}.

The sum $R_P + T_P = 1$ — energy conservation across a non-absorbing boundary.

For air-to-water (huge impedance ratio $\sim$ 3600), $R_P \approx 0.999$ — about 99.9% of incident sound is reflected at the air-water interface. This is why you can’t hear someone underwater from above the surface, or vice versa, and why the middle ear’s impedance-matching mechanism (lever ratio of 22:1 in human ossicles) is so important.

Looking ahead

We’ve done the simplest case: normal incidence. The general angle of incidence requires accounting for the direction of the wavevectors on both sides. That’s the next lesson — and the resulting law turns out to be a direct analogue of Snell’s law in optics.