L1 · Chapter 2

How much of a sound wave reflects at an interface and how much passes through

Sandbox Physics · 2026-10-29

An ultrasound probe does not see an object directly. It sees the part of the wave that turns back where two materials meet, and how strong that echo is depends only on how much the impedance changes across the boundary. This chapter derives the reflection coefficient from two continuity conditions, checks it against the energy flux, and works out how much sound a plastic plate and a glass plate send back.

Two things that must hold at the boundary

A plane wave in a medium of impedance $Z_1$ hits a medium of impedance $Z_2$ head on. On the left there are an incident and a reflected wave, on the right only a transmitted one. Call their pressure amplitudes $p_i$ , $p_r$ and $p_t$ .

Two conditions must hold at the boundary. The pressure is continuous, or the massless boundary layer would feel an infinite acceleration. The normal velocity is continuous, or the two media would separate or pass through each other.

Write each wave’s velocity with $v=\pm p/Z$ from the previous chapter, with a minus sign for the reflected wave travelling left. The two conditions give two linear equations, and their solution is the pair of coefficients below.

p_i+p_r=p_t,\qquad \frac{p_i-p_r}{Z_1}=\frac{p_t}{Z_2}

Left: pressure is continuous. Right: normal velocity is continuous.

R=\frac{p_r}{p_i}=\frac{Z_2-Z_1}{Z_2+Z_1},\qquad T=\frac{p_t}{p_i}=\frac{2Z_2}{Z_2+Z_1}=1+R

When the impedances match,

R=0

and the wave passes without noticing the boundary, however different the densities and sound speeds are.

A wave in water meets polystyrene at normal incidence. Arrow heights are pressure amplitudes: incident 1, reflected

R=0.246

, transmitted

T=1.246

. The transmitted wave carries only 93.9 % of the energy because the plastic’s impedance is higher. Original schematic.

Source: J. A. Jensen · R. Barnkob

More pressure on the far side, yet energy is conserved

Going from water into a stiffer material, $T>1$ : the transmitted pressure is larger than the incident pressure. Energy is still conserved, because the energy flux is $p^2/(2Z)$ and the larger impedance on the far side means the same pressure carries less energy.

Divide the reflected and transmitted fluxes by the incident flux and the two shares must add to one. The identity below holds for any $Z_1$ and $Z_2$ and can be checked by expanding it.

Going the other way, from a stiff material into water, $R$ changes sign: the reflected pressure is inverted. That is why, in the channel data of the pulse-echo experiment, the back-face echo of a plate arrives upside down.

R^2+\frac{Z_1}{Z_2}T^2=1

The first term is the reflected share of the energy, the second the transmitted share.

Source: R. Barnkob · M. Settnes and H. Bruus

A check without the formula: simulate the wave in one dimension

A derived formula can still be checked independently. The benchmark program behind the pulse-echo experiment splits a one-dimensional tube into many cells and updates pressure and velocity in each cell in turn, sending a pulse from water into plastic or glass. The program never evaluates a reflection coefficient; the boundary is just two kinds of cells side by side.

With 120 cells per wavelength the simulated reflection is 0.24633 for polystyrene against the formula’s 0.24621, and 0.78904 against 0.78898 for Pyrex. The energy on the two sides adds up to 1.0002 and 1.0001 of the incident energy. Halving the cell size cuts the error by about four, so the difference is a grid error that shrinks as the cells get finer.

Source: R. Barnkob · M. Settnes and H. Bruus

Worked example: the echoes of a 2 mm plastic plate

Water has $Z_1=1.49\,\mathrm{MRayl}$ and polystyrene $Z_2=1050\times2350\,\mathrm{Rayl}=2.47\,\mathrm{MRayl}$ . At the front face $R=0.246$ , so only 6.06 % of the energy is reflected and 93.9 % enters the plate.

At the back face the wave goes from plastic into water and reflects with $-0.246$ . Crossing back into the water it is transmitted with $T_{21}=2Z_1/(Z_1+Z_2)=0.754$ . The first back-face echo therefore has amplitude $T_{12}R_{21}T_{21}=1.246\times(-0.246)\times0.754=-0.231$ , nearly as bright as the front echo but inverted. Each extra round trip inside the plate multiplies the amplitude by $R_{21}^2=0.061$ , down to 0.014.

For Pyrex, $Z_2=12.7\,\mathrm{MRayl}$ and $R=0.789$ at the front face, so 62 % of the energy reflects straight away. Little is lost on each round trip inside, and the later echoes run $-0.298$ , $-0.185$ and onward: the plate rings.

In time, the front face sits at 28 mm and its echo arrives after $2\times28\,\mathrm{mm}/1497\,\mathrm{m/s}=37.4\,\mu\mathrm{s}$ . The back-face echo also crosses the plate twice, adding $2\times2\,\mathrm{mm}/2350\,\mathrm{m/s}=1.70\,\mu\mathrm{s}$ . Chapter 4 shows how this delay makes the plate look thinner in the image.

Source: R. Barnkob · M. Settnes and H. Bruus

Try it in the Lab

Compare the echoes of a plastic and a glass plate

In the “Three wires and a plate” preset, read the “Pressure reflection at the plate” value and compare it with $R=0.246$ .
Choose the “Glass rings with echoes” preset. Count how many echoes follow the plate in the channel data and check whether they are evenly spaced.
Compare the back-face echoes: for plastic it is nearly as bright as the front echo, for glass much weaker than the front echo. Explain both with the numbers of the worked example.

Exercises with explained answers

A wave goes from Pyrex into water at normal incidence. What are the pressure reflection and transmission coefficients, and what share of the energy is transmitted?
Show the answer
Now $Z_1=12.7$ and $Z_2=1.49\,\mathrm{MRayl}$ , so $R=-0.789$ and $T=0.211$ . The energy share is $(Z_1/Z_2)T^2=8.48\times0.0445=0.378$ , the same share that is transmitted going from water into glass.
Two materials differ in density and in sound speed, yet reflect no sound at all. What is the condition?
Show the answer
Equal impedance, $\rho_1c_1=\rho_2c_2$ , gives $R=0$ . A material twice as dense with half the sound speed is invisible to a wave at normal incidence. This no longer holds at oblique incidence.
Prove the energy relation $R^2+(Z_1/Z_2)T^2=1$ .
Show the answer
Insert $R=(Z_2-Z_1)/(Z_2+Z_1)$ and $T=2Z_2/(Z_2+Z_1)$ . The numerator becomes $(Z_2-Z_1)^2+4Z_1Z_2=(Z_2+Z_1)^2$ , exactly the denominator, so the sum is one.

Sources and what we recomputed

R. Barnkob, P. Augustsson, T. Laurell and H. Bruus, Acoustic radiation- and streaming-induced microparticle velocities determined by micro-PIV in an ultrasound symmetry plane, arXiv:1208.6534v1; Physical Review E 86, 056307 (2012)
Version 1, 31 August 2012, read in full. Eq. 1 (standing-wave field), Eq. 2 (radiation force), Eqs. 5–7 (streaming scale and Rayleigh streaming), Eqs. 9–10 (geometry and thermoviscous corrections), Eqs. 12–14 (velocity amplitude, velocity ratio, critical diameter), Eqs. 16–18 and Table I (wall drag), Table II (material data), Tables IV–V and Figs. 8–9 (measured energy densities, streaming coefficients and velocity ratios). We recompute the theory values; the measured data are quoted, not reproduced.
M. Settnes and H. Bruus, Forces acting on a small particle in an acoustical field in a viscous fluid, arXiv:1110.6037v1; Physical Review E 85, 016327 (2012)
Version 1, 27 October 2011, read in full. Eq. 1 (viscous penetration depth), Eqs. 4–5 (first-order pressure and compressibility), Eqs. 48a–e (Gor’kov potential with the viscous dipole coefficient) and Table I (Pyrex and cell parameters).
J. A. Jensen, Notes for the International Summer School on Advanced Ultrasound Imaging, Technical University of Denmark (21 May 2024)
Chapter 2, Eqs. 2.45–2.51 (linear acoustic equations and wave equation). Chapter 3, Eqs. 3.8–3.11 (line aperture, first zero, first sidelobe), Eqs. 3.15–3.17 (grating lobes) and Eq. 3.24 (F-number). Chapter 5, Eqs. 5.3–5.4 (pulse and round-trip delay). The impedance reflection coefficients are not taken from these notes.