The formulas
Z is acoustic impedance, ρ is density and c is the sound velocity of the wave mode in use. Z1 and Z2 belong to the material the wave leaves and the material it enters.
- Z = ρ × c. In kg/m³ and m/s the answer is in kg/(m²·s), also called rayl. One MRayl is 10⁶ of those. Steel at about 7,750 kg/m³ and 5,890 m/s gives about 45.6 MRayl.
- Pressure reflection r = (Z2 − Z1) ÷ (Z2 + Z1). It is negative when Z2 is lower than Z1, which is why an echo off a steel-to-air boundary is inverted.
- Intensity reflected R = r² = ((Z2 − Z1) ÷ (Z2 + Z1))². Intensity transmitted T = 1 − R, which also equals 4 × Z1 × Z2 ÷ (Z1 + Z2)².
- Pressure transmission t = 2 × Z2 ÷ (Z1 + Z2). It can be above 1 even though T never is.
- Loss in dB = 10 × log10 of an intensity share, the same as 20 × log10 of the matching pressure ratio. For reflection that is 10 × log10(R).
Evident's technical notes give the impedance equation and the dB loss at a boundary (Equations 6, 7a and 7b). This page writes the same physics in its own words. Reflection is the same in both directions; the pressure signs and the transmitted pressure are not.
Which velocity to use
Use the velocity of the wave mode that is actually crossing the boundary. A straight-beam probe sends a longitudinal wave, so use the longitudinal velocity. A shear wave travelling in the part has its own, lower velocity, so the same steel has a different Z for shear. Liquids and gases carry only longitudinal waves, so a shear wave does not cross into a liquid, and the model here does not cover that case. The presets are longitudinal values.
Worked example 1: water into steel
Water: 1,000 kg/m³ × 1,480 m/s = 1.48 MRayl. Steel 1020: 7,750 kg/m³ × 5,890 m/s = 45.65 MRayl. r = (45.65 − 1.48) ÷ (45.65 + 1.48) = 0.937, so R = 0.937² = 0.878 and T = 0.122. About 88 percent of the intensity reflects and about 12 percent enters the steel. The echo off the steel surface is 10 × log10(0.878) = −0.56 dB below the incident wave, and one pass into the steel loses 10 × log10(0.122) = −9.15 dB. Going from steel back into water gives the same R and the same dB.
Worked example 2: steel to air
Air at 20 °C: 1.204 kg/m³ × 343 m/s = 0.000413 MRayl. Against steel at 45.65 MRayl, r = (0.000413 − 45.65) ÷ (0.000413 + 45.65) = −0.99998, so R = 0.99996 and T = 0.000036. Almost nothing crosses, which is why an air gap under a probe blocks the sound and why a crack that is open and air-filled reflects so strongly. The minus sign means the echo is inverted. These numbers use preset values and are original arithmetic, not exam questions.
Couplant, wedges and flaws
A couplant film fills the gap between probe and part, and a thin film of a liquid passes far more sound than air does. A material closer to the part's impedance gives a smaller reflection at the boundary. A flaw filled with a material whose impedance is near the part's reflects less than an open one. The acoustic impedance guide explains these in context, and the UT formulas guide sets this next to the other UT sums. For refraction at an angle, use the Snell's law calculator; for probe wavelength and near field, the UT beam calculator.
Preset values
Water, glycerin, acrylic, aluminum and steel 1020 use the longitudinal velocities in Table 2 of the Olympus technical notes. Their densities are that table's impedance divided by its velocity, rounded to the nearest 10 kg/m³, so each preset lands on the table's impedance. Air uses 1.204 kg/m³ and 343 m/s, the 20 °C sea-level values. Source pages checked on 25 October 2026. Real materials vary with alloy, temperature and condition, so enter your own values when you have them.
Impedance questions
Which velocity do I use in Z = ρ × c?
The velocity of the wave mode that is travelling. For a straight-beam probe that is the longitudinal velocity. For a shear wave in the part, use the shear velocity, which is lower, so Z is lower. The presets here are longitudinal values.
Is the reflection coefficient the same as the percent reflected?
Only if it is the intensity one. The pressure reflection coefficient is (Z2 − Z1) ÷ (Z2 + Z1) and can be negative. Squaring it gives the share of intensity reflected, which is never negative. Read a question's wording to see which one it wants.
Why is the transmitted pressure ratio sometimes above 1?
Going from a low impedance into a high one, such as water into steel, the pressure on the far side of the boundary can exceed the incident pressure. The energy still adds up: the reflected and transmitted intensity shares always total 100 percent.
Does this cover an angle-beam wave?
No. These coefficients are for a wave arriving square to the boundary. At an angle the split depends on the angle and on mode conversion, and the shares change.