Reflection, transmission, refraction, and impedance

Start with a pressure boundary

A pressure wave reaches a flat boundary between two fluids. Some of the wave returns into the first fluid and some continues into the second. At the boundary their pressure changes must fit together. A transmitted pressure amplitude can exceed the incident amplitude without creating extra energy.

Start with supplied ideal values: medium 1 has density ρ₁ = 2 kg/m³ and sound speed c₁ = 3 m/s; medium 2 has ρ₂ = 3 kg/m³ and c₂ = 4 m/s. The incident peak pressure change is A = 0.02 Pa and its angular frequency is ω = 6 rad/s. These are artificial teaching parameters, not measurements of a real fluid pair.

Both fluids are homogeneous, isotropic, inviscid and at rest. Small linear disturbances meet a stationary, massless interface without surface tension, an added layer or an external source. The equilibrium pressures match. We use pressure and normal particle-velocity continuity for this stated model.

Units, standards, and physical constants and Algebra, functions, graphs, and logarithms prepare the quantities and two simultaneous equations. Geometry, trigonometry, and coordinate systems is a published return path for the oblique-angle part. The needed definitions are given here; first follow the normal-incidence pressure case.

Define pressure, velocity and impedance

Let x increase normally from medium 1 into medium 2, with the interface at x = 0. Medium 1 occupies x < 0 and medium 2 occupies x > 0. The coordinate z runs along the interface; t ≥ 0 s is time. The scalar p is the signed acoustic pressure change about equilibrium, not the ambient pressure. Normal particle velocity u_x is positive toward medium 2.

For one plane pressure wave travelling in the positive normal direction, p/u_x = Z, where the specific acoustic impedance is Z = ρc. For a wave travelling in the negative direction, u_x = −p/Z. The sign describes the direction of the velocity relative to a signed pressure perturbation; it does not make the positive material parameter Z negative.

Here Z₁ = 6 Pa·s/m and Z₂ = 12 Pa·s/m. Density times speed has units kg/(m²·s), equal to Pa·s/m. This is specific impedance, pressure divided by velocity; it is not pressure divided by a volume flow rate.

The supplied bulk moduli K = ρc² are 18 Pa and 48 Pa. The incident amplitude relative to K₁ is 1/900. The transmitted value derived below is 1/1800 of K₂. These small ratios are consistent with the stipulated linear model; they do not establish an experimental accuracy for a real interface.

Solve the two interface conditions

At normal incidence write p_i = A cos(k₁x − ωt), p_r = P_r cos(−k₁x − ωt) and p_t = P_t cos(k₂x − ωt). The subscripts mean incident, reflected and transmitted. P_r is a signed coefficient under this cosine convention; a negative value means the reflected pressure has opposite phase. The wave numbers k_j = ω/c_j are 2 rad/m and 3/2 rad/m.

At x = 0 the three cosine factors agree. Pressure continuity gives A + P_r = P_t. Normal velocity continuity gives (A − P_r)/Z₁ = P_t/Z₂. The minus sign on reflected velocity is essential. These identities hold at all times, including instants with zero cosine; no division by that time factor is needed.

Solving gives r_p = P_r/A = (Z₂ − Z₁)/(Z₂ + Z₁) and t_p = P_t/A = 2Z₂/(Z₂ + Z₁). With the supplied impedances, r_p = 1/3 and t_p = 4/3. Thus P_r = 1/150 Pa and P_t = 2/75 Pa. Both are pressure coefficients, not power fractions.

The common interface velocity coefficient is (A − P_r)/Z₁ = P_t/Z₂ = 1/450 m/s. Its actual value is that coefficient times cos(ωt). The velocity belongs to the fluid particles, while the phase pattern travels through the fluid. Neither is a material trajectory drawn by the later ray arrows.

Read the pressure snapshot

At t = 0 s the incident and reflected pressures add on the left. Their total is (80/3 mPa) cos((2 rad/m)x); on the right the transmitted field is (80/3 mPa) cos((3/2 rad/m)x). Both sides give 80/3 mPa at x = 0, although their spatial wave numbers differ.

The table records the same fixed spatial coordinates. Incident and reflected components are defined only in medium 1; the entries in medium 2 are marked not applicable. The total-field column uses the left sum or the right transmitted field, with the common interface value recorded once.

The wavelength is λ_j = 2π/k_j: λ₁ = π m and λ₂ = 4π/3 m. The common period is 2π/ω = π/3 s. These are whole monochromatic half-space fields. A finite drawing window does not turn them into a finite-energy pulse or a measured beam.

Compare amplitude with power

For one travelling plane wave with peak pressure amplitude P, its time-averaged intensity along the propagation direction is I = P²/(2Z). The factor 1/2 comes from averaging cos² over a full period. Using an RMS amplitude instead would change the amplitude convention; our supplied amplitudes are peak values.

The normal-incidence incident, reflected and transmitted intensity magnitudes are I_i = 1/30000 W/m², I_r = 1/270000 W/m² and I_t = 1/33750 W/m². The reflected wave carries its flux back toward medium 1. The power ratios are R = I_r/I_i = 1/9 and T = I_t/I_i = 8/9, so R + T = 1.

Transmission t_p = 4/3 is compatible with T = 8/9 because power also depends on impedance. At normal incidence R = r_p² and T = (Z₁/Z₂)t_p². Squaring the amplitude alone would miss the impedance change.

Do not insert the combined pressure on the left into the one-travelling-wave formula. The instantaneous normal flux there is p_left u_left = (p_i² − p_r²)/Z₁; the cross terms cancel because reflected velocity has the opposite sign. Its average is I_i − I_r, equal to I_t.

Match phase along the interface

Now use a separate oblique-incidence case with the same media and incident amplitude. Each angle is measured from the interface normal, not from the interface plane. Stipulate sin(θ₁) = 3/10. The reflected and transmitted rays show wave-vector directions, not fluid-particle paths.

A stationary boundary keeps the common frequency. Matching phase at every position along the plane requires k₁ sin(θ₁) = k₁ sin(θ_r) = k₂ sin(θ₂). On the chosen real angles 0 ≤ θ₁, θ_r, θ₂ < π/2, this gives θ_r = θ₁ and sin(θ₂) = (c₂/c₁)sin(θ₁) = 2/5.

The transmitted angle is larger than the incident angle because c₂ > c₁ in this example. The parallel wave number is 3/5 rad/m on all three waves. Their frequencies agree, while the transmitted speed and wavelength differ.

Normal particle velocity is the velocity along a ray multiplied by its cosine projection. Define the one-travelling-wave normal impedance Q_j = Z_j/cos(θ_j). Pressure and normal velocity continuity become A + P_r = P_t and (A − P_r)/Q₁ = P_t/Q₂. The combined incident-plus-reflected field generally has a different pressure/normal-velocity ratio from this single-wave Q₁.

Compare normal and oblique power

For this case Q₁ = 60/√91 Pa·s/m and Q₂ = 60/√21 Pa·s/m. The same two-equation solution now gives r_p = (√91 − √21)/(√91 + √21) and t_p = 2√91/(√91 + √21). The equal normal-velocity coefficient is √91√21/[1500(√91 + √21)] m/s.

The power crossing one common area of interface uses the normal intensity J = P² cos(θ)/(2Z), rather than just intensity along a ray. Consequently R = r_p² and T = [Z₁ cos(θ₂)/(Z₂ cos(θ₁))]t_p². These positive-cosine formulas apply to the real, non-grazing propagating branches used here.

The oblique ratios are approximately R = 0.12320128 and T = 0.87679872. Their exact sum is 1 because (a − b)² + 4ab = (a + b)² with a = √91 and b = √21. Normal incidence and oblique incidence use different projection factors even with the same two media.

Compare snapshots and directions

The upper panel is the normal-incidence pressure snapshot from the exact spatial table. The lower panel is the separate oblique case from the direction table. Their common media do not make them the same incidence case. Solid and dashed curves distinguish pressure components; arrows in the lower panel distinguish propagation directions.

The pressure panel uses 240 px/m horizontally and 4 px/mPa vertically. The ray panel uses the same 160 px/m scale for x and z, so its direction angles are preserved. A ray length of 1.7 m is a drawing choice, not a wavelength, travel-time measurement or beam width.

The propagating-angle condition follows directly from sin(θ₂) = (4/3)sin(θ₁). A positive real transmitted cosine requires sin(θ₁) < 3/4 for these media. At the threshold the transmitted ray is grazing and our positive-cosine formulas exclude it. With sin(θ₁) = 4/5 the required sine is 16/15, so no real transmitted ray exists.

That failure of a propagating ray does not prove that the transmitted field is zero. A treatment of an evanescent field requires additional complex-wave analysis; no numerical evanescent field is supplied here. This lesson has compared signed pressure, particle velocity, phase direction and normal power within its stated fluid-interface model.

Worked examples

Try it yourself

1. Use Z = ρc and the one-normal-travelling-wave relation p/u_x to check the units of specific acoustic impedance.

   Show the answer and reasoning

   Both give Pa·s/m = kg/(m²·s). The main impedances are 6 Pa·s/m and 12 Pa·s/m.

   Pressure divided by particle velocity differs from pressure divided by volume flow. For a reflected wave the signed ratio p/u_x is −Z because it propagates oppositely.

2. In the main normal case, compare t_p with T. Why may the pressure amplitude ratio exceed 1?

   Show the answer and reasoning

   t_p = 4/3 while T = (Z₁/Z₂)t_p² = 8/9. The larger impedance of medium 2 allows a larger peak pressure with less than the full incident power transmitted.

   Pressure amplitudes follow linear relations. Power depends on amplitude squared and on impedance, with the same peak convention on both sides.

3. Keep ρ₁ = 2 kg/m³, c₁ = 3 m/s and A = 0.02 Pa, but set ρ₂ = 1 kg/m³ and c₂ = 3 m/s. Find r_p, t_p, R and T at normal incidence.

   Show the answer and reasoning

   Z₂ = 3 Pa·s/m, so r_p = −1/3 and t_p = 2/3. R = 1/9 and T = (6/3)(2/3)² = 8/9.

   The negative reflected pressure coefficient reverses phase under the fixed cosine convention. It does not mean negative reflected power; R contains its square.

4. Keep medium 1 and the incident wave, but set ρ₂ = 3/2 kg/m³ and c₂ = 4 m/s. Is there normal-incidence pressure reflection, and are the two wavelengths equal?

   Show the answer and reasoning

   Both impedances are 6 Pa·s/m, giving r_p = 0, t_p = 1, R = 0 and T = 1 at normal incidence. The unequal speeds still give wavelengths π m and 4π/3 m at the same ω.

   Matching ρc removes normal-incidence reflection in this model. It does not match sound speeds or establish an oblique match for every angle.

5. For the main oblique case, compare incident and transmitted angles, frequency and wavelength. Measure angles from the normal.

   Show the answer and reasoning

   sin(θ₁) = 3/10 and sin(θ₂) = 2/5, so θ₂ > θ₁ on the stated branches. Both have ω = 6 rad/s, while λ₁ = π m and λ₂ = 4π/3 m.

   The stationary interface preserves frequency; parallel phase matching combines it with different speeds to change wavelength and direction.

6. In the main media, set sin(θ₁) = 4/5. Does a real non-grazing transmitted ray exist, and does the answer prove a zero transmitted field?

   Show the answer and reasoning

   Phase matching would require sin(θ₂) = 16/15, so no real transmitted ray exists. This does not prove that the transmitted field is zero.

   The real positive-cosine ray formulas do not cover that input. An evanescent field lies outside the supplied calculation; it must not be replaced by an unsupported zero-field claim.

What to carry forward

• Interface pressure and normal particle velocity determine signed pressure coefficients; reflected velocity has the opposite propagation sign.
• Pressure amplitude and normal power ratios differ. Impedance and, at oblique incidence, cosine projection enter the power balance.
• A stationary planar interface preserves frequency and parallel phase. Refraction formulas require their stated real propagating-angle domain.