Reflection, Transmission, and Refraction

Abstract

The behavior of one-dimensional waves propagating through media that are not homogeneous will be the focus of this chapter. We start with an examination of the behavior of planewaves impinging on a planar interface between two fluid media with different properties and then extend that analysis to multiple interfaces and to waves that impinge on such an interface from an angle that is not perpendicular to that surface. The extent of those boundaries separating regions with different acoustical properties will be much larger than the wavelength of the sound. Many cases to be examined here will assume that the extent of the boundary is infinite and the wave incident on such an interface will be both reflected back into the medium from which it originated and be transmitted into the second medium on the other side of the interface. This exploration concludes with consideration of wave propagation through a medium whose properties change slowly and continuously through space resulting in curved ray paths. If the variation of sound speed is linear with height or depth, then the ray paths are arcs of circles. Complicated sound speed profiles will be approximated by piecewise-linear segments that have constant sound speed gradients.

In Chap. 10, we transitioned from a lumped-element perspective to a formalism which treated fluids as continuous media that include the effects of both compressibility and inertia throughout. We described disturbances from equilibrium using a wave equation and focused attention on planewaves that could be described as propagating along one spatial dimension. In this chapter, we will examine the behavior of such one-dimensional waves propagating through media that are not homogeneous. We start with an examination of the behavior of planewaves impinging on a planar interface between two fluid media with different properties and then extend that analysis to multiple interfaces and to waves that impinge on the interface from an angle that is not perpendicular to that surface. This exploration concludes with consideration of wave propagation through a medium whose properties change slowly and continuously through space.

What does “slowly” mean in the previous sentence? To answer that question, we will break down the general problem of propagation through an inhomogeneous medium into two limiting cases. As before, such limiting cases suggest appropriate (usually simple) analytical approaches that will develop useful intuition for understanding cases that may be intermediate between the simpler limits. In this chapter, the interesting limits depend on the scale of the medium’s inhomogeneity relative to the wavelength of the sound that impinges on it (from afar, since we consider only planewaves). If a wave encounters a deviate region with a size that is on the order of its own wavelength or smaller, and that region has a different density, a different compressibility, or both, then we treat the inhomogeneous region as a “scatterer.” Those cases will be examined in Chap. 12 when the problem of radiation in three dimensions is analyzed and the scattering body will be driven by the impinging sound wave causing the ensonified region to behave as a radiating “source.” That “scattered” sound field will be superimposed on the incident sound field.

In this chapter, we will consider the opposite limit, where the size of the boundary that separates regions with different acoustical properties is much larger than the wavelength of the sound. In fact, many cases to be examined here will assume that the extent of the boundary is infinite. In those cases, the problem is treated as one where the wave incident on such an interface will be both reflected back into the medium from which it originated and be transmitted into the second medium on the other side of the interface.

For the case where the interface is not discontinuous, what we will mean by requiring that the medium’s properties (e.g., sound speed and density) “are varying slowly” will again be related to the rate at which the property changes in space relative to the wavelength of the sound. For example, if we specify the change in sound speed with position,dc/dz ≡ g, then g will have the units m/s per m, which is equivalent to a frequency. The wave also has a characteristic frequency, f = c/λ, that is the ratio of the sound speed to the wavelength. If g ≪ f, then any significant change in sound speed will occur over a very large number of wavelengths.

1 Normal Incidence

All of us are familiar with echoes, whether produced by a handclap reflected from a large building; a loud “hello” shouted from a precipice and reflected from a bluff, as diagrammed in Fig. 11.1; or a “flutter echo” produced by some impulsive sound source reverberating between long parallel walls. In those cases, a sound wave in air impinges on a rigid solid surface. The air’s particle velocity that accompanies the pressure wave (via the Euler equation) cannot penetrate the solid. To satisfy this rigid boundary condition, we can imagine a sound wave of equal amplitude, but propagating in the opposite direction, coming from within the solid toward the interface, just as was done when examining the reflection of a pulse propagating along a string in Sect. 3.2. As justified in Sect. 12.4.1, the superposition of the two waves cancels the velocity at the interface, satisfying the condition that the interface is impenetrable to the gas, while doubling the acoustic pressure amplitude on that surface.

Fig. 11.1

The sound produced by this person is nearly a planewave (wave fronts shown in blue), characterized by the wavenumber, ${\overrightarrow{k}}_i$, when it is reflected by a large flat rigid surface. The duration of the sound pulse is sufficiently short that the reflected wave, characterized by ${\overrightarrow{k}}_r$, and the incident wave only interact near the surface. To satisfy the condition that no fluid can enter or leave the solid interface, we imagine another planewave of equal amplitude traveling back (with wave fronts shown in ) toward the source. Where the incident and reflected wave superimpose at the rigid surface, the pressure is doubled, but the slope of the pressure, (dp/dx)_x = 0, evaluated at the interface, is zero. By Euler’s equation, the total acoustic particle velocity normal to the surface vanishes, as required

To compare how a real wave reflects in the “rigid boundary” case (and to determine how “rigidly” the boundary behaves), let’s say the magnitude of the incident planewave, $\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|$, approaching the boundary is 94 dB_SPL ⇒ $\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|=\sqrt{2}\kern0.5em \mathrm{Pa}=1.0\kern0.5em {\mathrm{Pa}}_{rms}$. Again, we focus on a single-frequency wave and indicate the incident wave’s amplitude and phase by the complex phasor, ${\hat{\mathbf{p}}}_i$. The time-averaged intensity of a 94 dB_SPL planewave in air is ${\left\langle {I}_{air}\right\rangle}_t={\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|}^2/{\left(2\rho c\right)}_{air}={10}^{\left(94/10\right)}\times \kern0.5em {10}^{-12}\;\mathrm{W}/{\mathrm{m}}^2\cong 2.5\kern0.5em \mathrm{mW}/{\mathrm{m}}^2$. (If this is not instantaneously obvious, you need to review Sect. 10.5.1.)

Let’s also say the wave was reflected from a concrete wall. Since the pressure amplitude at the wall is double that of the sound wave far from the wall, and that the pressure is continuous across the interface, we can calculate the intensity of the sound that entered the wall. The density of concrete is about 2600 kg/m³ and the speed of compressional waves in concrete is about 3100 m/s, so (ρc)_concrete = 8.06 MPa-s/m. Since the pressure at the wall is twice that of the wave in air, $\left|{\hat{\mathbf{p}}}_{\mathbf{t}}\right|=2\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|=2\sqrt{2}\kern0.5em \mathrm{Pa},$ the time-averaged intensity of the sound penetrating the concrete is ${\left\langle {I}_{concrete}\right\rangle}_t={\left|{\hat{\mathbf{p}}}_{\mathbf{t}}\right|}^2/{\left(2\rho c\right)}_{concrete}=0.5\;\upmu \mathrm{W}/{\mathrm{m}}^2$, less than 0.02% of the intensity of the sound in air.

The equations developed in this chapter can be used to show that the amplitude of a wave in air that is reflected from a concrete wall is nearly the same as the incident wave, but not quite. Our assumption that the echo had the same amplitude as the incident sound wave was just as solid as the concrete.

We can also compare the particle velocity of the sound in air with that of the compressional wave within the concrete. In dry air, a 94 dB_SPL sound wave would have a particle velocity amplitude of$\left|{\hat{\mathbf{v}}}_{\mathbf{i}}\right|=\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|/{\left(\rho c\right)}_{air}=3.3\kern0.5em \mathrm{mm}/\mathrm{s}$. In the concrete, the pressure of the transmitted wave is twice that in air (far from the wall), but$\left|{\hat{\mathbf{v}}}_{\mathbf{t}}\right|=2\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|/{\left(\rho c\right)}_{concrete}=0.35\kern0.5em \upmu \mathrm{m}/\mathrm{s}$. Again, we see that our assumption of a rigid and immobile boundary was good to about one part in ten thousand (0.01%). We can check our intensity results using Eq. (10.36), since we also know ${\left\langle I\right\rangle}_t=\left|{\hat{\mathbf{p}}}_{\mathbf{t}}\right|\left|{\hat{\mathbf{v}}}_{\mathbf{t}}\right|/2$, since ${\hat{\mathbf{p}}}_{\mathbf{t}}$ and ${\hat{\mathbf{v}}}_{\mathbf{t}}$ are in-phase for our assumed traveling wave. In dry air at one atmosphere, p_m = 101,325 Pa, and letting T_m = 7 °C, (ρc)_air = 423 Pa-s/m. This gives 〈I_air〉_t = 2.4 mW/m² and in the concrete 〈I_concrete〉_t = 0.5 μW/m², in good agreement with the earlier calculation, as must be the case.

Using Eq. (9.38), the power dissipated per unit area due to the thermal relaxation losses at the interface can be calculated and compared to the incident intensity, since the concrete wall will behave as an isothermal boundary.¹ Since this loss mechanism is frequency dependent, let’s do the calculation for f = 1 kHz. From the DeltaEC Thermophysical Properties of air at 300 K and 1 bar = 10⁵ Pa, δ_κ = 84.6 μm, corresponding to a thermal absorption of 0.76 μW/m². This is a miniscule loss (0.03%), although thermal relaxation at the interface is responsible for a slightly larger decrease in the reflected amplitude, at this frequency, than the intensity of the sound that enters the concrete. Why do we not need to calculate the viscous boundary layer dissipation?

Now that we have the behavior of a common reflection situation “under our belts,” we can start our formal investigation of reflection and transmission of planewaves impinging normally on an interface between any two dissimilar media. For these calculations, keep in mind a very simple case of two immiscible liquids in a uniform gravitational field. Let’s assume that we have a sea of oil floating on a sea of water, as shown in Fig. 11.2, with the “up” direction being +y and y = 0 at the interface between the two liquids. The mass density of the oil, ρ_oil, is less than the density of water. To simplify the specification of variables on either side of the interface, we will designate parameters of the wave and of the medium above the interface (y > 0) with the subscript, i (incident), and those below the interface (y < 0) with the subscript, t (transmitted).

Fig. 11.2

Coordinate system for two fluids that can be thought of as oil floating on water. The dashed line is normal to the plane interface between the two fluids at y = 0. In this diagram, it is assumed that the sound originates from the oil which has a specific acoustic impedance of z_i = ρ_ic_i. A wave will be transmitted across the interface into a second medium (water) with a specific acoustic impedance z_t = ρ_tc_t

Let’s imagine that a single-frequency plane sound wave originates far above the interface and is propagating in the –y direction. That wave excites the interface with a pressure amplitude, $\left|{\hat{\mathbf{p}}}_{\mathbf{i}}\right|$. We can express the pressure amplitude as a function of y and t above the interface

$$ {p}_i\left(y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{i}}{e}^{j\left(\omega\;t+{k}_iy\right)}\right]\kern1em \mathrm{for}\kern1em y>0. $$

(11.1)

That incident plane pressure wave also has an associated fluid particle velocity, ${\hat{\mathbf{v}}}_{\mathbf{i}}$_, that is related to the pressure by the Euler equation.

$$ {v}_i\left(y,t\right)=\mathit{\Re e}\left[\frac{{\hat{\mathbf{p}}}_{\mathbf{i}}}{\rho_i{c}_i}{e}^{j\left(\omega\;t+{k}_iy\right)}\right]=\mathit{\Re e}\left[\frac{{\hat{\mathbf{p}}}_{\mathbf{i}}}{z_i}{e}^{j\left(\omega\;t+{k}_iy\right)}\right]\kern1em \mathrm{for}\kern1em y>0 $$

(11.2)

In the right-hand term of Eq. (11.2), z_i is the specific acoustic impedance of the oil. Using our generic expression, Eq. (10.21), for the sound speed, c, in a fluid with an adiabatic bulk modulus, B_s, $c\equiv \sqrt{B_s/\rho }$, we see that the specific acoustic impedance is a combination of the fluid’s density and compressibility,$z=\rho c=\sqrt{\rho {B}_s}$.

As for the echo case, there will be a wave that is reflected at the interface that will travel in a direction opposite to that of the incident wave.

$$ {p}_r\left(y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{r}}{e}^{j\left(\omega\;t-{k}_iy\right)}\right]\kern1em \mathrm{for}\kern1em y>0 $$

(11.3)

The presence of the incident wave will excite motion in that interface which will also generate a wave propagating into the water as well as the wave that is reflected back into the oil. To demonstrate the necessity for the simultaneous existence of the incident, reflected, and transmitted waves, we need to consider the properties of the interface between the two media.

Since we will treat these media and the waves they contain as a linear system, we can guarantee that the reflected and transmitted waves both have the same frequency as the incident wave: ω = ω_i = ω_r = ω_t. As we have seen for the simple harmonic oscillator, at steady state, the forced linear system can only respond at the forcing frequency. It is the incident wave that is forcing the motion of the interface. Since the frequencies of all of the waves must be the same, and the sound speeds in the two media might differ, the wavelength of the sound in the water will be different than that in the oil: λ_t = c_t/f = 2π/k_t.

$$ {p}_t\left(y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{t}}{e}^{j\left(\omega\;t+{k}_ty\right)}\right]\kern1em \mathrm{for}\kern1em y<0 $$

(11.4)

Our second requirement at the interface is that the two fluids always remain in contact. There are cases where the amplitude of the incident wave is sufficiently large to produce a vapor cavity at the interface, but we will limit our attention here to wave amplitudes that are insufficient to create such cavitation effects [1]. In most practical cases, the boundaries do not separate, so the normal velocities of the two fluids must match at the interface for all times.

$$ {v}_i\left(0,t\right)+{v}_r\left(0,t\right)={v}_t\left(0,t\right) $$

(11.5)

Our final requirement is dictated by the fact that the interface, y = 0, is only a mathematical construct—it has no mass. Since that interface is massless, Newton’s Second Law of Motion would guarantee that any pressure difference across the interface would produce a non-zero force that would create infinite fluid accelerations. To guarantee that does not happen, we require that the pressure be continuous across that interface for all time.

$$ {p}_i\left(0,t\right)+{p}_r\left(0,t\right)={p}_t\left(0,t\right) $$

(11.6)

Those three conditions are sufficient to completely determine the behavior of the waves at the interface when we recognize that v_t(y, t) is related to p_t(y, t) by the specific acoustic impedance of the water, z_t. Taking the ratio of Eq. (11.6) to Eq. (11.5), the amplitude ratio of the reflected and transmitted waves to the amplitude of the incident wave can be calculated.

$$ \frac{p_i\left(0,t\right)+{p}_r\left(0,t\right)}{v_i\left(0,t\right)+{v}_r\left(0,t\right)}=\frac{p_t\left(0,t\right)}{v_t\left(0,t\right)}={z}_t $$

(11.7)

Recognizing that the incident and reflected waves are both in the oil, but are traveling in opposite directions, the magnitude of the specific acoustic impedance, |z_i|, is the same for both waves, but their signs are opposite. This leads to the further simplification of Eq. (11.7).

$$ {z}_i\frac{p_i\left(0,t\right)+{p}_r\left(0,t\right)}{p_i\left(0,t\right)-{p}_r\left(0,t\right)}=-{z}_t $$

(11.8)

Again, because we are limiting ourselves to linear acoustics, it is only the ratios of the amplitudes that have any significance. If we double the amplitude of the incident wave, the amplitudes of the reflected and transmitted waves must also double. It is therefore reasonable to define a pressure reflection coefficient, $\mathbf{R}\equiv {\hat{\mathbf{p}}}_{\mathbf{r}}/{\hat{\mathbf{p}}}_{\mathbf{i}}$, as we did in Eq. (10.105). With that definition and a little algebraic manipulation, Eq. (11.8) can be re-written in a compact form to represent the magnitude of the pressure reflection coefficient, R.

$$ R\equiv \frac{{\hat{\mathbf{p}}}_{\mathbf{r}}}{{\hat{\mathbf{p}}}_{\mathbf{i}}}=\frac{z_t-{z}_i}{z_t+{z}_i}=\frac{\left(\raisebox{1ex}{${z}_t$}\!\left/ \!\raisebox{-1ex}{${z}_i$}\right.\right)-1}{\left(\raisebox{1ex}{${z}_t$}\!\left/ \!\raisebox{-1ex}{${z}_i$}\right.\right)+1} $$

(11.9)

This result makes sense and is reminiscent of Eq. (10.105). If the two media have the same specific acoustic impedances, there is no reflection. Plugging in the specific acoustic impedances for air and concrete produces the same result as we obtained for the “echo” example that began this treatment and produced nearly perfect reflection: R = 0.9999.

In the case of oil over water, as shown in Fig. 11.2, then z_i < z_t so R is positive and the phase of the reflected wave will be the same as the incident wave, although their amplitudes will differ, similar to the in-phase pulse reflection in Fig. 3.4. If the situation were reversed so that the incident wave originated in the water, then z_i > z_t so R is negative and the phase of the reflected wave will be inverted, as shown for the pulses in Fig. 3.3.

Defining the pressure transmission coefficient in a similar way, $T\equiv {\hat{\mathbf{p}}}_{\mathbf{t}}/{\hat{\mathbf{p}}}_{\mathbf{i}}$, the pressure continuity boundary condition of Eq. (11.6) can now be written as 1 + R = T.

$$ T\equiv \frac{{\hat{\mathbf{p}}}_{\mathbf{t}}}{{\hat{\mathbf{p}}}_{\mathbf{i}}}=\frac{2{z}_t}{z_t+{z}_i}=\frac{2\left(\raisebox{1ex}{${z}_t$}\!\left/ \!\raisebox{-1ex}{${z}_i$}\right.\right)}{\left(\raisebox{1ex}{${z}_t$}\!\left/ \!\raisebox{-1ex}{${z}_i$}\right.\right)+1} $$

(11.10)

The factor of two is reassuring, since we expected pressure doubling at the interface in the echo example where z_t ≫ z_i, and we saw the same factor of two in Eq. (10.105).

For oil above water, we can use ρ_oil = 950 kg/m³, c_oil = 1540 m/s, z_oil = (ρc)_oil = 1.463 × 10⁶ Pa ‐ s/m, and ρ_water = 998 kg/m³, c_water = 1481 m/s, and z_water = (ρc)_water = 1.478 × 10⁶ Pa-s/m. The specific acoustic impedance of these two liquids is very close, so for a planewave originating in the oil, R = 0.005 and T = 1.005. If the planewave originated in the water, R = −0.005 and T = 0.995. The fact that R is negative for sound traveling from water into oil indicates that the wave reflected back into the water from the interface is reflected with a 180° phase shift with respect to the incident wave.

If Fig. 11.2 represented air over water, as on a lake, then the impedance contrast would be similar to the echo example, if we assume dry air at one atmosphere, p_m = 101,325 Pa, and T_m = 20 °C, then (ρc) _air = 413 Pa-s/m. If the wave originated in the air, the results would be similar to the echo example. If the wave originated in the water, R = −0.9994, so again the reflection is almost perfect, but the reflected wave is 180° out-of-phase with incident wave. The pressure transmission coefficient is T = 0.00056, so we call that interface, in such a situation (i.e., the wave originating in the medium of higher specific acoustic impedance), a pressure release boundary.

2 Three Media

If we had two gases in contact, then we would need a membrane to keep them from diffusing into each other. Similarly, we might be interested in the transmission of sound between two living spaces separated by a wall. There is no reason why we could not extend the same style of argument just employed with one interface between two media. We would need five waves instead of three since the central region, with specific acoustic impedance, z_c, can support both waves moving up and moving down, as shown in Fig. 11.3.

Fig. 11.3

For a planewave that propagates through three media, bounded by two parallel interfaces, two additional waves need to be included for propagation through the central region. In this diagram, two waves are “trapped” inside the central medium, between y = 0 and y = L, that has a specific acoustic impedance of z_c = ρ_cc_c. The wavevector of the incident planewave, ${\overrightarrow{k}}_i$, and the reflected wavevector, ${\overrightarrow{k}}_r$, are both in the upper medium with z_i = ρ_ic_i. The transmitted wave represented by the wavevector, ${\overrightarrow{k}}_t$, is in the lower medium, with specific acoustic impedance z_t = ρ_tc_t. All wave vectors are colinear and normal to both interfaces

$$ {p}_{up}\left(y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{up}}{e}^{j\left(\omega\;t-{k}_cy\right)}\right]\kern1em \mathrm{and}\kern1em {p}_{down}\left(y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{down}}{e}^{j\left(\omega\;t+{k}_cy\right)}\right] $$

(11.11)

Again, linearity guarantees that these two new waves also have a frequency ω = ω_up = ω_down, so their wavelengths and wavenumbers depend upon the speed of sound in the central medium: ${\lambda}_c={c}_c/f=2\pi /\left|{\overrightarrow{k}}_c\right|$. We would need to impose the continuity of normal particle velocity (11.5) and the continuity of pressure (11.6) on two planes, y = 0 and y = L, if the central medium has thickness, L. We are again only interested in the amplitude ratios (i.e., specification of the incident amplitude, ${\hat{\mathbf{p}}}_{\mathbf{i}}$, is still arbitrary) and can calculate the complex pressure reflection coefficient, R.

$$ \mathbf{R}=\frac{\left(1-\raisebox{1ex}{${z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_t$}\right.\right)\cos \left({k}_cL\right)+j\left(\raisebox{1ex}{${z}_c$}\!\left/ \!\raisebox{-1ex}{${z}_t$}\right.-\raisebox{1ex}{${z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_c$}\right.\right)\sin \left({k}_cL\right)}{\left(1+\raisebox{1ex}{${z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_t$}\right.\right)\cos \left({k}_cL\right)+j\left(\raisebox{1ex}{${z}_c$}\!\left/ \!\raisebox{-1ex}{${z}_t$}\right.+\raisebox{1ex}{${z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_c$}\right.\right)\sin \left({k}_cL\right)} $$

(11.12)

The power transmission coefficient will be a scalar, as it was in Eq. (10.109), and will be related to the reflection coefficient through energy conservation, as it was in Eq. (10.110).

$$ {T}_{\Pi}=\frac{4}{2+\left(\raisebox{1ex}{${z}_t$}\!\left/ \!\raisebox{-1ex}{${z}_i$}\right.+\raisebox{1ex}{${z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_t$}\right.\right){\cos}^2\left({k}_cL\right)+\left(\raisebox{1ex}{${z}_c^2$}\!\left/ \!\raisebox{-1ex}{${z}_t{z}_i$}\right.+\raisebox{1ex}{${z}_t{z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_c^2$}\right.\right){\sin}^2\left({k}_cL\right)} $$

(11.13)

In many cases, the incident and transmitted media are the same, z_i = z_t, simplifying the power transmission coefficient.

$$ {T}_{\Pi}={\left[1+\left(\frac{1}{4}\right){\left(\raisebox{1ex}{${z}_c$}\!\left/ \!\raisebox{-1ex}{${z}_i$}\right.-\raisebox{1ex}{${z}_i$}\!\left/ \!\raisebox{-1ex}{${z}_c$}\right.\right)}^2{\sin}^2\left({k}_cL\right)\right]}^{-1}\kern1em \mathrm{if}\kern1em {z}_i={z}_t $$

(11.14)

2.1 A Limp Diaphragm Separating Two Gases

If we apply Eq. (11.12) to the case of two different gases separated by a membrane with z_c ≫ z_i and z_c ≫ z_t, which is so thin (i.e., L ≪ λ_c) that (z_c/z_t) sin (k_cL) ≪ 1and cos(k_cL) ≅ 1, then the pressure amplitude reflection coefficient is the same as (11.9), so R ≅ (z_t − z_i)/(z_t + z_i), making the diaphragm effectively transparent and the pressure reflection coefficient dependent only upon the relative specific acoustic impedances of the gases.

Let us consider a limp latex diaphragm, having a thickness of 0.006″ = 150 μm, separating air from molecular hydrogen, H₂. At 1.0 kHz, with c_latex ≌ 1000 m/s, (k_cL) = ωL/c_latex ≅ 0.001. Using (ρc)_latex ≅ 50 kPa-s/m, (z_c/z_t) sin (k_cL) ≅ 0.1 ≪ 1, the simplified reflection coefficient of Eq. (11.9) will provide a decent approximation.

For ideal gases, ${\left(\rho c\right)}_{gas}=\sqrt{\gamma {p}_m{\rho}_m}$. Since the diaphragm is assumed to be limp, the pressures of the gases on either side must be identical (otherwise the limp membrane would bulge and then burst), so${\left({z}_t/{z}_i\right)}_{gas}=\sqrt{M_t/{M}_i}$, where M is the molecular mass of each gas. For air and hydrogen, the ratio of specific acoustic impedances is about $\sqrt{15}$, or its reciprocal, if the planewave originates on the air side, making R = ±0.59, with the sign dependent upon whether the planewave originates on the air side (+) or the H₂ side (−).

2.2 An Impedance Matching Antireflective Layer

An important special case covered by Eq. (11.12) occurs when the thickness of the intermediate layer is one-quarter of a wavelength to create a perfectly transmitting condition that repeats for integer multiples of half-wavelengths being added to the original quarter wavelength, resulting in an odd-integer multiple of one-quarter of the wavelength of sound, L = (2n − 1)λ_c/4; n = 1, 2, 3, …; hence, (k_c L) = (n−½)π. In that case, cos(k_cL) ≅ 0 and sin(k_cL) = 1. In addition, if ${z}_c=\sqrt{z_i{z}_t}$, then R = 0 for frequencies near f = (n − ½)c_c/2L, and the intermediate medium is known as a quarter-wavelength impedance matching layer.

In high-quality optics, like camera lenses, this effect is used to make antireflective coatings that optimize the transmission of light through the lens. Since this effect is wavelength-dependent, those optical coatings look magenta when they are optimized for transmission of green light (520 nm ≲ λ_green ≲ 570 nm), since red (620 nm ≲ λ_red ≲ 740 nm) and blue (450 nm ≲ λ_blue ≲ 495 nm) will have non-zero optical reflection coefficients.

Impedance matching layers are very useful for ultrasonic transducers used in medical imaging to optimize the transmission from the transducer material (typically a piezoelectric ceramic) to flesh and to maximize the return signal’s excitation of the transducer during detection.

2.3 The “Mass Law” for Sound Transmission Through Walls

If we consider a solid wall separating two rooms, both containing air, then z_c ≫ z_i = z_t. In that case, the power transmission coefficient of Eq. (11.14) is further simplified.

$$ {T}_{\Pi}\cong {\left[\frac{2{z}_i}{z_c\sin \left({k}_cL\right)}\right]}^2\kern1em \mathrm{if}\kern1em {z}_c\gg {z}_i={z}_t $$

(11.15)

For reasonably small wall thicknesses, L ≪ λ_c = c_c/f, sin(k_cL) ≅ k_cL, so the reduction of sound intensity transmission through the wall depends upon the surface mass density of the wall, ρ_S = ρ_cL.

$$ {T}_{\Pi}\cong {\left(\frac{2}{k_cL}\frac{z_i}{z_c}\right)}^2\cong {\left[\frac{z_i}{\pi f\left({\rho}_cL\right)}\right]}^2\kern1em \mathrm{if}\kern1em {k}_cL\ll 1 $$

(11.16)

For this reason, in architectural acoustics, this result is known as the “mass law” and is particularly important for multi-occupant dwellings (e.g., apartment buildings) because the transmission of sound depends inversely on the square of both the surface mass density of the wall and of the frequency of the sound.² [2] This inverse-square dependence upon frequency explains why the bass from a neighbor’s stereo in an adjoining apartment is more annoying than conversation.

2.4 Duct Constriction/Expansion Low-Pass Filters

We can exploit Eq. (11.14) for the power transmission coefficient to analyze the one-dimensional propagation through a duct that has an expansion chamber or constriction as shown in Fig. 11.4. Following the analysis in Sect. 10.8, and recognizing that such a system has the same sound speed and fluid density throughout, we can copy the five-wave procedure used in Sect. 11.2, but require the continuity of volume velocity, instead of particle velocity, to produce the analog of Eq. (11.14) where A replaces z_i = z_t and A_c replaces z_c [3].

Fig. 11.4

(Left) A compliance with volume, V = A_cL, is placed in a tube of otherwise uniform cross-sectional area, A, to create a low-pass filter. (Right) A constriction with cross-sectional area, A_c < A, is placed in a tube of otherwise uniform cross-sectional area, A, to produce a low-pass filter

$$ {T}_{\Pi}=\frac{1}{1+\frac{{\left[\left(\raisebox{1ex}{${A}_c$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)-\left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{${A}_c$}\right.\right)\right]}^2}{4}{\sin}^2(kL)} $$

(11.17)

It is worthwhile to notice that this coefficient for the power transmission is symmetric with respect to the area ratio, A_c/A, and its reciprocal, A/A_c. That means that the result does not depend upon whether the area change is produced by an expansion chamber, A_c/A > 1, shown in Fig. 11.4 (left), or by a constriction, A/A_c > 1, shown in Fig. 11.4 (right).

If we start by examining the limit where L ≪ λ or kL ≪ 1, we see that either the expansion chamber or the constriction acts as a low-pass filter as shown in Fig. 11.5 for A_c/A = 2, A_c/A = 5, and A_c/A = 10.

Fig. 11.5

The expansion/constriction duct transmission loss, T_Π, for small kL, in dB = 10log₁₀T_Π for area ratios of A_c/A = 2 or 0.5 (solid line), A_c/A = 5 or 0.2 (dotted line), and A_c/A = 10 or 0.1 (dashed line) plotted against a normalized frequency, f^∗, that is scaled by the length of the length of the expansion or constriction, L, and the sound speed, c: f^∗  = πf L/c, which is also f^∗  = πL/λ = kL/2

$$ {T}_{\Pi}={\left\{1+\frac{\pi^2{L}^2}{c^2}{\left[\left(\raisebox{1ex}{${A}_c$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)-\left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{${A}_c$}\right.\right)\right]}^2{f}^2\right\}}^{-1}\kern1em \mathrm{for}\kern1em kL\ll 1 $$

(11.18)

As expected from Eq. (11.18), the transmission for kL < 1 is reduced by 20 dB/decade, in accordance with the f ⁻² frequency dependence. It is clear from Fig. 11.5 that the frequency at which the transmission is reduced by 3 dB is dependent upon the area ratio. For A_c/A = 2 or 0.5, f^∗_-3dB = 0.665; for A_c/A = 5 or 0.2, f^∗_-3dB = 0.208; and for A_c/A = 10 or 0.1, f^∗_-3dB = 0.101.

As frequency increases, the requirement that kL < 1 will be violated. In Fig. 11.5, that means that f^∗  ≤ ½. This restriction indicates that the frequency bandwidth of the expansion/constriction strategy is rather limited as a low-pass filter. From Eq. (11.17), we see that the transmission coefficient becomes one (i.e., 0 dB, no attenuation), when kL = π or f^∗ = π/2, and returns to one for all successive kL = nπ as shown in Fig. 11.6. The maximum transmission loss, (T_Π)_max, occurs when L is equal to an odd multiple of quarter wavelength, (2n − 1)λ/4, making f = (2n − 1)c/4L.

$$ {\left({T}_{\Pi}\right)}_{\mathrm{max}}\cong -10{\log}_{10}\left[\left(\frac{1}{4}\right){\left(\frac{A_c}{A}+\frac{A}{A_c}\right)}^2\right] $$

(11.19)

For A_c/A = 5 or 0.2, (T_Π)_max= − 8.3 dB; and for A_c/A = 10 or 0.1, (T_Π)_max= − 14.1 dB.

Fig. 11.6

The expansion/constriction duct transmission loss, T_Π, in dB = 10log₁₀T_Π. Shown are area ratios of A_c/A = 2 or 0.5 (solid line), A_c/A = 5 or 0.2 (dotted line), and A_c/A = 10 or 0.1 (dashed line) plotted against a normalized frequency, 0 ≤ kL = (2πL/λ) ≤ π. This behavior repeats for integer multiples of π with no loss for kL = 0 or π and maximum loss for kL = π/2

As will be shown in Sect. 13.5.4, for frequencies above f_1,1 = 0.578 c/a, where $a=\sqrt{A/\pi }$ is the radius of the circular duct, the planewave nature of propagation in the duct can no longer be guaranteed. That further limits the utility of such a strategy. Despite these limitations, elaborate combinations of expansion chambers and constrictions are used to attenuate sound in automotive exhaust mufflers using techniques that go well beyond those discussed here [4].

3 Snell’s Law and Fermat’s Principle

The previous results for planewaves normally incident on an interface between two media can be extended to planewaves that impinge on the interface from some arbitrary angle, θ_i, with respect to the normal direction that characterized the surface. The incident wavevector will now have two components, ${\overrightarrow{k}}_i={k}_y{\hat{e}}_y+{k}_x{\hat{e}}_x$, where ${\hat{e}}_y$ and ${\hat{e}}_x$ are unit vectors in the +y and + x directions.

$$ {\displaystyle \begin{array}{l}{p}_i\left(x,y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{i}}\exp j\left(\omega t+\left|{\overrightarrow{k}}_i\right|y\cos {\theta}_i-\left|{\overrightarrow{k}}_i\right|x\sin {\theta}_i\right)\right]\\ {}{p}_r\left(x,y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{r}}\exp j\left(\omega t-\left|{\overrightarrow{k}}_i\right|y\cos {\theta}_r-\left|{\overrightarrow{k}}_i\right|x\sin {\theta}_r\right)\right]\\ {}{p}_t\left(x,y,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{t}}\exp j\left(\omega t+\left|{\overrightarrow{k}}_t\right|y\cos {\theta}_i-\left|{\overrightarrow{k}}_t\right|x\sin {\theta}_t\right)\right]\end{array}} $$

(11.20)

We can make a simple geometrical argument regarding the directions of the reflected and transmitted planewaves by adopting a perspective that was introduced by the Dutch physicist, Christiaan Huygens, in 1678,³ that makes a simple argument which matches boundary conditions at the interface. Figure 11.7 shows the incident wavevector, ${\overrightarrow{k}}_i$, with lines normal to ${\overrightarrow{k}}_i$, indicating the planes of constant pressure phase. We will imagine those lines represent the pressure maxima of the waves propagating toward the interface. Those wave fronts impinge on the interface, and we will consider each intersection of those incident wave fronts and the interface to be a source of pressure maxima for both the reflected and transmitted planewaves.

Fig. 11.7

Diagram showing an incident wave with wavevector, ${\overrightarrow{k}}_i$, intersecting a plane interface at an angle, θ_i, relative to the normal to the interface. Wave fronts of equal phase are shown perpendicular to the wavevector spaced by one wavelength of the incident sound, λ_i. That incident wave excites a reflected wave () and a transmitted wave (). Successive maxima of pressure along the interface are separated by the “trace wavelength,” λ_trace, that is common to all three waves. Based on the wavelengths in this figure, the sound speed in the lower medium, c_t, is greater than the sound speed in the upper medium, c_i < c_t. When the incident angle equals the critical angle, θ_i = θ_crit, then the direction of ${\overrightarrow{k}}_t$ is along the interface (i.e., θ_t = 90°), and we have total internal reflection for all θ_i ≥ θ_crit, so no energy is transmitted

The spacing between those maxima along the interface, which we will call the trace wavelength, λ_trace, is easy to calculate from trigonometry and Garrett’s First Law of Geometry⁴: sin θ_i = λ_i/λ_trace. For normally incident planewaves, the trace velocity c_trace = fλ_trace = ∞, since the wave fronts of equal phase intersect the interface so that the phase of the pressure on the interface is the same at all locations at every instant. Since λ_trace is the same for both the reflected and transmitted waves, the sound speeds in the two media will determine the angle that the reflected planewave, θ_r, and the angle transmitted planewave, θ_t, make with the direction of the normal to the interface.⁵

$$ {\lambda}_{trace}=\frac{\lambda_i}{\sin {\theta}_i}=\frac{\lambda_r}{\sin {\theta}_r}=\frac{\lambda_t}{\sin {\theta}_t} $$

(11.21)

The sound speed, c_i, is the same for both the incident and reflected planewaves, so θ_r = θ_i; the angle of incidence equals the angle of reflection. This case is called specular reflection. If the interface has a surface roughness that is characterized by random assortments of “bumps” that are all much smaller in height and extent than the wavelength of sound, the reflection will be diffuse.

Again, in a linear system, the frequency, ω, with which the incident planewave drives the interface, must also be the frequency of the driven reflected and transmitted planewaves. Since λ  = c/f, we can invert Eq. (11.21) to produce a result known as Snell’s law.⁶

$$ \frac{\sin {\theta}_i}{c_i}=\frac{\sin {\theta}_r}{c_i}=\frac{\sin {\theta}_t}{c_t} $$

(11.22)

Before moving on to calculation of the pressure reflection and transmission coefficients, it worth noticing that Snell’s law could have been derived by insisting that propagation time from one location in the upper medium to some other location in the lower medium is minimized by the same condition on the angles that is expressed in Eq. (11.22). This approach utilizes Fermat’s principle (1622)⁷: “The actual path between two points taken by a [wave] is the one that is traversed in the least time.”

For the actual path to be an extremum of the total transit time, a neighboring path must take the same time; their time difference being at most second order in the deviation of the alternate path from the path of least time. Figure 11.8 provides a diagram of the path of least time (ACB) and a nearby path (AXB).

Fig. 11.8

Three paths from point A in the upper medium to point B in the lower medium. In this diagram, it is assumed that the propagation speed of the wave in the lower medium, c_t, is slower than the propagation speed in the upper medium, c_i > c_t. The path AB () is the shortest distance, and the path ACB () takes the least time. The angles θ_i and θ_t can be determined by calculating the additional travel time beyond that required for ACB when going along the path through AXB () and setting that excess equal to zero [5]

In the upper medium of Fig. 11.5, the perpendicular XE makes CE the reduction in distance traveled by the nearby path AX. The time savings by traveling via AC is therefore t_i = CE/c_i. Similarly, in the lower medium, the perpendicular CF makes XF the additional distance traveled by the nearby path XB. The excess time in the lower medium is therefore t_t = XF/c_t. Since ∡EXC = θ_i and ∡FCX = θ_t, we can express the two transit times for the nearby path in terms of the common length XC using simple trigonometry.

$$ {t}_i=\frac{EC}{c_i}=\frac{XC\sin {\theta}_i}{c_i}={t}_t=\frac{XF}{c_t}=\frac{XC\sin {\theta}_t}{c_t} $$

(11.23)

Equation (11.23) also yields Snell’s law after dividing through by XC.

3.1 Total Internal Reflection

In passing from a faster medium into a slower medium, c_i > c_t, the angle that the transmitted wavevector makes with respect to the normal at the interface, θ_t, is smaller than the incident angle, θ_i, also measured from the normal. If the incident medium has a slower sound speed than the transmission medium, c_i < c_t, then it is possible that the angle of transmission, θ_t, dictated by Snell’s law, could become a complex number. In that case, we can define a critical angle, θ_crit, where θ_t = 90° (i.e., the transmitted wave travels along the interface but not into the second medium).

$$ \sin {\theta}_{crit}=\frac{c_i}{c_t};\kern1em \mathrm{if}\kern1em {c}_i<{c}_t $$

(11.24)

This is a consequence of the fact that the minimum trace velocity, c_trace = fλ_trace, shown in Fig. 11.7, in the lower medium, is limited to c_t, which equals c_trace when θ_t = 90^°. Using the fact that cos²θ + sin²θ = 1, the angle of transmission can be expressed in terms of the angle of incidence and the sound speed ratio.

$$ \cos {\theta}_t=\sqrt{1-{\sin}^2{\theta}_t}=\sqrt{1-{\left(\raisebox{1ex}{${c}_t$}\!\left/ \!\raisebox{-1ex}{${c}_i$}\right.\right)}^2{\sin}^2{\theta}_i}=-j\sqrt{{\left(\raisebox{1ex}{${c}_t$}\!\left/ \!\raisebox{-1ex}{${c}_i$}\right.\right)}^2{\sin}^2{\theta}_i-1} $$

(11.25)

If c_t > c_i, the first term under the square root becomes imaginary, which is expressed explicitly in the final term at the right of Eq. (11.25).

Substitution of Eq. (11.25) into the expression for the transmitted pressure in Eq. (11.20) produces an exponential decay of the sound with increasing distance away from the interface, for incident angles that exceed the critical angle, producing an inverse characteristic depth, (1/δ).

$$ {k}_z=\frac{1}{\delta }=\left|{\overrightarrow{k}}_t\right|\sqrt{{\left({c}_t/{c}_i\right)}^2{\sin}^2{\theta}_i-1} $$

(11.26)

$$ {p}_t\left(x,z,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{p}}}_{\mathbf{t}}{e}^{+\left(y/\delta \right)}{\mathrm{e}}^{j\left[\omega\;t-\left|{\overrightarrow{k}}_i\right|x\sin {\theta}_i\right]}\right];\kern1em \mathrm{for}\kern0.5em y\le 0 $$

(11.27)

This solution is a wave that is localized just below the interface. For angles of incidence, θ_i < θ_crit, the transmitted wave fills the half-space of the lower medium, as shown in the lower half of Fig. 11.7. When θ_i = θ_crit, all of that outgoing energy collapses to the surface in a layer with exponential thickness, δ. The thickness of that layer decreases monotonically as θ_i increases beyond θ_crit.

If we apply the Euler equation to Eq. (11.27) to determine the velocity, v_t, normal to the surface, we find ${\hat{\mathbf{v}}}_{\mathbf{t}}=j{\hat{\mathbf{p}}}_{\mathbf{t}}\left(\omega\;\rho\;\delta \right)$, so the pressure and particle velocity are 90° out-of-phase. The intensity leaving the interface is zero since there is no in-phase component of pressure and velocity. That does not mean that the layer contains no energy, just that none of the energy is propagating. If there were some scattering centers (i.e., defects, voids, inclusions of material with some compressibility, and/or density contrast), the energy trapped in that layer could be reradiated (i.e., scattered; see Sect. 12.6) and escape.

The concept of total internal reflection was first recognized in optics. One application of total internal reflection that I find particularly inspiring occurs in the compound eyes of insects and other arthropods. With thousands of lenses, the use of an equal number of irises, like those in avian and mammalian eyes, is not particularly efficient. In the arthropod’s compound eye, each lens is connected to the optic nerve by a transparent tube that contains the light by total internal reflection, due to the difference between the index of refraction (i.e., the optical equivalent of our specific acoustic impedance) between the tube and the surrounding fluid. The way that insects control the amount of light that reaches the optic nerve is by a chemical reaction in the fluid that surrounds the tube. Under bright lighting, the fluid will produce a precipitate with small particles that scatters light out of the trapped layer, thereby reducing the amount of light reaching the optic nerve. The physical structure of the compound insect eye is shown in Fig. 11.9.

Fig. 11.9

(Left) The compound eyes of the blue bottle fly. (Right) Each of the “tubes” is optical fibers, called an ommatidium. They guide light from the lens to the optic nerve endings (4 and 5) using total internal reflection to regulate the transmission of the light. Ant eyes have about six ommatidia and dragonfly eyes have as many as 25,000. This anatomical drawing of a compound eye shows the corneal lens (1) and the crystalline cone (2) which combine to form the dioptric apparatus. At the base is the light-sensitive rhabdom (5 and 6). The space between the tubes (3) contains the pigment cells that will form a precipitate that scatters some of the optical energy out of the trapped layer at the interface if the light is too bright, effectively reducing the field of view (known in optics as the “numerical aperture” of the lens). (Photo and diagram taken from Wikipedia)

Proof of this system’s efficacy is the fly’s ability to avoid getting swatted outdoors in broad daylight and indoors in a darkened room. (We should now pause briefly out of respect for Darwin and the power of natural selection.) I would be most interested in finding a similar system for controlling the sound amplitude that reaches the auditory nerve in any animal, although the ability to focus sound seems ubiquitous in marine mammals, as shown for the Cuvier’s beaked whale in Fig. 11.18.

3.2 The Rayleigh Reflection Coefficient

Although we have calculated the angles of reflection and transmission for planewaves of sound impinging on a planar interface from an arbitrary direction, we have yet to calculate the amplitude reflection coefficient in terms of the specific acoustic impedances, as we did for the case of normal incidence at the interface between two media in Eq. (11.9). To produce the equivalent of Eq. (11.9) for oblique incidence (i.e., θ_i ≠ 0°), we need to generalize the velocity boundary condition of Eq. (11.5), since it is only the normal component of the velocity that needs to be continuous across the boundary to avoid cavitation at the interface.⁸

$$ \left|{\hat{\mathbf{v}}}_{\mathbf{i}}\right|\cos {\theta}_i+\left|{\hat{\mathbf{v}}}_{\mathbf{r}}\right|\cos {\theta}_r=\left|{\hat{\mathbf{v}}}_{\mathbf{t}}\right|\cos {\theta}_t $$

(11.28)

Pressure continuity, expressed in Eq. (11.6), is unchanged since pressure is a scalar quantity.

Forming the ratio of those boundary conditions, as we did in Eq. (11.7), leads to the desired result for the ratio of the reflected pressure to the incident pressure $R\equiv {\hat{\mathbf{p}}}_{\mathbf{r}}/{\hat{\mathbf{p}}}_{\mathbf{i}}$.

$$ R\equiv \frac{{\hat{\mathbf{p}}}_{\mathbf{r}}}{{\hat{\mathbf{p}}}_{\mathbf{i}}}=\frac{\frac{z_t}{\cos {\theta}_t}-\frac{z_i}{\cos {\theta}_i}}{\frac{z_t}{\cos {\theta}_t}+\frac{z_i}{\cos {\theta}_i}} $$

(11.29)

Once again, we are confronted with an expression for the reflection coefficient that can vanish, providing perfect transmission, if the numerator of Eq. (11.29) is zero.

$$ \frac{\cos \kern0.5em {\theta}_t}{\cos \kern0.5em {\theta}_i}=\frac{z_t}{z_i} $$

(11.30)

Using Snell’s law to eliminate θ_t, this angle, called the angle of intromission, θ_intro, can be expressed in terms of both the impedance and sound speed or mass density ratios.

$$ {\sin}^2\left({\theta}_{intro}\right)=\frac{{\left({z}_t/{z}_i\right)}^2-1}{{\left({z}_t/{z}_i\right)}^2-{\left({c}_t/{c}_i\right)}^2}=\frac{1-{\left({z}_i/{z}_t\right)}^2}{1-{\left({\rho}_1/{\rho}_2\right)}^2} $$

(11.31)

Not every combination of media will have an angle of intromission that depends upon both sound speed and density ratios. For θ_intro to be real, both the numerator and denominator of Eq. (11.31) must have the same sign.

There are four combinations of impedance and sound speed ratios:

• (i) No θ_crit and no θ_intro If both z_t/z_i < 1 and c_t/c_i < 1, there is no critical angle, so 0 < R < −1 for all θ_i. For this case, θ_i > θ_t, so (cosθ_t/ cos θ_i) > 1 and z_t/z_i < 1, so there can be no angle of intromission.

• (ii) Real θ_crit but no θ_intro If both z_t/z_i > 1 and c_t/c_i > z_t/z_i, then there will be a critical angle above which there will be total internal reflection so R = 1 if θ_i > θ_crit. Since c_t/c_i > z_t/z_i, the numerator of Eq. (11.31) will be positive, but the denominator will be negative, so there can be no θ_intro.

• (iii) Real θ_intro but no θ_crit If z_t/z_i > 1 and c_t/c_i < 1, then there will be no critical angle, so 0 ≤ R < 1 for all θ_i. Since c_t/c_i < z_t/z_i, both the numerator and the denominator of Eq. (11.31) will be positive, so there will be an angle of intromission, θ_i = θ_intro < 90^°.

• (iv) Real θ_intro and real θ_crit If c_t/c_i > 1 and z_t/z_i < 1, then there will be both a critical angle, θ_crit < 90°, above which there will be total internal reflection so R = 1 if θ_i > θ_crit, and both the numerator and the denominator of Eq. (11.31) will be negative, so there will be an angle of intromission θ_i = θ_intro < θ_crit < 90^°.

Total transmission at the angle of intromission has been observed in high-porosity marine sediments, like silty clays, which exhibit a sound speed through its bulk which is lower than that of the interstitial fluid within its pores. When high-porosity sediment is at the water/sediment interface, there can be total transmission of sound into the seafloor. Measurements of the angle of intromission in coastal regions around Italy indicate that the properties of the high-porosity sediments are surprisingly uniform over large areas [6].

4 Constant Sound Speed Gradients

To this point in our discussion of the refraction of sound, we have only considered plane boundaries that separate media with a discontinuity in their sound speed, density, and/or specific acoustic impedance. There are many situations of interest where the refraction (or bending) of a planewave of sound can create interesting and significant effects resulting from a gradual change in the acoustical properties of the medium. These effects are easily calculable in the limit that such sound speed changes are a linear function of position when those changes occur over a distance that extends over very many acoustic wavelengths.

Figure 11.10 shows two examples of sound refraction in air that supports a temperature gradient that depends upon the time of day. As expressed by the quadratic dependence of sound speed on the mean temperature of an ideal gas in Eq. (10.23), the change in sound speed, Δc, with height, z, above the ground is related to the changes in the absolute temperature, T.

Fig. 11.10

(Left) During the day, the temperature of the atmosphere generally decreases with altitude above the ground causing sound waves to refract upward since the speed of sound is proportional to $\sqrt{T}$. (Right) The opposite conditions can occur at night when the temperature of the air in contact with the ground is colder that than the air above it (a “temperature inversion”). This causes sound to be refracted downward. If there is wind, it can also cause the sound speed to vary with height above the ground. (Figures courtesy of T. B. Gabrielson [7])

$$ \frac{dc}{dz}=\left(\frac{dc}{dT}\right)\left(\frac{dT}{dz}\right)=\frac{c}{2T}\left(\frac{dT}{dz}\right) $$

(11.32)

Similar refractive effects occur in the ocean where the speed of sound in seawater is a complicated function of temperature, salinity, and depth (pressure).

$$ {\displaystyle \begin{array}{c}c=\kern0.15em 1493.0+3\left(t-10\right)-0.006{\left(t-10\right)}^2-0.04{\left(t-18\right)}^2\\ {}+1.2\left(S-35\right)-0.01\left(t-18\right)\left(S-35\right)+D/61\end{array}} $$

(11.33)

In Eq. (11.33), the sound speed, c in m/s, is a function of the temperature, t, in degrees Celsius. The salinity, S, is measured in grams of salt per kilogram of water, and D is the depth below the surface measured in meters [8]. That formula is valid within ±0.2 m/s for −2 °C ≤ t ≤ +24.5 °C, 0.030 ≤ S ≤ 0.042, and 0 ≤ D ≤ 1000 m.

Figure 11.11 shows the sound speed variation with depth that was measured over 9 years in the ocean 15 miles (24 km) south-east of Bermuda. Although significant sound speed variation over the 9 years is apparent within roughly 2 km of the surface, it is possible to represent the change in sound speed with depth using a piecewise-linear approximation.

Fig. 11.11

(Left) These deep-ocean sound speed profiles were taken every 2 weeks over a 9-year period at a location 24 km SE of Bermuda. [9] The solid curve is the average and the dashed curves show the extremes. This typical deep-sea profile may be divided into a number of layers having different properties. At the top, the diurnal layer shows day-night variability and responds to weather changes. Below it, down to the depth of about 300 m, lays the seasonal thermocline that is above the main thermocline, which exhibits the strongest sound speed gradient. Below the thermocline, beneath 1200 m, the deep isothermal layer has a constant temperature of +4 °C. There, the sound speed’s increase with depth is dominated by the increasing pressure. (Right) The measured sound speed profile is replaced by a piecewise-linear approximation that provides a constant sound speed gradient for each layer, as indicated by the embedded table. At a depth of approximately 3700 m, the sound speed is the same as it is at a depth of approximately 560 m. Between those two depths, sound can be trapped in the “deep sound channel” in the same way that light is trapped in an optical fiber, except the core of an optical fiber that is about 10 μm and the sound channel that is about 1 km, a height ratio of about 10⁸

4.1 Constant Gradient’s Equivalence to Solid Body Rotation

The goal in this sub-section is to develop a simple formalism that will permit calculation of the path of planewaves in a medium with a sound speed that is a linear function of height or depth. Furthermore, we would like to be able to track the trajectory of the sound (i.e., follow the direction of the wavevector) through regions of changing sound speed gradient, a process known as ray tracing. To do this, we must first recognize that the propagation of sound through a region of constant sound speed gradient is isomorphic to the solid body rotation of a disk, since the tangential velocity, ${\overrightarrow{v}}_{\mathrm{tan}}$, of any point on the rotating disk is proportional to its (linear) distance, $\left|\overrightarrow{r}\right|$, from the axis of rotation.

$$ {\overrightarrow{v}}_{\mathrm{tan}}=\overrightarrow{r}\times \overrightarrow{\omega} $$

(11.34)

Since this equation refers to a rotating disk, the use of the cross-product might seem to be pedantic excess, but it is intended to remind you of your study of solid body rotational dynamics during your freshman physics course.

If we limit ourselves to linear changes in sound speed with height or depth, then the sound speed gradient is a constant: dc/dz ≡ g. In analogy with a rotating disk of radius, R, as diagrammed in Fig. 11.12, we can equate sound speed, c, with the tangential velocity of a point on the rim of the disk, ${\overrightarrow{v}}_{rim}$, and integrate our above definition; it is easy to see that g plays the same role as ω (and has the same units) in Eq. (11.34).

$$ {\int}_0^{v_{rim}} dc=g{\int}_0^R dz\kern0.5em \Rightarrow \kern0.5em R=\frac{\left|{\overrightarrow{v}}_{rim}\right|}{g} $$

(11.35)

Fig. 11.12

Shown at the right is a disk that is rotating. The dashed, colored arrows indicate the tangential velocity that increases linearly with distance from the axis of rotation. Such a velocity distribution in space is identical to the assumed constant sound speed gradient that will control the refraction of sound waves

I like to think of the refraction (bending) of the sound wave’s trajectory in a constant sound speed gradient by picturing the advance of the planewave fronts as represented schematically in Fig. 11.13.

Fig. 11.13

A conceptual sketch of planewave fronts moving through a medium with the velocity gradient shown at the left. The arrows at the top of the wave fronts are always longer than the arrows at the bottom. This turns the wave front downward until the wave front is nearly horizontal. Once horizontal, both ends of the wave front will move with the same speed, and the wave will continue downward without changing direction

To introduce this “disk rotation” approach to analysis of the propagation of planewaves through a medium with constant speed gradients, let’s consider the measurement of road noise during the day, when the temperature of the air is decreasing with increasing altitude, as shown in Fig. 11.10 (left).

Assume that you are asked to determine the sound pressure level created by traffic on a highway at a proposed housing location that is 150 m from the highway. By measuring the average temperature at a height of 0.5 m and at 5 m, you determine that the sound speed at 0.5 m is 342.8 m/s and at 5.0 m is 341.3 m/s. How high must your microphone be above the ground, at a distance of 150 m from the source, to ensure that you intercept sound that leaves the highway in an initially horizontal direction?

To begin, let’s calculate the sound speed gradient, dc/dz = g ≌ Δc/Δz = (2.8–1.3 m/s)/4.5 m = 0.333 s⁻¹. Since the sound speed decreases with height, we know that the wave will be refracted upward, as shown in Fig. 11.10 (left). We’ll assume that the noise is generated at the intersection of the tire and the road at z = 0, where c_o = 343.0 m/s (extrapolating down to the surface from the lowest measurement at 0.5 m above the ground).

We now need to calculate the radius, |R|, of this limiting ray’s circular path.

$$ \left|R\right|=\frac{c_o}{g}=1029\kern0.5em \mathrm{m} $$

(11.36)

A simple trigonometry identity can be used to determine the height, h = 11.0 m, above the ground that a microphone must be placed to receive the sound radiated by the tires. For me, drawing a sketch, like in Fig. 11.14, is always crucial.

Fig. 11.14

Sketch (not to scale) of the ray path of the sound generated by the tire noise on a road surface. The observation point is 150 m from the sound source. Between the ground and the circular arc is a “shadow zone” that is created by the upward refraction of the sound produced by the sound speed gradient, dc/dz = 0.333 s⁻¹, which was used to calculate the radius, |R| = 1029 m, applying Eq. (11.36). A microphone must be placed at least a distance, h, above the ground to intercept the tire noise. At the surface (h = 0), the sound speed is c_o = 343.0 m/s. At h = 11.0 m above the surface, the sound speed is 339.3 m/s

$$ {\displaystyle \begin{array}{c}\sin \theta =\frac{150}{1029}=0.146\kern1em \Rightarrow \kern1em \theta =0.146\;\mathrm{rad}\kern0.5em ={8.38}^{{}^{\circ}}\\ {}h=\left|R\right|-\left|R\right|\cos \theta =\left|R\right|\left(1-\cos \theta \right)\cong \left|R\right|\frac{\theta^2}{2}=11.0\kern0.5em \mathrm{m}\end{array}} $$

(11.37)

The final expression above made use of the small-angle expansion of cosine in Eq. (1.6).

Although the refractive process was not understood until the twentieth century [7], reports of the existence of an acoustic “shadow zone” appeared much earlier. Below is the account of R. G. H. Kean as he watched the Battle of Gaines’s Mill during the American Civil War [10]:

I distinctly saw the musket-fire of both lines … I saw batteries of artillery on both sides come into action and fire rapidly. Yet looking for near two hours, from about 5 to 7 P.M. on a midsummer afternoon, at a battle in which at least 50,000 men were actually engaged, and doubtless at least 100 pieces of field-artillery … not a single sound of the battle was audible to General Randolph and myself … [However, the] cannonade of that very battle was distinctly heard at Amhurst Court-house, 100 miles west of Richmond, as I have been most credibly informed.

The process of refraction in a sound speed gradient, as illustrated in a particularly simple form in Fig. 11.14, is entirely reversible. If we were to say that we had a sound source directed 8.4° below the horizontal direction, located 11.0 m above the ground, then we could just as easily say that it would intersect the ground 150 m away, based on Fig. 11.14 and the results of Eq. (11.37).

Since much of the interest in the propagation of sound in a sound speed gradient (linear or otherwise) tends to concern sound waves that are nearly grazing the horizontal axis, the grazing angle is more commonly used in Snell’s law than the angle measured with respect to the normal to the surface, which was our previous choice for discussion of reflection and transmission from a plane surface of discontinuity. If we now use the grazing angle, then we can rewrite Eq. (11.22) by letting θ be the angle below the horizontal and c_o be the sound speed at a location where the sound waves are propagating horizontally (i.e., cos 0° = 1).

$$ \frac{c}{\cos \theta }={c}_o $$

(11.38)

Let’s apply the above version of Snell’s law to the example in Fig. 11.14. That ray is horizontal at h = 0, where c_o = 343.0 m/s. Assuming, as before, a constant sound speed gradient, at the “source height” of 11.0 m, c = 339.3 m/s. Plugging directly into Eq. (11.38), cos θ = c/c_o, so θ = 0.147 radians = 8.4°. Of course, we could have used Eq. (11.37) to calculate the sound speed 11.0 m above the surface, since we already knew that θ = 8.4°.

4.2 Sound Channels

We will now apply this formalism to a particularly interesting case that occurs in a sound speed profile like the one shown in Fig. 11.11, describing the deep ocean off of Bermuda, which we have approximated by a series of contiguous line segments with constant (but different) sound speed gradients. There is a sound speed minimum of c_min = 1486 m/s, at a depth of 1112 m, between the main thermocline and the deep isothermal layer. As evident from Fig. 11.13, in such a constant gradient (i.e., linear sound speed profile), the sound will “go to the slow,” and therefore, we expect that some sound will become trapped around that sound speed minimum in a sound channel.

In this sub-section, we will trace several rays from a submerged source to develop an understanding of the channel’s propagation characteristics. Assume that there is a sound source at a depth of 800 m, where the sound speed (by interpolation) is 1507 m/s. We start by calculating the upward angle from the source that will result in a horizontal ray at the top of the upper gradient, z = 560 m, where the sound speed is c_max = 1523 m/s. By Snell’s law, as expressed in terms of the grazing angles in Eq. (11.38), θ₁ = cos⁻¹(1507/1523) = 0.145; hence, θ₁ = 8.31°. The radius of the circular path of the ray, |R₁| = c_o/g = (1523 m/s ÷ 0.067 s⁻¹) = 22.73 km. Based on the diagram in Fig. 11.15, the sound ray intersects the top of the channel at a horizontal distance that is r₁ = |R₁|sin θ₁ = 3.3 km from the source.

Fig. 11.15

The diagram of the sound speed vs. depth at the left is based on the piecewise-linear approximation of Fig. 11.11. There is a sound speed minimum at a depth of z = 1112 m that is the axis of a sound channel shown by the horizontal line. The velocity maxima are shown by the horizontal lines at depths of 560 m and 3700 m, corresponding to c_max = 1523 m/s. The ray paths, ranges, depths, and especially the ray path radii, |R₁| and |R₂|, are not drawn to scale. The limiting ray paths for a sound source located at a depth of 800 m, and horizontal distance, r = 0, are shown in , launched at an angle of θ₁ = 8.31° above the horizontal. The turning point for the ray in the upper layer occurs at a horizontal distance from the source of r₁, while the turning point for the ray in the lower layer occurs at a horizontal distance from the source of r₃. The ray crosses the channel axis at r₂ and enters the lower layer at an angle θ₂ below the horizontal

From the top of the channel (z = 560 m), the ray will return to the channel axis along the same circular path and will intersect the axis (z = 1112 m) at θ₂ = cos⁻¹ (1486/1523), so θ₂ = 0.221 rad = 12.66°. The horizontal distance, r₂ – r₁ = |R₁| sin θ₂ = 5.0 km.

The ray enters the deep isothermal layer where |R₂| = (1486 m/s ÷ 0.0143 s⁻¹) = 103.9 km. (The center of that circle is only a geometrical “construction point,” so the fact that it is located above the stratosphere should not be a cause for our concern.) Since we have defined the bottom of the channel as the location where the sound speed in the deep isothermal layer equals the sound speed at the top of the main thermocline, c_max = 1523 m/s, Snell’s law will guarantee that it will reach that depth with a ray that is again horizontal, after traveling a horizontal distance, r₃ – r₂ = |R₂| sin θ₂ = 22.8 km. From that point at horizontal distance from the source, r₃, the path should repeat indefinitely with the cycle distance from the top of the channel back to the top of the channel being 2(r₃ – r₁) = 55.6 km.

To calculate the angular width of the sound radiated by our source located at z = 800 m below the surface, which will be trapped in the channel, we need to follow a ray that leaves the source at an angle below horizontal, θ₋₁, that will also take it to the depth (z = 3700 m) where the sound speed is again c_max. That initially downward-directed ray will also have to cross the sound channel axis at θ₂ = 12.66° if it is to become horizontal at z = 3700 m. Again, Snell’s law guarantees that (1507 m/s ÷ cos θ₋₁) = (1486 m/s ÷ cos θ₂) = 1523 m/s, so | θ₋₁| = | θ₁| = 8.31°. That result is easy to visualize since we can imagine a “virtual source” that is located 2r₁ away from the original source, where the ray makes an angle of − 8.31° with respect to the horizontal. Any ray that leaves the source within ± 8.31° will be trapped in the sound channel created by the two sound speed gradients.

Again, Snell’s law makes it easy to see that an initially horizontal ray leaving our source at a depth of z = 800 m will go up and down in the channel from a depth of z = 800 m to a depth of z = 2581 m, where the sound speed will again be 1507 m/s and the rays will again be horizontal. Figure 11.16 diagrams ray paths from a more complicated sound channel for a source located on the channel axis [11].

Fig. 11.16

Ray diagram for transmission paths in the deep sound channel for a source located on the channel axis. Depth is in fathoms (1 fathom = 6 feet = 1.829 m), range is in miles (1 mi = 1.609 km), and sound speed is in feet/second (1 fps = 30.48 cm/s), as a function of depth is provided at the right side of the graph. For angles θ > θ_max = 12.2°, waves escaping the channel are reflected from the air-water interface [11]

The results illustrated in the previous example can be generalized by defining a maximum angle ±θ_max (above or below the horizontal) that will result in the trapping of a wave launched at the channel axis, in terms of the minimum sound speed at the axis, c_min, and the maximum sound speed, c_max, at the bottom and top of the channel.

$$ \cos {\theta}_{\mathrm{max}}=\frac{c_{\mathrm{min}}}{c_{\mathrm{max}}} $$

(11.39)

Expanding cos θ_max for small angles and defining Δc = c_max − c_min, Eq. (11.39) can be approximated as ${\theta}_{max}\cong \sqrt{2 Dc/{c}_{max}}$. When the source is above or below the height or depth of the channel by a distance, z_o, in a layer D thick, Snell’s law, and the requirement that a trapped ray at the channel boundary is horizontal, provides the range of trapping angles, ${\theta}_{\pm 1}=\pm {\theta}_{max}\sqrt{z_o/D}$, when the source is not located on the sound channel’s axis. Along the channel axis, z_o = D. At the extremes, z_o = 0, only the horizontal ray is trapped. The long-range transmission ability of the deep sound channel was used during World War II to rescue aviators that crashed at sea.⁹ A downed pilot would drop a small explosive charge that was rigged to detonate near the depth of the sound channel axis. If the sound were detected with two or more hydrophones, also located at the axis depth, the reception times could be used to localize the search by triangulation. Successful rescues were made this way using hydrophones connected to shore stations that were thousands of miles from the aircraft impacts. More recently, the same ability to make localizations at sea has been used for missile-impact location [12], and measurement of the time delays has been used to measure the mean temperature of the deep ocean [13].

4.3 Propagation Delay*

It is interesting to calculate the difference in the propagation times for sound traveling along different paths to reach the same horizontal distance from the source. Although the path along the axis is the shortest, it is also going through the medium with the minimum sound speed. The longer (curved) paths travel through water that has a faster sound speed. Does the shorter path beat the faster path?

Let’s first consider the ray that is generated on the sound channel axis and travels along a circular path to the top of the channel. The initial angle of such a ray above the horizontal is ${\theta}_{max}\cong \sqrt{2 Dc/{c}_{max}}$, as already shown. The transit time, T_upper, for that ray can be calculated if we integrate from θ = 0° to θ = θ_max, where the sound speed depends upon angle, c (θ ) = c_max cos θ.

$$ {T}_{upper}={\int}_0^{\theta_{\mathrm{max}}}\frac{\left|R\right| d\theta}{c\left(\theta \right)}={\int}_0^{\theta_{\mathrm{max}}}\frac{\left|R\right| d\theta}{c_{\mathrm{max}}\cos \theta }=\frac{\left|R\right|}{c_{\mathrm{max}}}{\int}_0^{\theta_{\mathrm{max}}}\frac{d\theta}{\cos \theta } $$

(11.40)

There is an analytical solution for the above definite integral, but simply writing the answer provides no useful physical insight.

$$ {\int}_0^{\theta_{\mathrm{max}}}\frac{d\theta}{\cos \theta }={\int}_0^{\theta_{\mathrm{max}}}\sec \theta \kern0.5em d\theta =\ln {\left[\csc \theta -\cot \theta \right]}_0^{\theta_{\mathrm{max}}} $$

(11.41)

In these problems, the angles are generally small, so the series expansion of the sine and cosine functions and the binomial expansion can both be employed to simplify the integration and its interpretation.

$$ {\int}_0^{\theta_{\mathrm{max}}}\frac{d\theta}{\cos \theta}\cong {\int}_0^{\theta_{\mathrm{max}}}\frac{d\theta}{\left(1-\frac{\theta^2}{2}\right)}\cong {\int}_0^{\theta_{\mathrm{max}}}\left(1+\frac{\theta^2}{2}\right) d\theta =\kern0.5em {\theta}_{\mathrm{max}}+\frac{\theta_{\mathrm{max}}^3}{6} $$

(11.42)

The transit time for the axial ray, T_axial, that goes the same horizontal distance, r = |R| sin θ_max, is just T_axial = r/c_min. Since c_min = c_max cos θ_max, the same approach can be used.

$$ {T}_{axial}=\frac{\left|R\right|\sin {\theta}_{\mathrm{max}}}{c_{\mathrm{max}}\cos {\theta}_{\mathrm{max}}}\cong \frac{\left|R\right|}{c_{\mathrm{max}}}\frac{\left({\theta}_{\mathrm{max}}-\frac{\theta_{\mathrm{max}}^3}{6}\right)}{\left(1-\frac{\theta_{\mathrm{max}}^2}{2}\right)}\cong \frac{\left|R\right|}{c_{\mathrm{max}}}\left({\theta}_{\mathrm{max}}-\frac{\theta_{\mathrm{max}}^3}{6}\right)\left(1+\frac{\theta_{\mathrm{max}}^2}{2}\right) $$

(11.43)

We have to be vigilant at this point to be sure we are evaluating Eq. (11.43) to the same level of approximation as we had in Eq. (11.42), which is correct to third order in θ_max. The product of the two binomials includes a term that is linear in θ_max as well as two terms that are third order.

$$ {T}_{axial}=\frac{\left|R\right|}{c_{\mathrm{max}}}\left({\theta}_{\mathrm{max}}-\frac{\theta_{\mathrm{max}}^3}{6}+\frac{\theta_{\mathrm{max}}^3}{2}\right)=\frac{\left|R\right|}{c_{\mathrm{max}}}\left({\theta}_{\mathrm{max}}+\frac{\theta_{\mathrm{max}}^3}{3}\right) $$

(11.44)

These calculations demonstrate that the (longer) curved path is traversed in less time than the shorter (axial) path. We can use Eq. (11.44) with Eqs. (11.40) and (11.42) to approximate the travel time difference, ΔT.

$$ \Delta T={T}_{axial}-{T}_{upper}\cong \frac{\left|R\right|}{c_{\mathrm{max}}}\left[\left({\theta}_{\mathrm{max}}+\frac{\theta_{\mathrm{max}}^3}{3}\right)-\left({\theta}_{\mathrm{max}}+\frac{\theta_{\mathrm{max}}^3}{6}\right)\right]=\frac{\left|R\right|}{c_{\mathrm{max}}}\frac{\theta_{\mathrm{max}}^3}{6} $$

(11.45)

4.4 Under Ice Propagation

Sound propagation under Arctic ice provides an interesting variation on the sound channel, since sound can be reflected specularly (i.e., θ_r = θ_i) from the ice sheet and the speed of sound under the ice increases monotonically with depth. The lack of solar heating at the surface causes the main thermocline, shown in Fig. 11.11, to be absent under Arctic ice. Typical ray paths under Arctic ice are shown in Fig. 11.17.

Fig. 11.17

Typical sound speed profile (right) and ray paths (left) under Arctic ice. Depth is in fathoms (1 fathom = 6 feet = 1.829 m), range is in kiloyards (1 kyd = 0.9144 km), and sound speed is in feet/second (1 fps = 30.48 cm/s) [14]

4.5 Sound Focusing

A final illustration of these refractive processes in constant sound speed gradients, at a much different scale than the global sound propagation in the deep sound channel, comes from the evolution of marine mammals. The Cuvier’s beaked whale (Ziphius cavirostris) is a member of the Ziphiidae family of toothed whales. It relies on echolocation, but in an aqueous environment, the excessive hydrodynamic drag and turbulence noise that would be produced by external “ears” (i.e., pinna), like those found on land mammals, is unacceptable. In this whale, the “ear” is located internally behind the jaw (in ), as shown in Fig. 11.18.

Fig. 11.18

The sound-sensing organ of the Cuvier’s beaked whale is shown by the small area in the drawing of the whale’s head at the right. At the left is a tomographic slice through the mandibular fat body which acts as a sound channel (i.e., lens) to focus sounds from the water to the whale’s sound-sensing organ. The sound speed of the mandibular fat body (in yellow at the right) shows the sound speed as a function of location ( corresponding to 1700 m/s and corresponding to 1300 m/s). Diagrams from Soldervilla, et al. [15]

Figure 11.19 demonstrates that this region acts like a lens. Parallel rays that enter the “channel” are focused to a single point (see Fig. 11.20). As we have done many times now, we can represent the distribution of sound speeds in the whale’s mandible using a piecewise-linear approximation. In this case, the sound speed gradient g = 60 m/s ÷ 0.06 m = 1000 s⁻¹; hence, |R| = c_o/g = 1.4 m. Using Snell’s law, cos θ_max = 0.957; hence, θ_max = 16.8°. This makes the “focal length” of the whale’s acoustic lens, d_focal = |R| sin θ_max = 40.5 cm, just about the extent of the mandibular fat bodies.

Fig. 11.19

A simplified approximation to the sound channel (i.e., lens) created by the mandibular fat body that assumes that the sound speed gradient is constant, with its minimum value (1340 m/s) along the axis, and its maximum value (1400 m/s) at the upper and lower extremes, 6.0 cm above and below the sound channel axis

Fig. 11.20

(Left) Ray paths for an ordinary lens. (Right) Ray paths for a graded-index (GRIN) lens. In both cases, parallel rays enter the lens and are focused to a single point

The strategy of using a propagation speed profile to act as a lens is very popular in fiber-optic telecommunication systems and fiber-optic sensors. As shown in Fig. 11.20, this application is intended to capture light from an LED or solid-state laser and focus that light into the core of single-mode optical fibers that typically have waveguide (core) diameters on the order of 10 microns or less. In sensor applications [16,17,18], it usually is used to spread the light from an optical fiber over the surface of some sensing element after which a second GRIN lens injects the modulated light back into the optical fiber.

The graded-index lens was patented by Nippon Sheet Glass in 1968 and given the trademark SELFOC^®. According to the accepted evolutionary timeline [19], marine mammals had produced the acoustical equivalent of GRIN lenses for at least 32 million years before the NSG patent was issued.

Talk Like an Acoustician

Cavitation effects	Fermat’s principle
Pressure reflection coefficient	Critical angle
Pressure transmission coefficient	Total internal reflection
Pressure release boundary	Angle of intromission
Impedance matching layer	Refraction
Antireflective coating	Piecewise-linear approximation
Trace wavelength	Ray tracing
Specific acoustic impedance	Grazing angle
Specular reflection	Sound channel
Diffuse reflection	SOFAR channel
Snell’s law

Exercises

• 1. Air-water interface. A 1.0 kHz planewave in water with pressure amplitude, $\left|\hat{\mathbf{p}}\right|=100\kern0.5em \mathrm{Pa}$, is normally incident on the air-water interface (i.e., the propagation direction is at a right angle to the air-water surface; hence, θ_i = 0°). You may assume the speed of sound in water is 1500 m/s and the speed of sound in air is 343 m/s. The density of water can be taken as 1000 kg/m³ and of air to be 1.19 kg/m³.

  • (a) Transmitted amplitude. What is the amplitude of the pressure that is transmitted into the air?

  • (b) Trace velocity. What is the “trace velocity” of the waves along the interface?

  • (c) Intensities. What is the intensity of the incident wave in the water and the transmitted wave in air?

  • (d) Transmission loss. Express the reduction in the intensity of the wave when it crosses the interface from the water into the air in decibels.

  • (e) Oblique incidence. If the angle of incidence is increased to 45°, what is the amplitude of the pressure transmitted into the air?

  • (f) Trace velocity. What is the “trace velocity” of the waves along the interface?

  • (g) Transmitted pressure. If the planewave originates in that air and makes an angle of 45° to the normal, what is the magnitude of the acoustic pressure of the sound in the water?

• 2. Matching layer. What is the density and sound speed required for a 1.0-cm-thick layer of material that produces complete transmission of sound from water into steel if the frequency of the sound wave is 20 kHz? Let ρ_water = 1000 kg/m³ and c_water = 1500 m/s, ρ_Fe = 7700 kg/m³, and c_Fe = 6100 m/s.

• 3. Alcohol on water. An acoustics experiment is performed in an aquarium partially filled with water (c = 1500 m/s and ρ_m = 1000 kg/m³). At the bottom of the aquarium, an ultrasonic transducer shoots a narrow beam of sound up to the surface of the water at an angle of 45°, and the sound beam reflects back down to the bottom of the aquarium where it hits a distance along the bottom that is 16.0 cm from the transducer. This situation is diagrammed in the upper portion of Fig. 11.21. You may assume the bottom of the aquarium is perfectly absorbing, so the beam does not bounce back up again and that the problem is entirely two dimensional.

  After making the required measurement, the experimenter comes back from lunch and finds that a layer of ethyl alcohol (c = 1150 m/s and ρ_m = 790 kg/m³) has carefully been poured on top of the water in the aquarium. (Good grief!) Now a second beam of sound comes down and hits the bottom at a distance of 22.0 cm from the ultrasonic transducer, as diagrammed in the lower portion of Fig. 11.21. How thick is the layer of alcohol on top of the water, assuming that the two liquids do not mix?

• 4. Oblique incidence at a water-sediment boundary. A sinusoidal planewave with effective pressure p_rms = 100 Pa is incident at 45° on a silt bottom with ρ_silt = 2000 kg/m³ and c_silt = 2000 m/s.

  • (a) Angle of refraction. What angle does the planewave make with respect to the normal once it enters the silt bottom?

  • (b) Transmitted pressure. What is the effective pressure of the wave in the silt bottom?

  • (c) Angle of intromission. At what angle does the planewave have to approach the silt from the water so that there is 100% transmission into the silt and no wave reflected back into the water?

• 5. Critical ray and skip distance in a “mixed layer.” The sound speed at the water’s surface is 1490 m/s and increases linearly to c_o = 1491.5 m/s with depth to the bottom of the mixed layer at D = 100 m. Below that depth, there is a constant negative sound speed gradient g₂ = −0.045 s⁻¹. For an underwater sound source (projector) at a depth, z_s = 60.0 m, determine the critical (grazing) angle, θ_o, for the limiting ray that remains trapped in the mixed layer and the “skip distance,” r_s, between the successive locations where the critical ray intersects the surface.

• 6. Arctic refraction. The water in the Arctic is nearly isothermal. Sound speed increases with depth because of the pressure effect. Assume the surface temperature is 0 °C, the salinity is 35 ppt, and the sound speed, in meters/second, as a function of depth, is given by the Eq. (11.46), where z is the depth below the surface expressed in meters [20].

  $$ c(z)=1449+0.016z $$

  (11.46)

The turning depth of the ray is 2.0 km where the sound speed is c(2 km) = 1481 m/s.

• (a) Initial grazing angle. What is the initial angle that ray makes with the water-ice interface? Report the angle which the ray makes with the water’s surface, not the angle with respect to the normal to the surface.

• (b) Return to the surface. What is the distance along the surface at which the ray returns to the ice-water interface?

  Fig. 11.21

  

  (Above) The sound beam of an ultrasonic transducer bounces of the free surface of a water-filled aquarium. (Below) the beam is displaced after a layer of alcohol is added to the surface