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13.6. Reflection of a point source from a plane

Here we consider what happens when a point source is placed near an infinite reflective planar boundary. Essentially, a hard reflecting surface is the acoustic equivalent to a mirror in optics whereby each reflecting element on its surface acts as a light source. A mirror can be regarded as a perfect hologram because it produces an intensity that varies with direction in accordance with the law of reflection. That is, the angle of reflection is equal to the angle of incidence. Hence, when you walk past a mirror, the view changes, whereas when you walk past a picture or video screen, it does not. This is because the latter provides only intensity information and no directional information. The directional information comes from the phase of each point source on the surface. For analytical purposes, it is often convenient to replace the reflecting plane with a transparent plane of symmetry, which has a symmetrically identical source behind it as shown in Fig. 13.2.

This property of symmetry has already been applied in previous examples such as the domes in Sections 12.9 and 12.10. The source and its image both have the same perpendicular distance d from the plane. With the image source present, the pressure fields with and without the reflecting plane can be shown to be identical if we consider that in both cases the field is symmetrical to either side of the plane. Therefore, the pressures must be equal on opposite faces of the plane, in which case the pressure gradient in the plane must be zero. Because it takes a pressure gradient to generate a particle velocity, this also satisfies the boundary condition of zero particle velocity normal to the plane. Of course, there is no physical image source, so this model is only valid on the source side of the surface.

Figure 13.2 Reflection of a point source from (a) a plane and (b) equivalent source and image.

The principle can be expressed by the superposition of fields as follows. If the point source is located at a distance z ₀ from an infinite screen at z = 0, there will be an extra field superimposed on the original because of a virtual source behind the screen. Using Eqs. (13.2), (13.31), (13.48), and (13.49) for a point source, the field

{\tilde{p}}_{S} (w, ϕ, z)

${\tilde{p}}_{S} (w, ϕ, z)$

produced by the source in cylindrical coordinates is given by

{\tilde{p}}_{S} (w, ϕ, z) = j k ρ_{0} c {\tilde{U}}_{S} \frac{e^{- j k R_{S}}}{4 π R_{S}},

${\tilde{p}}_{S} (w, ϕ, z) = j k ρ_{0} c {\tilde{U}}_{S} \frac{e^{- j k R_{S}}}{4 π R_{S}},$

(13.87)

where

R_{S} = \sqrt{w^{2} + w_{0}^{2} - 2 w w_{0} \cos (ϕ - ϕ_{0}) + {(z + z_{0})}^{2}} .

$R_{S} = \sqrt{w^{2} + w_{0}^{2} - 2 w w_{0} \cos (ϕ - ϕ_{0}) + {(z + z_{0})}^{2}} .$

(13.88)

The field

{\tilde{p}}_{I} (w, ϕ, z)

${\tilde{p}}_{I} (w, ϕ, z)$

produced by the image is then

{\tilde{p}}_{I} (w, ϕ, z) = j k ρ_{0} c {\tilde{U}}_{S} \frac{e^{- j k R_{I}}}{4 π R_{I}},

${\tilde{p}}_{I} (w, ϕ, z) = j k ρ_{0} c {\tilde{U}}_{S} \frac{e^{- j k R_{I}}}{4 π R_{I}},$

(13.89)

where

R_{I} = \sqrt{w^{2} + w_{0}^{2} - 2 w w_{0} \cos (ϕ - ϕ_{0}) + {(z - z_{0})}^{2}},

$R_{I} = \sqrt{w^{2} + w_{0}^{2} - 2 w w_{0} \cos (ϕ - ϕ_{0}) + {(z - z_{0})}^{2}},$

(13.90)

which produces a resultant field

\begin{matrix} \tilde{p} (w, ϕ, z) = {\tilde{p}}_{S} (w, ϕ, z) + {\tilde{p}}_{I} (w, ϕ, z) \\ = j k ρ_{0} c {\tilde{U}}_{S} (\frac{e^{- j k R_{s}}}{4 π R_{S}} + \frac{e^{- i k R_{I}}}{4 π R_{I}}) . \end{matrix}

$\begin{matrix} \tilde{p} (w, ϕ, z) = {\tilde{p}}_{S} (w, ϕ, z) + {\tilde{p}}_{I} (w, ϕ, z) \\ = j k ρ_{0} c {\tilde{U}}_{S} (\frac{e^{- j k R_{s}}}{4 π R_{S}} + \frac{e^{- i k R_{I}}}{4 π R_{I}}) . \end{matrix}$

(13.91)

Let us now recast this equation in the form

\tilde{p} (w, ϕ, z) = j k ρ_{0} c {\tilde{U}}_{S} G (w, ϕ, z | w_{0}, ϕ_{0}, z_{0}),

$\tilde{p} (w, ϕ, z) = j k ρ_{0} c {\tilde{U}}_{S} G (w, ϕ, z | w_{0}, ϕ_{0}, z_{0}),$

(13.92)

where

G (w, ϕ, z | w_{0}, ϕ_{0}, z_{0}) = \frac{e^{- j k R_{S}}}{4 π R_{S}} + \frac{e^{- i k R_{I}}}{4 π R_{I}}

$G (w, ϕ, z | w_{0}, ϕ_{0}, z_{0}) = \frac{e^{- j k R_{S}}}{4 π R_{S}} + \frac{e^{- i k R_{I}}}{4 π R_{I}}$

(13.93)

is a bounded Green's function. Notice that we use the upper case G. An interesting feature of this bounded Green's function is that its normal derivative with respect to the plane (i.e., with respect to z) is zero. Now suppose that part of the plane is in motion and radiating sound. Points on the plane can be represented by G if we let z ₀ → 0 so that the source and its image coalesce. Hence

G (w, ϕ, z | w_{0}, ϕ_{0}, 0) = \frac{e^{- j k R}}{2 π R} = 2 g (w, ϕ, z | w_{0}, ϕ_{0}, 0),

$G (w, ϕ, z | w_{0}, ϕ_{0}, 0) = \frac{e^{- j k R}}{2 π R} = 2 g (w, ϕ, z | w_{0}, ϕ_{0}, 0),$

(13.94)

where

R = \sqrt{w^{2} + w_{0}^{2} - 2 w w_{0} \cos (ϕ - ϕ_{0}) + z^{2}} .

$R = \sqrt{w^{2} + w_{0}^{2} - 2 w w_{0} \cos (ϕ - ϕ_{0}) + z^{2}} .$

(13.95)

We can use this Green's function in the monopole Rayleigh integral to represent a planar source in an infinite baffle. Because the normal derivative of G is zero, the dipole Rayleigh integral vanishes, and the point sources on the surface become monopole point sources of double strength.

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13.6. Reflection of a point source from a plane

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Acoustics