## Abstract

In a previous paper, the special visual appearance of art glazes was explained using the auxiliary function method (AFM) for solving the radiative transfer equation. Glazes are made of low concentrated colored scattering centers embedded in a transparent medium and the artist modulates the color by varying the number of glaze layers. A simple model of glazes and the new solving method have both been validated by comparison between flux measurements and modeling. The color of art glazes is analyzed here, and the study shows a spectacular maximum of saturation (purity) of the color that is never reached, to the best of our knowledge, with other techniques, such as pigment mixtures. This phenomenon is explained once more using the AFM that allows separation of the different contributions to the scattered fluxes. It is then shown that, on the one hand, single scattering never induces a maximum of saturation. On the other hand, multiple scattering has a typical increasing and decreasing behavior with an increasing number of glaze layers and thus participates to the maximum of saturation, just as the scattering by the diffuse base layer. A comparison between glazes and pigment mixtures, where the proportion of colored pigments with white pigments varies instead of the number of layers, shows that this maximum of saturation is much smaller with the second technique. To the best of our knowledge, we present a new development of the AFM that allows separation of the different origins of light scattering. We also show that it is possible to determine the optical properties of the scattering centers and of the base layer to create the required visual effect of a scattering medium.

© 2006 Optical Society of America

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### Equations (23)

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(1)
$$\frac{\mathrm{d}{w}^{\pm}\left(\mu ,\text{\hspace{0.17em}}\tau \right)}{\mathrm{d}\tau}=\mp \frac{{w}^{\pm}\left(\mu ,\text{\hspace{0.17em}}\tau \right)}{\left|\mu \right|}\pm \frac{q}{4\pi}{\displaystyle {\int}_{{\mu}_{1}=0}^{1}\left[{w}^{+}\left({\mu}_{1},\text{\hspace{0.17em}}\tau \right)+\text{\hspace{0.17em}}{w}^{-}\left({\mu}_{1},\text{\hspace{0.17em}}\tau \right)\right]\text{\hspace{0.17em}}\frac{2\pi \mathrm{d}{\mu}_{1}}{{\mu}_{1}}}\pm \frac{q}{4\pi}\frac{{W}^{+}\left(\tau \right)}{{\mu}_{o}},$$
(2)
$$\frac{\mathrm{d}{w}^{\pm}\left(\mu ,\text{\hspace{0.17em}}\tau \right)}{\mathrm{d}\tau}=\mp \frac{{w}^{\pm}\left(\mu ,\text{\hspace{0.17em}}\tau \right)}{\left|\mu \right|}\pm \frac{q}{2}\text{\hspace{0.17em}}\left[f\left(\tau \right)+g\left(\tau \right)\right]$$
(3)
$${w}^{-}\left(\mu ,\text{\hspace{0.17em}}0\right)=\frac{\rho}{\pi}\text{\hspace{0.17em}}A\mu \text{\hspace{0.17em}}\mathrm{exp}\left(-h/\mu \right)+\frac{q}{2}{\displaystyle \text{\hspace{0.17em}}{\int}_{s=0}^{h}t\left(s\right)\times \text{\hspace{0.17em}}\mathrm{exp}\left(-s/\mu \right)\mathrm{d}s},$$
(4)
$$A=\frac{\text{\hspace{0.17em}}T\left({\mu}_{i}\right)\text{\hspace{0.17em}}\mathrm{exp}\left({-h/\mu}_{o}\right)+\pi q\text{\hspace{0.17em}}{\displaystyle {\int}_{s=0}^{h}t\left(s\right)B\left(s\right)\mathrm{d}s}}{1-\rho K},$$
(5)
$$B\left(\tau \right)={\displaystyle {\int}_{\mu =0}^{1}\left[\mathrm{exp}\mathbf{\left(}\left(\tau -h\right)/\mu \mathbf{\right)}+R\left(\mu \right)\mathrm{exp}\mathbf{\left(}-\left(h+\tau \right)/\mu \mathbf{\right)}\right]\text{\hspace{0.17em}d}\mu},$$
(6)
$$K\text{\hspace{0.17em}}=\text{\hspace{0.17em}}2{\displaystyle {\int}_{\mu =0}^{1}R\left(\mu \right)\mathrm{exp}\left(-2h/\mu \right)\mu \mathrm{d}\mu}.$$
(7)
$${w}^{-}\left(\mu ,\text{\hspace{0.17em}}0\right)={w}_{B}^{\text{\hspace{0.17em} \hspace{0.17em}}-}\left(\mu ,\text{\hspace{0.17em}}0\right)+{w}_{\mathrm{M}\mathrm{S}}^{\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}-}\left(\mu ,\text{\hspace{0.17em}}0\right)+{w}_{\mathrm{S}\mathrm{S}}^{\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}-}\left(\mu ,\text{\hspace{0.17em}}0\right)\mathrm{.}$$
(8)
$$C\left(x\right)={\displaystyle {\int}_{\mu =0}^{1}R\left(\mu \right)\mathrm{exp}-\left(x/\mu \right)\mathrm{d}\mu ,}$$
(9)
$$D\left(x\right)={\displaystyle {\int}_{\mu =0}^{1}R\left(\mu \right)\mathrm{exp}-\left(x/\mu \right)\mu \mathrm{d}\mu ,}$$
(10)
$$E\left(x\right)={\displaystyle {\int}_{\mu =0}^{1}\mathrm{exp}-\left(x/\mu \right)\mathrm{d}\mu \mathrm{.}}$$
(11)
$${w}_{B}^{\text{\hspace{0.17em} \hspace{0.17em}}-}\left(\mu ,\text{\hspace{0.17em}}0\right)=\Lambda \text{\hspace{0.17em}}\frac{T\left({\mu}_{i}\right)\mathrm{exp}\left({-h/\mu}_{o}\right)}{2\pi}\text{\hspace{0.17em}}\mu \text{\hspace{0.17em}}\mathrm{exp}\left(-h/\mu \right)$$
(12)
$$\Lambda =\frac{2\rho}{1-2\rho D\left(2h\right)}.$$
(13)
$${{w}_{\mathrm{MS}}}^{-}\left(\mu ,\text{\hspace{0.17em}}0\right)=\frac{q}{2}{\displaystyle \text{\hspace{0.17em}}{\int}_{0}^{h}\left[\mathrm{exp}-\left(s/\mu \right)+\Lambda \mu \text{\hspace{0.17em}}\left[C\left(h+\tau \right)+E\left(h-\tau \right)\right]\text{\hspace{0.17em}}\mathrm{exp}-\left(h/\mu \right)\right]\text{\hspace{0.17em}}f\left(s\right)\mathrm{d}s}.$$
(14)
$${{w}_{\mathrm{SS}}}^{-}\left(\mu ,\text{\hspace{0.17em}}0\right)=\frac{q}{2}{\displaystyle {\int}_{0}^{h}\left[\mathrm{exp}-\left(s/\mu \right)+\Lambda \mu \left[C\left(h+\tau \right)+E\left(h-\tau \right)\right]\text{\hspace{0.17em}}\mathrm{exp}-\left(h/\mu \right)\right]\text{\hspace{0.17em}}g\left(s\right)\mathrm{d}s}\mathrm{.}$$
(15)
$${k}_{m}=c{k}_{g}+\left(1-c\right){k}_{w},$$
(16)
$${s}_{m}=c{s}_{g}+\left(1-c\right){s}_{w},$$
(17)
$$\begin{array}{c}X=K{\displaystyle \text{\hspace{0.17em}}\int S\left(\lambda \right)\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}}\overline{x}\left(\lambda \right)R\left(\lambda \right)\mathrm{d}\lambda ,\end{array}$$
(18)
$$\begin{array}{c}Y=K{\displaystyle \text{\hspace{0.17em}}\int S\left(\lambda \right)\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}}\overline{y}\left(\lambda \right)R\left(\lambda \right)\mathrm{d}\lambda ,\end{array}$$
(19)
$$\begin{array}{c}Z=K{\displaystyle \text{\hspace{0.17em}}\int S\left(\lambda \right)\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}}\overline{z}\left(\lambda \right)R\left(\lambda \right)\mathrm{d}\lambda ,\end{array}$$
(20)
$$\begin{array}{c}K=\frac{100}{{\displaystyle \int S\left(\lambda \right)\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}}\overline{y}\left(\lambda \right)\mathrm{d}\lambda}.\end{array}$$
(21)
$$\begin{array}{l}L*=f\text{\hspace{0.17em}}\left(\frac{Y}{{Y}_{B}}\right),\\ a*=\frac{500}{116}\text{\hspace{0.17em}}\left[f\text{\hspace{0.17em}}\left(\frac{X}{{X}_{B}}\right)-f\text{\hspace{0.17em}}\left(\frac{Y}{{Y}_{B}}\right)\right],\\ b*=\frac{200}{116}\text{\hspace{0.17em}}\left[f\text{\hspace{0.17em}}\left(\frac{Y}{{Y}_{B}}\right)-f\text{\hspace{0.17em}}\left(\frac{Z}{{Z}_{B}}\right)\right],\end{array}$$
(22)
$$f\left(A\right)=116{A}^{1/3}-16\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em} if \hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}f\left(A\right)\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}8,$$
(23)
$$f\left(A\right)={\left(\frac{29}{3}\right)}^{3}A\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} if \hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}f\left(A\right)\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}8\text{\hspace{0.17em}}\left(\mathrm{Pauli}\text{\hspace{0.17em}}\mathrm{correction}\right)\mathrm{.}$$