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Theory and Closed-Form Solutions for Three- and n-Layer Flat Refractive Geometry

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Abstract

We develop in this paper a new formulation for the calibration and pose estimation of three-layer flat-refractive geometry. We then extend it into the general n-layer case. We show that contrary to state-of-the-art theory, our new forms yield stable solutions under the presence of high levels of noise while improving the accuracy estimates by two orders of magnitude or more. Our closed-form expressions facilitate a direct and linear solution of the pose parameters and need no a priori knowledge. In developing our forms, we provide new insights into the nature of such systems, specifically showing that the effect of refraction on the system is captured by a single surface replacing all layers. By characterizing this surface form, we solve the n-layer problem using a fixed set of parameters, irrespective of the number of layers. Finally, we identify a new configuration in which the camera obtains central-perspective properties. Such configuration contrasts the axial nature of this system and can simplify subsequent processing. Our analyses demonstrate that reaching the levels of accuracies we record requires only simple means, e.g., a single image of a planar target. Hence, further to advancing the theory of flat-refractive geometry, we also provide a viable framework for its application.

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Notes

  1. For the general model in which the optical axis and the normal to the interface do not coincide, c.f. Agrawal et al. (2012).

  2. This fails to hold in the plate-refractive scenario, where \( \mu _n= \mu _0 \). This can be treated with little effort (Sect. 8.4).

  3. Alternative formulations, e.g., \( {\textbf {M}}={\textbf {M}}_x{\textbf {M}}_y\) are possible, but high correlations remain.

  4. In a standard imaging system \(\eta \) is bound to be small, but even if not, it would not reach \(90^{\circ }\) as such an angle suggests the interface is orthogonal to the optical axis, and no image will be formed.

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Acknowledgements

The authors would like to thank Prof. Tali Treibitz, head of the marine imaging lab at the University of Haifa, Israel, for providing the resources for the real-world experiments, and Aviad Avni for the support and assistance in their performance. Funding was provided in part by the Neubauer family foundation.

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Correspondence to Bashar Elnashef.

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Appendices

Surface of Refraction

For a better understanding of the nature of \( d_v \), the surface of refraction, we evaluate Eq. (21), which can be given by:

$$\begin{aligned} {{d}_{v}}={{d}_{0}}+\frac{\tan {{\theta }_{1}}-\tan {{\theta }_{2}}}{\tan {{\theta }_{0}}-\tan {{\theta }_{2}}}d_1 \end{aligned}$$
(48)

For clearer insight, we use first the paraxial ray assumption, where \( \tan \theta \approx \sin \theta \). Dividing the numerator and denominator by \(\sin \theta _0\) allows to express \(d_v\) in terms of the indices of refraction as follows:

$$\begin{aligned} {{d}_{v}}={{d}_{0}}+\frac{{{\xi }_{1}}-{{\xi }_{2}}}{1-{{\xi }_{2}}}d_1= {{d}_{0}}+\frac{{{\mu }_{2}}-{{\mu }_{1}}}{\left( {{\mu }_{2}}-1 \right) {{\mu }_{1}}}d_1 \end{aligned}$$
(49)

The size of \(d_v\) is primarily governed by \( d_0 \), and the second term modifies its distance. As the multiplier of \( d_1 \) is smaller than 1, the contribution of the second term to \( d_v \) is a fraction of the intermediate layer thickness. Sign-wise, the denominator is always positive (refraction indices are greater than 1), but the numerator can change signs, depending on the relations between the indices of refraction of the second and third layers. The typical case is when \(\mu _1 > \mu _2\), and here the numerator is negative. Hence, \(d_v\) should be located within layer #0 (Fig. 3). In the atypical case where \( \mu _1 < \mu _2 \), the numerator becomes positive, and \(d_v\) becomes bigger than \(d_0\), but smaller than \(d_0+d_1\). In such a case, it would be located within the first layer (Fig. 3).

Second-order analysis—a second-order approximation of \( \tan \theta _i \) would yield:

$$\begin{aligned} \tan {{\theta }_{i}}=\frac{r}{f}\left( 1+\frac{1}{2}{{\xi }_{i}}\left( {{\xi }_{i}}-1 \right) {{\left( \frac{r}{f} \right) }^{2}} \right) \end{aligned}$$
(50)

Substituting Eqs. (50) into (48) and with further derivations, we have,

$$\begin{aligned} {{d}_{v}}={{d}_{0}}+{{\left( \frac{r}{f} \right) }^{2}} \left( 1-\frac{\mu _{2}^{2}\left( 1-{{\mu }_{1}} \right) }{\mu _{1}^{2}\left( 1-{{\mu }_{2}} \right) } \right) d_1 \end{aligned}$$
(51)

Here again, the multiplier of \(d_1\) is smaller than 1, and its contribution is only a fraction of its dimension. Its sign is dictated by the relation between \(\mu _1\) and \(\mu _2\). For example, for the typical, air, glass, water setup, we have \(\mu _0 = 1,~\mu _1 = 1.5,~\text {and}~\mu _2 = 1.33\), and \( d_v \) would have the form, \( d_v=d_0-0.3\cdot \left( \nicefrac {r}{f} \right) ^2 d_1 \). Therefore, the contribution of \( d_1 \) would be as big as 30% of its actual thickness. The obtained expression allows us also to draw on the convexity of \(d_v\), which is directly obtained by taking the second-order derivative. Extension of these forms to n-layers follows directly, where the indices of refraction and their relation to \( \mu _n \) dictate the form of \( d_v \) in reference to \( d_0 \).

Pose Estimation and Calibration Under Non-orthogonality

The transformed ray in object-space (last layer) will be of the form:

$$\begin{aligned} {{\textbf {P}}}_R{\textbf {X}}_h= & {} \left[ {\textbf {M}}\textbf{R}\,|\,{\textbf {M}}{} {\textbf {t}} + k{\textbf {n}}\right] {\textbf {X}}_h= {\textbf {MP}}{} {\textbf {X}}_h + k^\eta {\textbf {n}}\\= & {} \left( \begin{matrix} {{\textbf {p}}}^{1T}{} {\textbf {X}}_h\cos \eta + {{\textbf {p}}}^{3T}{} {\textbf {X}}_h \sin \eta \\ {{\textbf {p}}}^{2T}{} {\textbf {X}}_h \\ - {{\textbf {p}}}^{1T}{} {\textbf {X}}_h\sin \eta + {{\textbf {p}}}^{3T}{} {\textbf {X}}_h\cos \eta + k^\eta \end{matrix}\right) \nonumber \end{aligned}$$
(52)

where \( {\textbf {p}}^{iT} \) are the rows of \( {\textbf {P}} \). Enforcing collinearity between both rays and eliminating \( \lambda ^\eta \) from the cross product, we arrive at the following constraint:

$$\begin{aligned} {{\textbf{x}}^{\eta }}\times {{\textbf{P}}_{R}}{{\textbf{X}}_{h}}=\left( \begin{matrix} y\left( -{{\textbf{p}}^{1T}}{{\textbf{X}}_{h}}\sin \eta +{{\textbf{p}}^{3T}}{{\textbf{X}}_{h}}\cos \eta +k^\eta \right) -{{f}^{\eta }}\left( 1+\tilde{k}^\eta \right) {{\textbf{p}}^{2T}}{{\textbf{X}}_{h}} \\ -{{x}^{\eta }}\left( -{{\textbf{p}}^{1T}}{{\textbf{X}}_{h}}\sin \eta +{{\textbf{p}}^{3T}}{{\textbf{X}}_{h}}\cos \eta +k^\eta \right) +{{f}^{\eta }}\left( 1+\tilde{k}^\eta \right) \left( {{\textbf{p}}^{1T}}{{\textbf{X}}_{h}}\cos \eta +{{\textbf{p}}^{3T}}{{\textbf{X}}_{h}}\sin \eta \right) \\ {{x}^{\eta }}\left( {{\textbf{p}}^{2T}}{{\textbf{X}}_{h}} \right) -y\left( {{\textbf{p}}^{1T}}{{\textbf{X}}_{h}}\cos \eta +{{\textbf{p}}^{3T}}{{\textbf{X}}_{h}}\sin \eta \right) \\ \end{matrix} \right) =\textbf{0} \end{aligned}$$
(53)

Developing the third component of Eq. (53) yields:

$$\begin{aligned}{} & {} \left( x\cos \eta + f\sin \eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad - y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\cos \eta + {\textbf {p}}^{3T}{} {\textbf {X}}_h \sin \eta \right) =0 \end{aligned}$$
(54)

As the value of \(\eta \) is bound to be small and will not approach \(90^{\circ }\), we divide Eq. (54) by \(\cos \eta \), and with further simplification,

$$\begin{aligned}{} & {} x\left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) - y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad = \tan \eta \left[ y\left( {\textbf {p}}^{3T}{} {\textbf {X}}_h\right) - f\left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) \right] \end{aligned}$$
(55)

By adding and subtracting \(y k^\eta \tan \eta \) to the right-hand side of Eq. (55), and adding and subtracting \(\tilde{k}^\eta f\) to f in the bracketed term on the right-hand side, we have,

$$\begin{aligned}{} & {} \left( x-\tilde{k}^\eta f\tan \eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) - y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h-{k^\eta }\tan \eta \right) \nonumber \\{} & {} \quad = \tan \eta \left[ y\left( {\textbf {p}}^{3T}{} {\textbf {X}}_h+k^\eta \right) - f\left( 1+\tilde{k}^\eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) \right] \nonumber \\ \end{aligned}$$
(56)

The bracketed term on the right-hand side is equivalent to \(v_y\), the image-space residual for the y ordinate factored by the depth to the object-space point and by \(\tan \eta \). Hence, it is negligible in magnitude and only adds a random error term to the algebraic residual of Eq. (56)

$$\begin{aligned}{} & {} \left( x-\tilde{k}^\eta f\tan \eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) - y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h-k^\eta \tan \eta \right) \nonumber \\{} & {} \quad =0+\tan \eta v_y \end{aligned}$$
(57)

where \(v_y\) is the (total) residual. The contribution of \(k^\eta \tan \eta \) to the second term in Eq. (57) is three orders of magnitude smaller than \(\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) \) and contributes a millimeter term to a few meters measure. Its removal leads to:

$$\begin{aligned} \left( x-\tilde{k}^\eta f\tan \eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) - y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) =0 \end{aligned}$$
(58)

Developing the first component of Eq. (52) and dividing by \(\cos \eta \), yields:

$$\begin{aligned}{} & {} y\left( -{\textbf {p}}^{1T}{} {\textbf {X}}_h\tan \eta + {\textbf {p}}^{3T}{} {\textbf {X}}_h + k^\eta \right) \nonumber \\{} & {} \quad - \left( f- x\tan \eta \right) \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) =0 \end{aligned}$$
(59)

thus,

$$\begin{aligned}{} & {} y \left( {\textbf {p}}^{3T}{} {\textbf {X}}_h + k^\eta \right) - f \left( 1 + \tilde{k}^\eta -\frac{ x\tilde{k}^\eta \tan \eta }{f}\right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad = \tan \eta \left[ \left( x - f \tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) -y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) \right] \nonumber \\ \end{aligned}$$
(60)

The term in square brackets on the right-hand side of Eq. (60) is the left-hand-side of Eq. (57). Substituting its right-hand-side-equivalent in Eqs. (57) into  (60) transforms Eq. (60) to the form:

$$\begin{aligned}{} & {} y \left( {\textbf {p}}^{3T}{} {\textbf {X}}_h+k^\eta \right) - f \left( 1 + \tilde{k}^\eta -\frac{x\tilde{k}^\eta \tan \eta }{f}\right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad =\tan ^2\eta \left( {\textbf {p}}^{3T}{} {\textbf {X}}_h\right) v_y \end{aligned}$$
(61)

The bracketed term on the right-hand side of Eq. (61) is equivalent to the in-air residual of the y image plane ordinate, \(v_y\), factored by \(\left( {\textbf {p}}^{3T}{} {\textbf {X}}_h\right) \) and by \(\tan ^2 \eta \). Being negligible in size (e.g., for \(\eta =1^{\circ }\), \(\tan ^2 \eta =3e^{-4}~rad\)) we can write,

$$\begin{aligned} y = f \left( 1 + \tilde{k}^\eta -\frac{x\tilde{k}^\eta \tan \eta }{f}\right) \frac{{\textbf {p}}^{2T}{} {\textbf {X}}_h}{{\textbf {p}}^{3T}{} {\textbf {X}}_h + k^\eta } \end{aligned}$$
(62)

Finally, we develop the second component of Eq. (53) and divide it by \(\cos \eta \), yielding:

$$\begin{aligned}{} & {} \left( x + f\tan \eta \right) \left( -{\textbf {p}}^{1T}{} {\textbf {X}}_h\tan \eta + {\textbf {p}}^{3T}{} {\textbf {X}}_h + k^\eta \right) \nonumber \\{} & {} \quad - \left( f- x\tan \eta \right) \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{1T}{} {\textbf {X}}_h + {\textbf {p}}^{3T}{} {\textbf {X}}_h \tan \eta \right) =0\nonumber \\ \end{aligned}$$
(63)

further simplification of Eq. (63) yields:

$$\begin{aligned}{} & {} \left( x + f\tan \eta \right) \left( {\textbf {p}}^{3T}{} {\textbf {X}}_h + k^\eta \right) -\tan \eta \left( x + f\tan \eta \right) \left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad -f \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) -f \tan \eta \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{3T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad +x \tan \eta \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) +x\tan ^2\eta \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{3T}{} {\textbf {X}}_h\right) \nonumber \\{} & {} \quad =0 \end{aligned}$$
(64)

and

$$\begin{aligned}{} & {} \left( {{x}}+f\tan \eta - f\tan \eta \left( 1+{{{\tilde{k}^\eta }}} \right) \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta \right) \nonumber \\{} & {} \quad -x\tan \eta \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) -f\tan ^2\eta \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \nonumber \\{} & {} \quad -f\left( 1+{{{\tilde{k}^\eta }}} \right) \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \nonumber \\{} & {} \quad +x\tan \eta \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{1T}{\textbf{X}_h} \right) \nonumber \\{} & {} \quad -x\tan ^2\eta \left( 1 + \tilde{k}^\eta \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} \right) =0 \end{aligned}$$
(65)

with further simplification Eq. (65) becomes:

$$\begin{aligned}{} & {} \left( {{x}}-f\tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta \right) -f\left( 1+{{{\tilde{k}^\eta }}} \right) \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \nonumber \\{} & {} \quad +x\tilde{k}^\eta \tan \eta \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) -\tan ^2\eta \left[ -{{x}}\left( {\textbf {p}}^{3T}{{\textbf{X}_h}} \right) +f\left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \right] =0\nonumber \\ \end{aligned}$$
(66)

thus,

$$\begin{aligned}{} & {} \left( {{x}}-f\tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta \right) \nonumber \\{} & {} \quad -f\left( 1+{{{\tilde{k}^\eta }}}-\frac{x\tilde{k}^\eta \tan \eta }{f} \right) \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \nonumber \\{} & {} \quad -\tan ^2\eta \left[ -{{x}}\left( {\textbf {p}}^{3T}{{\textbf{X}_h}} \right) +f\left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \right] =0 \end{aligned}$$
(67)

and,

$$\begin{aligned}{} & {} \left( {{x}}-f\tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta \right) \nonumber \\{} & {} \quad -f\left( 1+{{{\tilde{k}^\eta }}}-\frac{x\tilde{k}^\eta \tan \eta }{f} \right) \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \nonumber \\{} & {} \quad =\tan ^2\eta \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} \right) \left[ {{x}} -f\frac{{\textbf {p}}^{1T}{{\textbf{X}_h}}}{{\textbf {p}}^{3T}{{\textbf{X}_h}}} \right] \end{aligned}$$
(68)

By adding and subtracting \(\tilde{k}^\eta f\) to f in the bracketed term on the right-hand side, we have,

$$\begin{aligned}{} & {} \left( {{x}}-f\tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta \right) \nonumber \\{} & {} \quad -f\left( 1+{{{\tilde{k}^\eta }}}-\frac{x\tilde{k}^\eta \tan \eta }{f}+ \frac{ \tilde{k}^\eta \tan ^2\eta }{f}\right) \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) \nonumber \\{} & {} \quad =\tan ^2\eta \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} \right) \left[ {{x}}-f\left( 1+\tilde{k}^\eta \right) \frac{{\textbf {p}}^{1T} {{\textbf{X}_h}}}{{\textbf {p}}^{3T}{{\textbf{X}_h}}} \right] \end{aligned}$$
(69)

hence,

$$\begin{aligned}{} & {} \left( {{x}}- f\tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta \right) \nonumber \\{} & {} \quad -f\left( 1+\tilde{k}^\eta -\frac{x\tilde{k}^\eta \tan \eta }{f} + \frac{\tilde{k}^\eta \tan ^2\eta }{f}\right) \nonumber \\{} & {} \quad \left( {\textbf {p}}^{1T}{{\textbf{X}_h}} \right) =\tan ^2\eta \left( {\textbf {p}}^{3T}{{\textbf{X}_h}} \right) {{v}_{x}}+0 \end{aligned}$$
(70)

Finally, we arrive at:

$$\begin{aligned}{} & {} {{x}}- f\tilde{k}^\eta \tan \eta \nonumber \\{} & {} \quad = f\left( 1+{{{\tilde{k}^\eta }}} -\frac{x \tilde{k}^\eta \tan \eta }{f}+ \frac{\tilde{k}^\eta \tan ^2\eta }{f}\right) \frac{{\textbf {p}}^{1T}{{\textbf{X}_h}}}{{\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta }\nonumber \\ \end{aligned}$$
(71)

The first two equations represent the varifocal model, and the third is the refraction invariant model. All three an be rewritten more explicitly as:

$$\begin{aligned} x-f\tilde{k}^\eta \tan \eta= & {} f\left( 1+{{{\tilde{k}^\eta }}}-\frac{x \tilde{k}^\eta \tan \eta }{f}\right) \frac{{\textbf {p}}^{1T}{{\textbf{X}_h}}}{{\textbf {p}}^{3T}{{\textbf{X}_h}} + k^\eta }\nonumber \\ \end{aligned}$$
(72)
$$\begin{aligned} y= & {} f \left( 1 + \tilde{k}^\eta -\frac{x \tilde{k}^\eta \tan \eta }{f}\right) \frac{{\textbf {p}}^{2T}{} {\textbf {X}}_h}{{\textbf {p}}^{3T}{} {\textbf {X}}_h + k^\eta }\nonumber \\ \end{aligned}$$
(73)
$$\begin{aligned} 0= & {} \left( x-f \tilde{k}^\eta \tan \eta \right) \left( {\textbf {p}}^{2T}{} {\textbf {X}}_h\right) - y\left( {\textbf {p}}^{1T}{} {\textbf {X}}_h\right) \nonumber \\ \end{aligned}$$
(74)

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Elnashef, B., Filin, S. Theory and Closed-Form Solutions for Three- and n-Layer Flat Refractive Geometry. Int J Comput Vis 131, 877–898 (2023). https://doi.org/10.1007/s11263-022-01729-y

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