Watching a goldfish swimming in a fishbowl

Even if we were to describe ‘two lost souls swimming in a fishbowl’ [1], we might not always choose to rely on the small-angles approximation in geometrical optics [2]. This is especially true for goldfish whose dimensions are not small with respect to the radius of the fishbowl. Fishbowls are readily available at pet stores, and small plastic goldfish can be found in toy stores. The goldfish can be attached to a wooden stick (as shown in figure 1(a)) to allow movement though the water, allowing easy adjustment of its position along the $x$ -axis (see figure 1(b)).

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This simple setup is well-suited for use in a lesson based on Inquiry-Based Learning, or more specifically, the Inquiry-Based Science Education (IBSE) approach [3], implemented through the BSCS 5E instructional model [4]. In particular, the Engage and Explore phases can be effectively supported using this apparatus. For the Explain phase, one could follow the traditional route of applying the small-angle approximation in geometrical optics, as commonly presented in textbooks [4, 5], or alternatively, extend the learning by introducing a more realistic approach based on trigonometry and analytic geometry (Extend phase)

In the present work the second approach is adopted, and both the position of the image ${x_I}$ and the transverse magnification $m$ are determined as functions of the actual position $x$ of the goldfish in the spherical fishbowl. The small-angles limit is recovered as a consistency check. It is found that the quantities ${x_I}$ and $m$ are monotonically increasing and decreasing functions of $x$, respectively. This behaviour can be confirmed through experiments using the simple apparatus illustrated in figures 1(a) and (b).

By schematizing the problem as in figure 2, the image of the goldfish (object) located at ${x_0}$, with height $h = {z_0}$, is seen to form at position ${x_{I}}$. The image is virtual and erect, with height ${z_{I}}$. Here we make the assumption that the fish is entirely positioned along the positive $z$-axis. Notice that the observer is placed, as in figure 2, on the $x$-axis at point $A$.

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To determine ${x_{I}}$ we consider the intersection of two rays: the prolongation of the refracted ray emerging from the bowl in $P$, which crosses the $x$-axis at point $A$, and the ray that passes undeflected through the centre of curvature $C$. For simplicity reasons, the bowl is assumed to be very thin, so that the interface between the whole fishbowl and air is given by a water-air effective interface. Therefore, the contribution due to the thin glass, whose index of refraction is about 1.5, is neglected. A more detailed analysis should include the presence of the glass of the bowl. However, by adding a thin spherical shell in the model, the qualitative results would not change, while the quantitative analysis would become more involved. No further approximation is made to solve the problem.

Given that the circumference shown in figure 2, cantered at point $C$, is described by the equation ${x^2} + {z^2} = {R^2}$, a ray coming from infinity and grazing the tip of the object located at ${x_0}$, with height $h = {z_0}$, intersects the bowl at point $P$, whose coordinate are $\left( {\sqrt {{R^2} - {\text{z}}_0^2} ,0,{\text{ }}{z_0}} \right)$. According to Snell’s law, this ray, which is incident at an angle ${\theta _1}$ with respect to the normal at point $P$, is refracted at an angle ${\theta _2}$, such that:

$$\begin{equation}{n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}\end{equation} \tag{ 1 }$$

where ${n_1}$ is the index of refraction of water, and ${n_2}$ is the index of refraction of air. In the following we shall take the ratio between the index of refraction of water and that of air equal to $4/3$. To provide a more general treatment of the problem, we postpone assigning specific values to the indices of refraction until a later stage. By considering triangle $CPA$, we find through elementary geometry that the angle $\alpha$ between the refracted ray and the horizontal is ${\theta _2} - {\theta _1}$. In this way, the equation of the extended line passing though points $A$ and $P$ is:

$$\begin{equation}z = {z_0} - \tan \left( {{\theta _2} - {\theta _1}} \right)\left[ {x - \sqrt {{R^2} - {\text{z}}_0^2} } \right].\end{equation} \tag{ 2 }$$

On the other hand, the line passing through point $C$ and the tip of the object is given by:

$$\begin{equation}z = \frac{{{z_0}}}{{{x_0}}}x.\end{equation} \tag{ 3 }$$

The intersection ${x_I}$ is found by equating the right-hand sides of equations (2) and (3), yielding:

$$\begin{equation}{x_I} = \frac{{1 + a\sqrt {1 - {{\left( {\frac{{{z_0}}}{R}} \right)}^2}} }}{{1 + a{\text{ }}\frac{{{x_0}}}{R}}}{x_0}\end{equation} \tag{ 4 }$$

where:

$$\begin{equation}a = \frac{{\tan \left( {{\theta _2} - {\theta _1}} \right)}}{{\left( {{z_0}/R} \right)}}.\end{equation} \tag{ 5 }$$

From figure 2 we observe that $\tan {\theta _1} = {z_0}/\sqrt {{R^2} - {\text{z}}_0^2}$ and $\sin {\theta _1} = \frac{{{z_0}}}{R}$. Furthermore, using equation (1) we obtain:

$$\begin{align}\tan {\theta _2} & = \frac{{\sin {\theta _2}}}{{\sqrt {1 - {{\sin }^2}{\theta _2}} }} = \frac{{\frac{{{n_1}}}{{{n_2}}}\sin {\theta _1}}}{{\sqrt {1 - {{\left( {\frac{{{n_1}}}{{{n_2}}}} \right)}^2}{{\sin }^2}{\theta _1}} }}\nonumber\\ & = \frac{{\frac{{{n_1}}}{{{n_2}}}{ }\frac{{{z_0}}}{R}}}{{\sqrt {1 - {{\left( {\frac{{{n_1}}}{{{n_2}}}} \right)}^2}{{\left( {\frac{{{z_0}}}{R}} \right)}^2}} }}.\end{align} \tag{ 6 }$$

Therefore, both $\tan {\theta _1}$ and $\tan {\theta _2}$ depend on the dimensionless ratios $\nu = \frac{{{n_1}}}{{{n_2}}}$ and $\rho = \frac{{{z_0}}}{R}$. Recalling basic trigonometric identities, the quantity $a = a\left( {\nu ,\rho } \right)$ can be expressed as:

$$\begin{equation}a\left( {\nu ,\rho } \right) = { }\frac{{\nu \sqrt {1 - {\rho ^2}} - \sqrt {1 - {\nu ^2}{\rho ^2}} }}{{\sqrt {\left( {1 - {\nu ^2}{\rho ^2}} \right)\left( {1 - {\rho ^2}} \right)} + \nu {\rho ^2}}}.\end{equation} \tag{ 7 }$$

Note that $a\left( {\nu ,\rho } \right)$ is always positive. Furthermore, while $\rho < 1$ (since the object is inside the spherical bowl), we must also require that $\nu \rho < 1$ to prevent total internal reflection. In fact, when $\frac{{{n_1}}}{{{n_2}}}\sin {\theta _1} = 1$ (for ${n_1} > {n_2})$, no refracted ray is observed for incidence angles ${\theta _1} > {\sin ^{ - 1}}\left( {\frac{{{n_2}}}{{{n_1}}}} \right)$. In such cases, the analysis must be carried out with much greater care. However, in what follows we assume that the condition $\nu \rho < 1$ is satisfied.

Returning to equation (14), the quantity ${x_0}$ can now be replaced by the general position $x$ of the object on the $x$ -axis, so that:

$$\begin{equation}{x_I} = \frac{{1 + a\left( {\nu ,\rho } \right)\sqrt {1 - {\rho ^2}} }}{{1 + a\left( {\nu ,\rho } \right)\frac{x}{R}}}x.\end{equation} \tag{ 8 }$$

Taking the derivative with respect to $x$ of the above expression, it can be shown that ${x_{I}}$ is a monotonously increasing function of $x$. The function ${x_{I}}/R$ (black full line) is plotted in figure 3 for $\nu = 4/3$ and for $\rho = 2/15$ as a function of the dimensionless variable $x/R$ for $- \sqrt {1 - {\rho ^2}} < x < + \sqrt {1 - {\rho ^2}}$. For comparison, the function $\frac{{{x_{I}}}}{R} = \frac{x}{R}$ is also plotted in the same range.

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From figure 3, we observe that, for $0 < x < + \sqrt {{R^2} - {\text{z}}_0^2}$ the goldfish appears closer to the observer than its actual position. Conversely, for $- \sqrt {{R^2} - {\text{z}}_0^2} < x < 0$ the image appears more distant than the object. At $x = 0$ the image position coincides with that of the object.

Referring again to figure 2, the transverse magnification $m$ can be found by setting $m = \frac{{{z_I}}}{{{z_0}}} = \frac{{{x_I}}}{x}$, so that:

$$\begin{equation}m\left( x \right) = \frac{{1 + a\left( {\nu ,\rho } \right)\sqrt {1 - {\rho ^2}} }}{{1 + a\left( {\nu ,\rho } \right){ }\frac{x}{R}}}.\end{equation} \tag{ 9 }$$

This function is monotonically decreasing with respect to the dimensionless position $x/R$. Therefore, as the goldfish moves away from the observer (i.e. as $x$ decreases) it appears increasingly larger. This is illustrated in figure 4, where the apparent size of the goldfish doubles when it reaches the farthest point from the observer.

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As a consistency check and to compare the small-angles approximation with the exact results, we begin by examining the limit for $\rho \to 0$ of the function $a\left( {\nu ,\rho } \right)$ in equation (7). In this limit we find that

$$\begin{equation}a\left( {\nu ,0} \right) = { }\nu - 1.\end{equation} \tag{ 10 }$$

This leads to the following predictions under the small-angles approximation are the following:

$$\begin{equation}{x_{I0}} = \frac{\nu }{{1 + \left( {{\text{ }}\nu - 1} \right)\frac{x}{R}}}x\end{equation} \tag{ 11a }$$

$$\begin{equation}{m_0}\left( x \right) = \frac{\nu }{{1 + \left( {{ }\nu - 1} \right)\frac{x}{R}}}.\end{equation} \tag{ 11b }$$

Let us now employ the standard equation for the concave spherical diopter:

$$\begin{equation}\frac{{{n_1}}}{p} + \frac{{{n_2}}}{q} = \frac{{{n_2} - {n_1}}}{{\left( { - R} \right)}}.\end{equation} \tag{ 12 }$$

By setting $p = R - x$, $q = {x_{I0}} - R$ and $\frac{{{n_1}}}{{{n_2}}} = \nu$, equation (12) becomes:

$$\begin{equation}\frac{\nu }{{R - x}} + \frac{1}{{{x_{I0}} - R}} = \frac{{\nu - 1}}{R}.\end{equation} \tag{ 13 }$$

Solving for ${x_{{I}0}}$, we retrieve the same result as in equation (11).

Now consider the transverse magnification given by ${m_0} = - \frac{{{n_1}q}}{{{n_2}p}} = \nu \frac{{R - {x_{I0}}}}{{R - x}}$. Substituting ${x_{{I}0}}$ form equation (11) in the latter expression, we obtain:

$$\begin{equation}{m_0}\left( x \right) = \nu \frac{{R - \frac{{\nu x}}{{1 + \left( {{ }\nu - 1} \right)\frac{x}{R}}}}}{{R - x}} = \frac{\nu }{{1 + \left( {{ }\nu - 1} \right)\frac{x}{R}}}\end{equation} \tag{ 14 }$$

which coincides exactly with the expression in equation (11).

Naturally, the difference between ${m_0}\left( x \right)$ and the exact value $m\left( x \right)$ from equation (9) becomes more significant when the size of the goldfish is not negligible with respect to the radius of the bowl. A direct comparison between the curves ${m_0}\left( x \right)$ and $m\left( x \right)$ is shown in figure 5 for $\nu = \frac{4}{3}$ (approximately the index of refraction of water) and for $\rho = 1/2$ (i.e. the height of the goldfish is half the radius of the bowl). The difference between the two curves is clearly noticeable in this case. Specifically, not only is the domain of the function $m\left( x \right)$ smaller than the interval $- 1 < x/R < + 1$, but the actual magnification $m\left( x \right)$ is also less than ${m_0}\left( x \right)$ at small distances from the observer and greater than ${m_0}\left( x \right)$ at larger distances. This underscores the importance of the exact approach in accurately capturing optical behaviour when the small-angles approximation breaks down.

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A realistic approach, based on trigonometry and analytic geometry rather than the small-angle approximation, is adopted to investigate the characteristic optical features of a goldfish in a fishbowl. The virtual, erect image of the goldfish, as seen by an external observer, forms in front of the actual position $x$ (measured from the centre of the bowl) when $x > 0$. Conversely, for $x < 0$, the image appears farther from the observer.

The transverse magnification is found to be a monotonically decreasing function of the variable $x$, indicating that the goldfish appears larger as it moves away from the observer. While these qualitative features are consistent with those predicted by the small-angle approximation, the exact analysis provides more accurate quantitative information justifying, in the author’s view, the additional computational effort.

This analysis may serve either as a critical review of the lens equation for spherical transparent surfaces under the small-angle approximation or as the foundation for an interdisciplinary, IBSE experiment, suitable for advanced high school or undergraduate physics students.

The author wishes to thank I. D’Acunto, V. Lamberti, and E. Benedetto for stimulating discussions.

All data that support the findings of this study are included within the article (and any supplementary files).