# Goldfish perspective

What does the world look like from the Goldfish point of view, from inside a spherical aquarium? If our eyes were inside, would we be able to see the straight lines, focus on different objects and what would a light point source looks like?

(elaborate on the curvatures, with or without the flat water above)

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Light point source looks like only point source. – Inquisitive Dec 29 '12 at 9:48

Compound Fish Bowl/Fish Eye Lens System

The bowl would make a compound lens system with the fish's eye. First, I will assume that we have the following indices of refraction : $$n_{\text{air}}\\ n_{\text{glass}}\\ n_{\text{water}}.$$ Usually $n_{\text{air}} \approx 1$, $n_{\text{glass}} \approx 1.5$, and $n_{\text{water}} \approx 1.33$, but I'll work the general problem here. I will assume first order geometric optics in the paraxial approximation for simplicity. If you wanted a more robust answer this system could be modeled and analyzed in an optical ray-tracing program like Zemax or Code5.

We have the following variables besides the indices of refraction shown in the figure below: $t_1$ is the distance from the object to the first spherical glass interface with radius of curvature $r_1$, $t_2$ is the distance from the glass interface to the spherical glass/water interface with radius of curvature $r_2$, and finally $t_3$ is the distance from the the glass/water interface to the front principal plane of the eye of the fish. Note that the green color denotes the glass and the blue color denotes the water.

The following Gaussian optics equations will be used: $$\phi = \frac{n_2-n_1}{r}\\ \phi_{\text{tot}} = \phi_1 + \phi_2 - \tau\phi_1\phi_2\\ f_R' = n_{\text{water}}f_e$$ where $\phi$ is the power of a single surface, $\phi_{\text{tot}}$ is the power of two combined surfaces separated by the reduced distance $\tau = \frac{t}{n}$, and $f_e = \frac{1}{\phi}$ is the effective focal length.

The first step is to calculate the power of the air/glass interface and the power of the glass/water interface: $$\phi_1=\frac{n_{\text{glass}}-n_{\text{air}}}{r_1}\\ \phi_2=\frac{n_{\text{water}}-n_{\text{glass}}}{r_2}$$

This implies that: $$\phi_{\text{3}}=\frac{n_{\text{glass}}-n_{\text{air}}}{r_1}+\frac{n_{\text{water}}-n_{\text{glass}}}{r_2}-\frac{t_2}{n_{\text{glass}}}\cdot\frac{(n_{\text{glass}}-n_{\text{air}})(n_{\text{water}}-n_{\text{glass}})}{r_1 r_2}\\ =\frac{n_{\text{glass}}-1}{r_1}+\frac{n_{\text{water}}-n_{\text{glass}}}{r_2}-\frac{r_1-r_2}{n_{\text{glass}}}\cdot\frac{(n_{\text{glass}}-1)(n_{\text{water}}-n_{\text{glass}})}{r_1 r_2}\\ =\frac{n_{\text{water}}n_{\text{glass}}-n_{\text{water}}}{n_{\text{glass}}r_1}+\frac{n_{\text{water}}-n_{\text{glass}}}{n_{\text{glass}}r_2}\\ \approx \frac{0.443}{r_1}-\frac{0.113}{r_2}$$ since $t_2 = r_1-r_2$, with the last line using the numerical values of the indices of refraction specified above.

Now the fish eye will have power $\phi_{\text{fish}}$, so we can use the Gaussian equations once again: $$\phi_{\text{tot}} = \phi_3 + \phi_{\text{fish}} - \frac{t_3}{n_{\text{water}}}\phi_3\phi_{\text{fish}}$$ Plugging in $\phi_3$ yields: $$\phi_{\text{tot}} = \frac{n_{\text{water}}n_{\text{glass}}-n_{\text{water}}}{n_{\text{glass}}r_1}+\frac{n_{\text{water}}-n_{\text{glass}}}{n_{\text{glass}}r_2} + \phi_{\text{fish}} - \frac{t_3}{n_{\text{water}}}\left(\frac{n_{\text{water}}n_{\text{glass}}-n_{\text{water}}}{n_{\text{glass}}r_1}+\frac{n_{\text{water}}-n_{\text{glass}}}{n_{\text{glass}}r_2}\right)\phi_{\text{fish}}$$ Which can probably be simplified but I'm not going to spend the time to do so. If we plug in some numerical values, including the estimated indices of refraction listed above, we can get an answer: $$r_1=100mm\\ r_2=97mm\\ t_2=3mm\\ t_3=97mm$$ corresponding to a fish bowl 200mm (8in) in diameter, the fish is at the center, and the glass thickness is 3mm. I compute the following values: $$\phi_1=0.0050/mm\\ \phi_2=-0.0018/mm\\ \phi_3=0.0033/mm\\ \phi_{\text{tot}}=0.0033+0.7619\cdot\phi_{\text{fish}}$$ This implies the the effective focal length of the compound system (fish bowl + fish eye) will be $$f_{e,\text{tot}}=\frac{1}{0.0033+0.7619\cdot\phi_{\text{fish}}}mm$$

I found the average focal length for a goldfish eye in

Role of the lens and vitreous humor in the refractive properties of the eyes of three strains of goldfish from Seltner et al (1989)

to be about $3mm$. Then we obtain $$f_{e,\text{tot}}=\frac{1}{0.0033+\frac{0.7619}{3}}mm=3.89mm.$$

What if the fish was really close to the glass, say $t_3=10mm$? Then $$\phi_{\text{tot}}=0.0033+0.9752\cdot\phi_{\text{fish}}=0.3284/mm,$$ i.e., $$f_{e,\text{tot}}=3.05mm.$$ or if the fish is looking the other way, i.e., $t_3 = 187mm$ then $$f_{e,\text{tot}}=5.56mm.$$

The above is the most extreme case. What is the effect? Note that the above focal lengths are the air (or optical) equivalent lengths and not the actual focal lengths.

Effect of Changing the Focal Length

Most focusing systems (including cameras and eyes) are set up such that the focal length doesn't change a whole lot, and for far away objects not much adjustment is needed to attain focus. We will assume this here (i.e., that the focal plane or lens system, or crystalline lens changes such that the basic focal length remains unchanged to maintain focus). We shall assume that the fish can focus on something normally only $10mm$ away, this would set the range of $z'$ to be $3mm < z' < 4.3mm$.

We now need to calculate the offset of the combined system principal plane from the principal plane of the original fish eye using: $$\frac{d'}{n{\text{water}}} = -\frac{\phi_3}{\phi_{\text{tot}}}\frac{t_3}{n_{\text{water}}}\\ \implies d' = -\frac{\phi_3}{\phi_{\text{tot}}}t_3$$ we assume above that the rear principal plane of the fish eye system is actually in an index of refraction of water. When the fish is at the extreme $f_{e,\text{tot}}=5.56mm$, then $$d' = -3.39mm.$$ When the fish is in the center, then $$d' = -1.25mm.$$

The air equivalent original fish eye had $f_{e,\text{fish}}=3mm$, and the extreme part of the fish on one side of the bowl changed the effective focal length to $f_{e,\text{tot}}=5.56mm$. Let's assume that the fish is viewing an object at $1000mm$ away. We can use the lens equation to get the difference between the two focal lengths. $$\frac{1}{z'}=\frac{1}{z}+\frac{1}{3mm}\text{vs.}\frac{1}{z'}=\frac{1}{z}+\frac{1}{5.56mm}\\ \implies \frac{1}{z'}=-\frac{1}{1000mm}+\frac{1}{3mm}\text{vs.}\frac{1}{z'}=-\frac{1}{1000mm}+\frac{1}{5.56mm}\\ \implies z'=3.009mm\quad\text{vs.}\quad z'=5.59mm$$ which means the magnifications would be : $$m_{\text{fish}}=-0.003009\\ m_{\text{tot}}=-0.005591$$ The new effective focal length must take into account the shift in the principal plane, the focal plane would be located $5.59mm-3.39mm=2.2mm$. This would result in the fish seeing objects at 1000mm as blurry (defocused) outside the bowl, when the fish is near the edge of the bowl and looking through the other side of the bowl, because the fish cannot shift the retina to the 2.2mm position (infinity focus is at the 3mm position). The fish could begin to see things when the compound system has a $z'=3mm+3.39mm = 6.39mm$ which corresponds to $z \approx 43mm$, i.e., when objects are about $43mm$ from the front principal plane in air (I don't know if this would be inside or outside the bowl in physical space, likely inside, so it cannot actually happen).

What about when the goldfish is in the middle of the bowl, i.e., $f_{e,\text{tot}}=3.89mm$? Then something at a distance of $1000mm$ has $z'=3.91mm$, and the focal plane is at $2.66mm$ from the rear principal plane of the original fish eye system. The fish could begin to see things when the compound system has a $z'=3mm+1.25mm = 4.25mm$ which corresponds to $z \approx 45mm$, nearly the same as the extreme case.

In both cases it is likely the gold fish will see objects as blurry.

If the goldfish is only $20mm$ from the edge, then objects closer than about $122mm$ can be seen without being blurry. If he gets right against the edge then the effect is minimized and he can probably see quite far.

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There also could be TIR, but it would be at very wide angles if at all... – daaxix Dec 30 '12 at 1:04
I can't follow your deductions, so I won't dispute them. But there's something fishy about at least one of your results: from the very center of the fishbowl, the shortest light path to any point is straight to it. So I have trouble figuring out how it could affect in anyway what the goldfish sees, because light can only get there from outside the bowl in a straight line, as if there was no bowl at all... – Jaime Dec 30 '12 at 4:23
@Jaime, the shortest light path must take into account the index of refraction of the material. It is the shortest in "optical space" not physical space. In order for the shortest path to coincide with the center of the fishbowl the incident light would have to be in the form of a converging spherical wave, centered exactly on center of the fish bowl. You can think of a single point, somewhere outside of the fish bowl, generating an expanding spherical wave (equivalently a set of radial rays from the source). The ray connecting the point to the center of the fish bowl, will be the shortest – daaxix Dec 30 '12 at 7:14
but, the other rays will refract, since they won't be perpendicular to the fish bowl surface. – daaxix Dec 30 '12 at 7:14
We are ray tracing, diffraction doesn't play any role here, so there's no need to consider spherical waves. So let me insist: the shortest light path, taking into account refractive indices and all, from any point outside the bowl to its center, is a straight line. And so any light reaching the center of the bowl does so geometrically as if there was no bowl at all, and an eye located exactly there will see objects outside undistorted. My guess is that the paraxial approximation breaks down in this particular case, although I am not sure why. – Jaime Dec 30 '12 at 9:58

The only weird effect from inside the bowl will be total internal reflection from the glass-air interface. This is what causes the distinctive fisheye effect when looking at a water-air interface from underwater, where you see the full $2\pi$ steradians of the air half-space concentrated on a smaller disk on the surface. There is significant distortion in this case, but only for objects outside the water.

But having a spherical bowl would actually lessen, rather than increase, this effect, and a fish located at the exact center of the bowl would not see any of it, as it would be looking perpendicular to the glass-air interface in every direction.

EDIT

If the glass is sufficiently thin, it will have a negligible refraction effect, since the refraction angle of the ray entering the glass, and the incidence angle of that same ray leaving the glass will be virtually the same, and their effects cancel out when applying Snell's law twice. So optically this is the same as being inside a sphere of water floating in space. Lets say this sphere has radius $r$, and we are looking at the water air interface from a distance $x$.

A point on the sphere that we see $\varphi$ degrees away from the normal, would be seen at an angle $\sin \theta = \frac{r}{x} \tan \varphi$ from the center of the sphere, and the angle between a ray with angle $\varphi$ and the normal to the interface will be $\theta_3 = \varphi - \theta$. The same ray in air will have angle with the normal calculated from Snell's law, $\sin \theta_1 = \frac{n_3}{n_1} \sin \theta_3$.

If you do the paraxial approximation, and take first order approximations for all trigonometric functions, you eventually get that, when looking with a small angle $\varphi$ from the normal, you are actually seeing light coming from an angle $\varphi'$ that can be calculated as:

$$\varphi' = \varphi (n - \frac{x}{r} n + \frac{x}{r})$$

where $n=\frac{n_3}{n_1} = 1.33$ for water.

So if you are really close to the glass, you will see things outside smaller than they are, by a factor of $\frac{1}{n} = 0.75$, if you are at the center of the bowl you will see everything undistorted, and if you are looking across the whole bowl, you will see things magnified by a factor of $\frac{1}{2-n} = 1.5$.

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Jaime this is incorrect, see my answer below. – daaxix Dec 30 '12 at 0:14
There will be a lens effect, I'm working out what the fish would actually see now...I have to compute the principal planes of the combined system... – daaxix Dec 30 '12 at 0:56
Jaime I think you are conflating what the fish will see being inside and what we will see outside. Your analysis works for a point source at the center of the sphere, such that the rays are exiting. Then an imaging system will will, of course, see the point located at the center of the bowl (no change of location). But a point source emitting from outside the bowl has many rays which intersect at non-normal angles of incidence! – daaxix Dec 30 '12 at 21:35
@daaxix As far as the geometry goes, direction of propagation makes no difference, and a ray going out of the fish bowl will follow the exact same path as a ray going in. That's why ray tracing traces rays from the location of the observer through an image plane, to light sources, rather than the other way around, which would have most of the rays never reaching the observer. Come up with one single example of a ray reaching the center of a spherical bowl that intersected the sphere at a non-normal angle, and I'll quit arguing. But I am pretty sure that example does not exist... – Jaime Dec 31 '12 at 1:26
I agree, but you have to think about an integral of an uncountable number of point sources, not just a point source at the center. The fish eye is not a delta function, it has an area. The fish eye will be imaging rays not going through the center of the bowl. – daaxix Dec 31 '12 at 5:36