The dome knows its offset in units of itself

The port that is not there establishes the claim underwater photographers are sold, and finds it to be exactly true rather than nearly true. A spherical port centred on the camera’s entrance pupil presents every ray with a normal interface, so Snell’s law bends nothing, and the camera behind the glass is the camera it was in air. The departure at zero offset is $10^{-15}$ degrees over a fan of rays, which is arithmetic.

It also establishes the other half. Move the sphere’s centre off the pupil and the bending starts, at a rate that essay quotes per millimetre.

This essay asks what a reader can do with that. Given a picture taken through an unknown dome, what can be recovered about the dome?

The answer is a ratio and no more, and the reason is a scale invariance so complete that the two pictures below are drawn on top of each other.

Why the invariance is exact #

The argument is short and it is worth having in full, because a scale invariance that holds approximately and one that holds exactly are different claims and only the second closes the question.

Three things are in the problem: a sphere, a straight ray, and Snell’s law at the surface.

A sphere scaled about the pinhole is a sphere. A straight ray through the pinhole is unchanged by scaling — it is the same line. And Snell’s law relates two angles by two refractive indices, with no length in it anywhere.

So scaling the radius and the offset together is a similarity of the whole configuration, and a similarity maps rays to rays and angles to angles. The bent ray leaves in the same direction. Not nearly the same; the same.

Measured across factors of 2, 5 and 0.4, the worst disagreement over a fan from 2° to 60° is $8.9\times10^{-16}$ degrees, which is the last bit of a double.

And the half that stops it being a triviality #

A scale invariance stated on its own is not a finding. Every optical system has one; the question is always what the invariant quantity turns out to be.

So the pair. Take the same two domes — ten and twenty centimetres — and give them the same offset, six millimetres. Now they are different instruments, and the difference is 0.35° over the same fan.

That is the check that turns the first claim into a measurement. The picture is a function of the ratio, not of the offset and not of the radius; and the way to know it is a function of the ratio rather than of nothing is to change the ratio and watch the picture move.

What the bending actually looks like #

Before the consequences, the shape of the curve is worth a paragraph, because it is not what a reader would guess from “the dome bends the ray”.

At zero angle from the axis the bending is zero for any offset, because a ray along the axis meets the sphere at a point where the normal is the axis whatever the centre is. The bending grows from there, roughly linearly at first, and then faster as the ray meets the glass at a steeper obliquity.

So the departure is a radial effect, largest at the edge of the frame and absent at its centre. It looks, to a photographer, exactly like a lens’s own barrel or pincushion distortion, and that resemblance is the practical problem: it is a departure that a lens-distortion fit will happily absorb into its own coefficients.

Which puts a dome’s decentring in the same category as everything a lens fitted from straightness alone can and cannot separate. A fit with enough radial terms will make the pictures straight; what it will not do is tell the photographer whether the correction it found belongs to the lens, the dome, or the shim behind it.

What the ratio does to the picture, in the calibration’s own terms #

The bend has a closed form, and running it through to the image says precisely which calibration parameters a mis-centred dome hides in — which is a sharper statement than “a distortion fit will absorb it”.

A ray leaving the pupil at $\theta$ meets a sphere whose centre is $\delta$ away, and the sine rule gives its angle of incidence as $\sin i = (\delta/R)\sin\theta$. Refracting bends it by $i(1 - 1/n)$ for small $i$, so the bend is $(\delta/R)(1 - 1/n)\sin\theta$: proportional to the ratio, and to nothing else about the glass. Put the essay’s own pair into it — ratios of 0.06 and 0.03, over a fan reaching 60° — and the predicted separation is 0.37°, against the 0.35° the figure measures.

Now carry it to the image. The mark sits at radius $r = f\tan\theta$, so an angular change $\Delta$ moves it by $f\Delta/\cos^{2}\theta$, and dividing through:

$$\frac{\Delta r}{r} = \frac{\delta}{R}\left(1 - \frac{1}{n}\right)\frac{1}{\cos\theta} \approx c\left(1 + \frac{r^{2}}{2f^{2}}\right), \qquad c = \frac{\delta}{R}\left(1 - \frac{1}{n}\right).$$

The leading term is a magnification, not a distortion. A displacement proportional to $r$ is a uniform scaling of the picture, which a calibration reads as a focal length — so a dome six millimetres off a hundred-millimetre radius, in fresh water, reports a focal length 1.5% wrong and nothing else at first order.

The second term is a genuine pincushion of coefficient $c/2$. For that same dome, an apparent $k_{1}$ of about $+0.0075$: real, small, and exactly the size of the mild pincushion a longer lens carries anyway.

So the worry the section above raises is right and can be made exact. A calibration that frees the focal length and one radial coefficient absorbs a mis-centred dome completely, into two parameters that both have perfectly plausible values afterwards, and its residual says nothing. What it does not do is absorb it into $k_{1}$ alone — the profile is wrong for that, since Brown–Conrady’s lowest radial term is cubic and the dome’s is linear — so a fit that pins the focal length and frees only distortion will fail to fit, and the failure is the one diagnostic available.

That is the same discriminator a pane of glass turns out to have, and for the same reason: both are interfaces whose leading effect is a scaling rather than a bending, and a scaling is the one departure a distortion polynomial has no term for.

What a specification sheet has and a picture has not #

A dome is sold on two numbers. It is an eight-inch dome, or a six-inch one, and it is shimmed to sit some number of millimetres from the pupil. Those are the two numbers an owner has.

The picture has their quotient.

That is a genuinely awkward position to be in, and it is worth stating the practical consequence rather than leaving it as arithmetic. A photographer who measures the distortion in their own pictures and works backwards learns how far off centre the dome is as a fraction of its own radius. To turn that into millimetres they need the radius, and the picture does not contain it.

The radius can of course be measured with a ruler out of the water. That is the point: the length comes from the workshop rather than from the photograph, and it is imported into the measurement from outside.

The same shape of answer, three times in one row #

A mirror ball’s outline gives the ratio of its radius to its distance and refuses to separate them. A pane of glass gives the product of its thickness and a function of its index. A dome port gives its decentring over its radius.

Three instruments, three ratios, and in each case the missing length has to come from somewhere the instrument is not — a near room point, a wide fan of sightlines, a ruler in the workshop.

The exception is the caustic, and the reason it is an exception sorts the whole row. A caustic is a mark left on the furniture and is therefore a length in the room; every other instrument here measures an angle at the eye, and an angle is a ratio of two lengths with the ratio being the only survivor.

What the ratio does buy #

It would be easy to read the above as bad news and it is not, quite. The ratio is the quantity that governs the image, so it is the quantity worth knowing.

An eight-inch dome at six millimetres and a four-inch dome at three millimetres shoot identically. A photographer choosing between them for image quality is choosing between the same instrument twice, and the difference between them lies entirely elsewhere — in how close the virtual image sits, in how much water the dome displaces, in whether a lens will focus that near.

So the picture answers the question that governs the picture and refuses the question that does not. That is an unusually clean division of labour, and it is more common than it looks: an instrument’s own output is generally a function of the dimensionless groups its physics contains, and the dimensioned quantities survive only where a length has been supplied.

The one number a photographer is actually shopping for #

There is a length in the dome problem that a picture does carry, and it is not the radius. It is worth putting here because it is the number the equipment is chosen on, and it is a different number from the one this essay has been refusing.

A dome forms a virtual image of the underwater world, close in front of the glass, and the lens behind has to focus on that. Where the virtual image sits depends on the dome’s radius in millimetres — a small dome puts it close, a large one further out — so a lens that cannot focus near enough will not work behind a small dome.

That is a length, it matters, and it is a property of the glass rather than of the picture. So the sorting is not “the dome has no lengths in it”; it is that the departures from a pinhole, which are what a picture of a scene reports, are governed by the ratio while the focusing geometry, which is what the equipment has to satisfy, is governed by the radius.

An owner therefore needs both numbers and gets one of them from the picture. Which is the same division this collection keeps finding: an observable answers the question it is a function of, and the questions it is not a function of stay open however carefully it is measured.

The centre that is not there either #

There is a second reading of a dome’s picture and it belongs beside this one, because it is the reading that decides whether “the camera behind the glass” means anything.

A picture through water has no viewpoint: the rays of a refracted picture, continued back, do not pass through any common point, and how badly they miss is a length in millimetres. A centred dome is the one configuration in the whole refraction field where that miss is zero — the bundle really does have a centre, and the centre really is the pupil.

Push the dome off centre and the miss appears. So the same ratio that governs the bending also governs whether the picture is a projection at all, and the two questions have one answer.

An aside on what “concentric” is doing #

The computation above models the dome as a single glass-to-water interface at the inner surface’s normal. That is not a simplification and it is worth saying why, because it looks like one.

A real dome is a shell with two surfaces. If the shell is concentric — inner and outer spheres sharing a centre — then a ray entering the glass and leaving it meets two surfaces whose normals are the same line through that centre, so whatever the first bends the second unbends, and the shell’s thickness cancels exactly.

That is why domes are made concentric, and it is why the thickness never appears in any of the numbers here. A shell that was not concentric would have a third parameter, and the picture would carry some function of all three rather than one clean ratio.

The cancellation is worth checking rather than believing, and the check is the one the computation already runs. Model the shell as a single interface at the inner normal and the centred case bends nothing at all; model it as two concentric surfaces with a real thickness and the answer is the same to the last bit, because the two refractions are inverse. A shell whose surfaces did not share a centre would be a meniscus lens, would have optical power, and would not be a dome port in any useful sense.

The invariance survives the second index #

One more thing scales out, and it is the one a reader is most likely to worry about.

The computation carries two refractive indices — glass and water — and a ratio invariance that held only for one particular pair of them would be a coincidence rather than a symmetry. It holds for any pair, because neither index has a length in it either. Snell relates two angles; the indices are pure numbers; and a similarity of the geometry leaves every angle where it was.

So the statement is stronger than the measured factors show. It is not that a ten-centimetre acrylic dome in seawater behaves like a twenty-centimetre one; it is that any dome behaves like any other of the same ratio in the same two media, and changing the media changes the curve without touching the invariance.

Where the recovery would go next #

Suppose a reader wants the radius in millimetres from the picture alone, with no ruler. What would have to be in the frame?

A length in the water at a known distance — a scale bar, a ruler, an object of known size at a measured range. That supplies the missing dimension directly, and everything follows.

What will not work is more pictures of the same scene through the same dome. Every one of them is the same instrument, and an instrument that has thrown a dimension away does not recover it by being used again. This is the same reason a second photograph from the same eye adds nothing to a scale that one has already lost.

What will also not work is a wider fan. The invariance holds across the whole range of angles the dome admits — it is exact at 2° and exact at 60° — so widening the fan improves the estimate of the ratio and does nothing at all for the radius. That is a different situation from the pane of glass, where a wider fan genuinely does separate the two unknowns, and the difference between the two cases is worth holding onto: a degeneracy that is exact cannot be broken by better data, and one that is merely severe can.

The general rule this row keeps arriving at #

Four of the five instruments in this row return a ratio. That is not a coincidence about mirrors and glass; it is what an instrument does when nothing in its own physics carries a length.

Reflection is scale-free. Refraction is scale-free. A pinhole is scale-free. So an optical arrangement made only of those three has a similarity group acting on it, and every observable is invariant under that group — which is to say, every observable is a function of the dimensionless ratios and of nothing else.

A length enters only where something breaks the similarity. The room breaks it, because the room did not scale with the ball. The table breaks it, because a caustic is thrown onto something at a fixed distance. A wavelength would break it, if the arrangement were small enough for diffraction to matter, which is why an optical measurement at the scale of the light itself behaves differently from every measurement in this collection.

So the question to ask of any instrument, before asking what it measures, is: what in this arrangement has a length in it? If the answer is nothing, the instrument returns ratios, and no amount of care with the picture will change that.

That test is quick and it is the one this row was assembled to demonstrate. A shadow has a length in it, because the floor is somewhere. A caustic has one, because the table is. An anamorph on a floor has one, because the marks are laid down in the room. And a dome port, photographed against water, has none — which is why the answer it gives is a fraction and why that is not a shortcoming of the reading.