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The most general form of a mirror in one dimension is a differentiable function $f:\mathbb R \rightarrow \mathbb R$, for which $f(x)$ determines the vertical position of the mirror at displacement $x$ and $f'(x)$ determines the slope and hence the angle of the mirror. The simplest examples of such mirrors are plane mirrors defined by $f(x)=c$ where $c$ is a constant, and parabolic mirrors defined by $f(x) = x^2/4a$ where $a$ is the focal length of the mirror.

Furthermore, for an object at some position in front of the mirror, an image is only formed if all rays that reflect once off the mirror intersect at one point (here I am referring to the continuations of the rays into lines, as is done when finding the position of a virtual image). It is relatively easy to prove that, for any object position, a plane mirror forms a valid image. However, I have not yet been able to prove so for a parabolic mirror, and for a general mirror the task seems formidable. Also, a general mirror would not necessarily form an image for any object; consider the image below, for example, where three rays from one point reflect and intersect at three distinct points. Hence, we should not be able to prove that every mirror always forms an image.

As such, I was wondering if there is a general set of criteria that determines whether or not some mirror will form an image for any object distance. It seems like a pretty important problem, and I haven't been able to find any discussion of it anywhere (then again, I'm not great at googling so I might have just missed it).

enter image description here

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    $\begingroup$ Do rays that bounce off a plane mirror intersect in one point? $\endgroup$
    – AccidentalTaylorExpansion
    Commented Jun 14, 2022 at 11:40
  • $\begingroup$ My mistake; their continuations do, in the sense of a virtual image. Good point, I'll correct the question now. $\endgroup$
    – Baylee V
    Commented Jun 14, 2022 at 11:41
  • $\begingroup$ You are referring to objects not as points but as the set of some points right? $\endgroup$
    – Kani Pen
    Commented Jun 14, 2022 at 12:15
  • $\begingroup$ @KaniPeni I'm actually only defining objects as singular points, since it's easier to phrase what the image of such an object is. $\endgroup$
    – Baylee V
    Commented Jun 14, 2022 at 12:23
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    $\begingroup$ Quick idea, with no guarantee of success: for conics, you might try to use their geometrical definition (in terms of focus and directrix) and relate it to Fermat's principle. I would guess the natural formalism is projective geometry, but I don't know any... $\endgroup$
    – Doru Constantin
    Commented Jun 14, 2022 at 12:27

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There is a caveat for the formation of images (virtual and real). I think you were talking about rigorous stigmatism, ie exact intersection of the light rays. With this definition, if you want every point to have an image, the only possibility is the plane mirror. If you want rigorous stigmatism between two points, the mirror will be a conic section (from the bifocal definition), and if it’s not a plane, they will be the only two conjugate points.

You’ll need to loosen the definition to approximate stigmatism, for any hope to work in general. It is this notion you will need for the parabolic mirror for example (which is why you were having a hard time proving it I imagine).

In general, if you look at the light rays emanating from a point which are reflected by an arbitrary mirror, you’ll observe caustics, the envelope of the rays. These caustics can have a cusp which corresponds to an image in the context of approximate stigmatism. This is how you can have multiple images.

The existence of images thus becomes the study of singularities of ray systems. The general study involves some serious maths (symplectic geometry etc) check out the relavant appendix in Arnold’s Mathematical Method to Classical Mechanics.

If you want to explicitly compute things, you'll need to assume better regularity. I'll assume your mirror to be smoother, at least $\mathcal C^2$ to define an osculating circle at every point. Locally approximating the mirror by the osculating circle, you'll have a cusp whenever the tangent plane of the mirror is perpendicular to the incident ray. Furthermore, the cusp will be on the line corresponding to the coinciding incident and reflected ray. The position along this line is given by the thin lens formula with the focal given by $$ f = r/2 $$ with $r$ the local curvature radius, and the sign depending on whether the mirror is locally convex or concave. If it's concave, you'll have a real image, and if it's convex you'll have a virtual image.

If you want to experiment around, you can use Geogebra (or any similar software) to plot out the envelope and test out the ideas.

Hope this helps and tell me if you need more details.

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To form an image it is not necessary for all rays coming from a source to meet in one place. Even in the case of spherical mirrors the position of the image of a point source of light depends on where the eye is. All the places where the image can bee seen form a curve called caustics. In the picture below each eye looking at a convex mirror has its own image of the point. When we look with two eyes (stereoscopic view) our perception is very complicated. enter image description here

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