What causes blurriness in an optical system?

Question

The way I understand the purpose of a typical optical system is that it creates a one to one mapping between each possible incident ray and a point on a sensor plane. This is like a mathematical function. If there was no mapping, and each ray was free to strike any point on a sensor there would be no image formed on it and it would be just blurry average light. This would be like having a camera sensor without a lens.

Now there's a very simple concept that creates this one to one mapping, pin-hole camera. In a pin-hole camera there's no blurriness possible, as long as the hole is small enough, each point opposing the hole is mapped onto one specific ray. This means this type of camera can never have a blurry image, no matter where it's focal point is. This can be proven, geometrically.

In an optical system that uses lenses to create this mapping however things are not always ideal, because blurriness does happen. Which indicates that mapping is not one to one, and that some sensor points share rays with each other creating local averages, i.'e blurriness. It is often claimed that it happens because the focal point is not at the "right place". If you consider the pin-hole model as the ideal you will understand that this is not true. Changing focal point alone will only make image seem smaller or larger. From geometrical optics alone I don't see what could possibly cause the sharing of rays. It seems to me there's more to it and that i'm not the only one confused.

So what does REALLY create the blurriness? Is there some sort of imperfection in lenses that causes them to send multiple rays to the same point on a sensor and that somehow becomes more visible at certain focal distances? This is the only explanation i have.

Answer 1

To add some details to Eoin's answer.

Your description of imaging as a mapping is a good one to begin with and it will get yoi a long way. However, even in ray optics (see my answer here for more info), i.e. when the propagation of light can be approximated by the Eikonal equation (see here for more info), the mapping of points one-to-one between the object and image plane as you describe can only happen in very special conditions. In general, a bundle of rays diverging from one point will not converge to a point after passing through an imaging system made of refracting lenses and mirrors. One has to design the system so that the convergence is well approximated by convergence to a point. As Eoin said, this non-convergence is the Ray theory description of aberration: spherical, comatic, astigmatic, trefoil, tetrafoil, pentafoil and so forth are words that are used to describe aberration with particular symmetries (spherical aberration is rotationally symmetric about the chief ray, coma flips sign on a $180^o$ rotation about the chief ray, trefoil flips sign on a $120^o$ rotation and so forth). There is also chromatic aberration, where the image point position depends on wavelength so that point sources with a spectral spread have blurred images. Lastly, the imaging surface, comprising the points of "least confusion" (i.e. those best approximating where the rays converge to a point) is always curved to some degree - it is often well approximated by an ellipsoid - and so even if convergence to points is perfect, the focal surface will not line up with a flat CCD array. This is know as lack of flatness of field: microscope objectives with particularly flat imaging surfaces bear the word "Plan" (so you have "Plan Achromat", "Plan Apochromat" and so forth).

Only very special systems allow for convergence of all ray bundles diverging from points in the object surface to precise points in the image surface. Two famous examples are the Maxwell Fisheye Lens and the Aplanatic Sphere: both of these are described in the section called "Perfect Imaging Systems" in Born and Wolf, "Principles of Optics". They are also only perfect at one wavelength.

An equivalent condition for convergence to a point is that the total optical path - the optical Lgagrangian is the same for all rays passing between the points.

Generally, lens systems are designed so that perfect imaging as you describe happens on the optical axis. The ray convergence at all other points is only approximate, although it can be made good enough that diffraction effects outweigh the non-convergence.

And of course, finally, if everything else is perfect, there is the diffraction limitation described by Eoin. The diffraction limit simply arises because converging plane waves with wavenumber $k$ cannot encode any object variation that varies at a spatial frequency greater than $k$ radians per second. This, if you like, is the greatest spatial frequency Fourier component that one has to build an image out of. Images more wiggly than this Fourier component of maximum wiggliness cannot form. A uniform amplitude, aberration-free spherical wave converges to an Airy disk, which is often taken as defining the "minimum resolvable diffraction limited distance". However, this minimum distance is a bit more complicated than that. It is ultimately defined by the signal to noise as well, so an extremely clean optical signal can see features a little bit smaller than the so-called diffraction limit, but most systems, even if their optics is diffraction limited, are further limited by noise to somewhat less than the "diffraction limit".

Answer 2

A perfect optical system has to accept a ray coming in at any angle and bend it to exactly the correct point on the image plane. Unfortunately it isn't possible to make a lens which can do this for all possible rays because the shape would have to be different for rays of different angles hitting the same part of the lens (or mirror).

A pinhole does this by limiting the rays to a very small range of ray angles and having a curved image plane. Systems of lenses approximate the ideal behaviour by adding more and more optical surfaces of different types of curvature - each to correct a specific range of problems. But (at least with a finite number of components) you can please all the rays.

Then there are other technical differences such as chromatic aberration (different wavelengths bend different amounts in the same glass)

Answer 3

I'm not sure previous posts answer your final question. I want to give you an intuitive picture of the situation. Provided the detail pointed by @WetSavannaAnimal that the "true" mapping is

{object plane point} to {image plane point}

when talking of conventional optical systems, and not ray to point, as always many rays passing through your lens ends in the same point (you may think then in

{every ray coming from the same object point} to {image plane point}),

you can understand blurriness causes with a ray diagram. I will also neglect diffraction, which reduces sharpness in your pinhole camera.

If your system is a single lens you may think of light as a cone:

If your image is forming in B and you put there a screen, a film, a sensor... you get your point-point mapping. If you put it in A or C you get a point-disk mapping, giving overlapping disks when your object have more than one point... These disks are usually gaussian or Airy-like, due to diffraction. As you point with the pinhole limit, finite aperture (not collecting the $2\pi \text{ srad}$ of light but a cone) is the key. If now we reduce the aperture with a diaphragm:

as we do in photography to enlarge Depth of Field, i.e., get every plane "focused", or get a picture focused wherever the plate is, as in the pinhole camera. In this limit, you can also understand how the lens does not matter at all, as it is locally flat in its center. We usually work in terms of transfer functions of black boxes to connect object and screen planes.

So if you need glasses, you can make some cheap ones with some aluminum foil and a pin. That arrangement allows a wide set of interesting phenomena observation... (like watching floaters, your own fundus, test for dust in your camera's lenses or sensors, etc.)

Answer 4

A fundamental limit on the image resolution will be the diffraction limit. Here, the ray approximation is breaking down, and, in a nutshell, the finite size of the optical elements causes the waves to 'spread out' so that there is not a one-to-one mapping any more.

However, I would guess that other things, like spherical and chromatic aberration will affect image quality more than the diffraction limit for most real-world cases.

Answer 5

No optical system exists, that can form (in the ray optics approximation) a sharp and geometrically undistorted REAL image of a REAL three dimensional object. By "ray optics approximation", I mean ignoring diffraction; NOT limited to just "paraxial" situations.

It is well known that an imaging optical system that forms an image with a LATERAL magnification of (m), also has a LONGITUDINAL or AXIAL magnification that is m^2. So if (m) is not equal to one (1) then the three d object must be distorted.

An object stretching along the axis from a distance 2f out to infinity, yields an image that is compressed between f and 2f on the image side. Since the lateral magnification varies along the axis, a three d object must give a distorted image.

So what about planar objects and images ? No optical system can form a sharp undistorted image of a plane object for two different object / image conjugate ratios (not simply object-image reversals). Any optical system that could do that for two different conjugate ratios, could also do so, at all possible conjugates; but it isn't possible. To achieve that would require that sin(A) / sin(B) = Tan(A) / Tan(B) for non paraxial ray angles A and B. The Tan ratio affects geometric distortion, while the sin ratio affects image sharpness (Aplanicity). So you can have sharp images, or you can have geometrically undistorted (plane) images, but you can't have both.

Only ONE case exists where a three d image of a three d object can be formed, that is both sharp in the geometrical case, and also a correct undistorted geometry at other than 1:1 magnification.

BUT ! The image and object cannot both be real; one must be virtual; and moreover, only for one specific 3d object and image. They are portions of spheres concentric with the aplanatic refracting surface of radius R. One has a radius R/N and the other has a radius NR, so the magnification is N^2. The spheres of course must have zero shell thickness. The radius R can be either convex (most common) or concave (most unusual, in fact generally unknown.)

Well all of this can be found in standard texts, and has been known for well over 100 years (except for the concave aplanatic case, which is quite new).