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Gaussian Optics

is another term for paraxial optics.

Carl Friedrich Gauss (1755-1855) laid in 1840 the foundations for this subject. In his work “Dioptrische Untersuchungen” Gauss showed, that the behaviour of any lens system can be determinedform the knowledge of it’s six cardinal points (also called Gauss-Points), namely two focal points, two nodal points (of unit angular magnification)  and two principal points (of unit linear magnification) .

Included in the paper were recipes for experimentally determining the positions of these points and iterative methods for calculating them in terms of the surface curvatures, separations and refractive indices of the lens system.

He explicitely mentioned  two equations where ray height r and ray angle \theta of an input and output ray are linked, as used in the the ABCD-Matrix.

    \[r' = A r + B \theta \]

    \[\theta' = C r + D \theta \]

\iff

    \[ \left( \begin{array}{c}r'\\\theta'\end{array} \right) = \begin{pmatrix} A & B \\ C & D \end{pmatrix} \left( \begin{array}{c}r\\\theta\end{array} \right) \]

But Matrix notation wasn’t known at that time. So instead he used an algorithm he learnt from Leonard Euler. It was a shorthand form of continuous fraction, now known as Gaussian Brackets.

It’s only allowed to use ABCD Matrices in the paraxial range, with \sin{\theta} \approx \theta

The angles \theta and \theta' are measured in radians!

For a table of ray transfer matrixes for simple optical components see ABCD matrix

gnomonic

gnomonic lenses (German : gnomonische Objektive) use an image mapping of the typer = f tan \theta
Type : gnomonical weak medium strong max
angles 54° 75° 102° < 180°
Gnomonic lenses can provide a wide viewing angle , but are no fisheye lenses

see:
fisheye types

hyperfocal distance

When a lens is focussed to the hyperfocal distance H, the DOF of the lens is maximized: The range of acceptable sharpness then extends from \frac{H}{2} to infinity

There are two Formulas in use:

H = \frac{f^2}{F \cdot CoC} + f

For f=50mm, F2, and CoC = 0.03mm we get
H = \frac{50mm^2}{2 \cdot 0.03mm} + 50mm = 41666.67mm + 50mm = 41716.67mm \approx 41.72m

This is the formula we use here. The results just differ in the focal length of the lens.

and

H = \frac{f^2}{F \cdot CoC}

For f=50mm, F2, and CoC = 0.03mm we get
H = \frac{50mm^2}{2 \cdot 0.03mm} = 41666.67mm \approx 41.67m

Where CoC is the circle of confusion, F is the F-number and f is the focal length of the lens.

The hyperfocal distance has curious mathematical properties:

The hyperfocal distance H is the distance at which you have to focus an object to receive the largest depth of field. If, namely, a lens is focussed to H, it is focused from \frac{H}{2} to infinity.

When focussed to \frac{H}{2}, so everything from \frac{H}{3} to \frac{H}{1} focused.
When focussed to \frac{H}{3}, so everything from \frac{H}{4} to \frac{H}{2} focused.
When focussed to \frac{H}{2}, so everything from \frac{H}{5} to \frac{H}{3} focused.
…
When focussed to \frac{H}{n}, so everything from \frac{H}{n+1} to \frac{H}{n-1} focused.

The distance

DOF = \frac{H}{n-1} - \frac{H}{n+1}

is the Depth of field.

Notice:

The depth of field is getting smaller, the larger is n, say the shorter the working distance is!

Image Circle

Although monitors and paper photos are rectangular, lenses are (usually) round.

So why are the images generated by round lenses not round? 
 Well .. images of lenses ARE round.

The diameter of these rounds images is called image or image circle. Outside the image circle the picture (hopefully) is dark. If the image circle is smaller than the diagonal of the sensor, the image has dark corners.
When the Image Circle is smaller than the height of the sensor, the image is round and outside of the image circle it’s dark.
Images of Fisheye lenses are typically like that and are then called “circular fisheye image”. Such image are especially helpful for checking whether the axis of sensor and lens are aligned:
The image circle must be in the middle of the (rectangular) screen image.

Maximum (diameter of the) circle that receives (good quality) image information.

The image circle limits the maximum sensor size for which a lens can be used.

An image circle of 6mm limits the use to maximum sensor to 1/3″.

An image circle of 8mm limits the use to maximum sensor to 1/2″.

An image circle of 16mm limits the use to maximum sensor to 1″.

Image Distance

Distance between the image side principal plane and the image (measured on the optical axis).

IR corrected

When lenses are designed, one of the important parameters is the spectrum for which the lenses are to be used.
Most lenses on the market are “corrected” (read : “designed”) for the visible range of the light, the part that humans can see. These are wavelengths between about 420nm and 720nm, deep violett to deep red colors.

Lenses are called IR corrected, if they are designed for near Infrared light (NIR .. roughly 800 to 1100nm), so you _could_ use them if your Illumination contains these wavelengths.

If lenses are not IR corrected, they will typically have a so called Focus Shift (or “longitudinal color aberration”) , say the focal length of an IR image is different from the focal length of the color image, thus either the color image or the IR image is focussed, but in general not both.

Lenses that _are_ IR corrected usually have a special antireflection coating , suited for the infrared spectrum.

One thing you have to take into account however is the quantumefficiency of your sensor, say, how well it can receive NIR light. Often sensor can receive light in this range of wavelengths one one third as good as they do for the visible range, say the brightness of the image is just 1/3 .. and lower. Ask your supplier about sensors dedicated for IR light.

If a lens is designed for VIS (= visible light) and also for NIR, you have to keep in mind that you in general can NOT get a nice color image _and_ a nice NIR image at the same time.

This is because color pixels let various NIR wavelengths pass say, the nice color images are overlayed with IR light.

According to the Rayleigh criterionlight of, say, 850nm wavelength can achieve only half the resolution of a lens optized for half the wavelength (425nm)
This means, even a cristal clear color image is overlayed by a slightly blurred IR image.

Say, if you _can_ choose, go for Visible light, not for NIR light.

If you need IR light AND Visible light at the same time, for example door cameras have this challenge, then go for a special filter that lets VIS pass plus a small IR range, for example 940 +/- 20nm

Lens

A lens element is a component made of an optical transparent material (usually glass) with two optical active surfaces.
“optical transparent” does not necessarily mean that the component is transparent for human eyes!
For example are lenses made of Germanium not transparent for humans. For certain cameras they are transparent however.
Lenses can be used to focus light
Lens_and_wavefronts[1]
The main effect of lenses is that they map object points to image points.
Let say we have two object points, for example the Top T0 and the bottom B0 of a vase. A two element compound lens works like this :
The object points T0 an B0 are mapped to image points T1 and B1 by lens element number 1. The lens elements are numbered starting from object side.
The images T1 and B1 then become the new “object points” of lens element 2 and map T1 and B1 to “images of image” of T0 and T0, called T2 and B2″.

A lens with 10 lens elements accordingly generates sequences
T0->T1->T2->… -> T10
and
B0->B1->B2->….-> B10

The image produced by a lens can be in front or behind the lens.
It can be smaller or bigger than the object.
Of course we want images of trees to be smaller than trees.
Image of microscopic details we of course want to be larger than the details.

lens equation

The “Gauss lens equation,” as it is known, goes like this:

\frac{1}{f} = \frac{1}{g} + \frac{1}{b}

where
f = focal length,
g = distance to object (measured from object side principal plane H).
b = distance to image (measured from image side principal plane H’).

It can be solved for focal length, object distance and image distance.

We may find the focal length by this equation:

f = \frac {1} {\frac {1} {g} + \frac {1} {b}}= \frac {1}{\frac {g + b}{g \cdot b}}=\frac {g \cdot b}{g + b}

We may find the object distance by this equation:

g = \frac {1} {\frac {1} {f} - \frac {1} {b}}=\frac{f \cdot b}{b - f}

We may find the image distance by this equation:

b = \frac {1} {\frac {1} {f} - \frac {1} {g}} = \frac {f \cdot g}{g - f}

Interpretation of the lens equation

What can be derived from this formula?

  • The focus plane contains the image of infinitely far away objects.
  • When the object moves to infinity (= when g gets infinitely large), then

    \frac {1} {g} \rightarrow 0
    And therefore b = f

  • On the object side, an object with a double focal length distance is mapped to an image with a double focus length distance on the image side.
  • In other words : If we want a 1:1 mapping in distance X from the sensor you have to choose
    • f = \frac {X}{4}

    (but subtract the distance of the principal points from X)

  • The picture is at infinity when the object distance equals the focus length.
  • When an object nears to the focus length, the image distance rises. Therefore \frac {1} {b} \rightarrow 0 holds.
  • For an object closer to the lens than the focal length \frac {1} {DistanceToImage} < 0 holds: the image is generated on the object side (!) of the lens.
    • The object and image distances are calculated from the primary planes on the object and image sides, respectively.
    As a result, the object distance differs from the working distance in most cases.
    The focal length is a paraxial concept. Therefore the Gauss lens equation is only valid in the paraxial region of the lens, the region where sin(x) \approx tan(x) \approx x holds. Most calculators don’t care (including ours), As a result, anytime larger viewing angles are involved, the results are essentially a “informed guess.”