2 A B C D E F G H I K L M N O P R S T U V W
Le Li Lo


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
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
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 so called “lens equation” reads

\frac{1}{focal length} = \frac{1}{distance to object} + \frac{1}{distance to image}

Solving for focal length

focal_length = \frac {1} {\frac {1} {distance to object} + \frac {1} {distance to Image}}= \frac {1}{\frac {distance to object + distance to image}{distance to object * distance to image}}=\frac {distance to object*distance to image}{distance to object+distance to image}

Solving for distance to object

distance to object = \frac {1} {\frac {1} {fokal length} - \frac {1} {distance to image}}=\frac{focal length * distance to image}{distance to image - focal length}

Solving for distance to Image

distance to image = \frac {1} {\frac {1} {fokal length} - \frac {1} {distance to object}} = \frac {focal length * distance to object}{distance to object - focal length}

Interpretation of the lens equation

What can be derived from this formula?

  • The Image of infinite distant objects in in the focal plane.

When thenobject moves to infinity (= when the distance_to_object get infinite large), then

\frac {1} {DistanceToObject} \rightarrow 0 Say,

distance to image = focal length

  • An object that is at double focal length distance ob object sideis mapped to an image that is in double focal length distance on image side.
  • In other words : If we want a 1:1 mappingin distance X  so choose
f = \frac {X}{4}
  • If the object distance equals the focal length then the image is at infinity.
  • The closer an object coming from infinity approaches the focal length, the closer get image distance and focal length. So \frac {1} {DistanceToImage} \rightarrow 0 holds.
  • An object in focal length in front of the lens is mapped to infinity on image side.
  • For an Object closer to the lens than the focal length \frac {1} {DistanceToImage} < 0 holds, say, the image is generated on the object side (!) of the lens.


Light that we can see (“visible light”, “VIS”) is a small part of a spectrum of a thing called “electromagnetic radiation”, distinguished by something we call “wavelength”.
As the wavelength varies in the visible spectrum, the light appearantly changes color from violet to red.

image of Frauhofer-Lines

There are no actual boundaries between one range of wavelengths and another. So numbers associated with a certain range are only approximate.

If we explore the spectrum from long wavelengths to shorter wavelengths, we meet :

  • “radio waves” : regular broadcast wavelengths are for example 500 meters long, but even longer radio waves exist
  • “short waves” :  (“Radar waves”, “Millimeter waves”)
  • “infrared”
  • “visible light”
  • “ultraviolet light”
  • soft x-rays
  • x-rays
  • hard x-rays
  • gamma rays

BuildingsHumansButterfliesNeedle PointProtozoansMoleculesAtomsAtomic Nuclei104108101210151016101810201 K100 K10,000 K10,000,000 KPenetrates Earth'sAtmosphere?RadioMicrowaveInfraredVisibleUltravioletX-rayGamma ray10310−210−50.5×10−610−810−1010−12Radiation TypeWavelength (m)Approximate Scaleof WavelengthFrequency (Hz)Temperature ofobjects at which this radiation is themost intensewavelength emitted−272 °C−173 °C9,727 °C~10,000,000 °C
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