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

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.