lens equation | OptoWiki Knowledge Base

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.