lens equation | OptoWiki Knowledge Base

With
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’)

The so called “lens equation” reads

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

Solving for focal length

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

Solving for distance to object

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

Solving for distance to Image

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

Interpretation of the lens equation

What can be derived from this formula?

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

When the object moves to infinity (= when g gets infinitely large), then

$\frac {1} {g} \rightarrow 0$ Say,

$b = f$

An object that is at double focal length distance on object side is mapped to an image that is in double focal length distance on 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)

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 $\frac {1} {b} \rightarrow 0$ 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.