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

lens equation

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.