Editing Scheimpflug principle (section)

=== Proof of the Scheimpflug principle ===
[[File:ScheimpflugProof.png|thumb|Figure 6. Object plane inclined to the lens plane]]

In a two-dimensional representation, an object plane inclined to the lens
plane is a line described by

: <math>y_u=au+b</math> .

By optical convention, both object and image distances are positive for real images, so that in Figure&nbsp;6, the object distance ''u'' increases to the left of the lens plane LP; the vertical axis uses the normal Cartesian convention, with values above the optical axis positive and those below the optical axis negative.

The relationship between the object distance ''u'', the image distance ''v'', and the lens focal length ''f'' is given by the thin-lens equation

: <math>\frac 1 u + \frac 1 v = \frac 1 f \,;</math>

solving for ''u'' gives

: <math>u=\frac{vf}{v-f} \,,</math>

so that

: <math>y_u=a \, \frac {vf} {v-f} +b</math> .

The magnification ''m'' is the ratio of image height ''y<sub>v</sub>'' to object height {{nowrap|''y<sub>u</sub>''&thinsp;:}}

: <math>m=\frac{y_v}{y_u} \,;</math>

''y<sub>u</sub>'' and ''y<sub>v</sub>'' are of opposite sense, so the magnification is negative, indicating an inverted image. From similar triangles in Figure 6, the magnification also relates the image and object distances, so that

: <math>m=-\frac{v}{u}=-\frac{v-f}{f}</math> .

On the image side of the lens,

: <math>\begin{align}
  y_{v} & =my_{u} \\ 
        & =-\frac{v-f}{f}\left( a \, \frac{vf}{v-f}+b \right) \\
        & =-\left( av+\frac{v}{f}b-b \right) \,,
\end{align}</math>

giving

: <math>y_{v}=-\left( a + \frac{b}{f} \right)v+b</math> .

The [[locus (mathematics)|locus]] of focus for the inclined object plane is a plane; in two-dimensional representation, the [[y-intercept]] is the same as that for the line describing the object plane, so the object plane, lens plane, and image plane have a common intersection.

A similar proof is given by Larmore (1965, 171–173).