Editing Scheimpflug principle (section)

==Derivation of the formulas==

=== 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).

=== Angle of the PoF with the image plane ===
[[File:ScheimpflugFormulaDerivations.png|thumb|Figure 7. Angle of the PoF with the image plane]]

From Figure 7,

: <math>\tan \psi = \frac {u' + v'} {S} \,,</math>

where <var>u′</var> and <var>v′</var> are the object and image distances along the line of sight and <var>S</var> is the distance from the line of sight to the Scheimpflug intersection at S. Again from Figure 7,

: <math>\tan \theta = \frac {v'} {S} \,;</math>

combining the previous two equations gives

: <math>\tan \psi = \frac {u' + v'} {v'} \tan \theta = \left ( \frac {u'} {v'} + 1 \right ) \tan \theta \,.</math>

From the thin-lens equation,

: <math>\frac {1} {u} + \frac {1} {v} = \frac {1} {u' \cos \theta} + \frac {1} {v' \cos \theta} = \frac {1} {f} \,.</math>

Solving for <var>u′</var> gives

: <math>u' = \frac {v' f} {v' \cos \theta - f} \,;</math>

substituting this result into the equation for {{nowrap|tan&thinsp;<var>ψ</var>}} gives

: <math>\tan \psi = \left ( \frac {f} {v' \cos \theta - f} + 1 \right ) \tan \theta
 = \frac {f + v' \cos \theta - f} {v' \cos \theta -f} \tan \theta \,,</math>

or

: <math>\tan \psi = \frac {v'} {v' \cos \theta - f} \sin \theta \,.</math>

Similarly, the thin-lens equation can be solved for <var>v′</var>, and the result substituted into the equation for {{nowrap|tan&thinsp;<var>ψ</var>}} to give the object-side relationship

: <math>\tan \psi = \frac {u'} {f} \sin \theta \,.</math>

Noting that

: <math>\frac {u'} {f} = \frac {u} {f} \frac {1} {\cos \theta}
                      = \frac {m + 1} {m} \frac {1} {\cos \theta} \,,</math>
the relationship between <var>ψ</var> and <var>θ</var> can be expressed in terms of the magnification <var>m</var> of the object in the line of sight:

: <math>\tan \psi = \frac {m + 1} {m} \tan \theta \,.</math>

=== Proof of the "hinge rule" ===
From Figure 7,

: <math>\tan \psi = \frac {u'} {J} \,;</math>

combining with the previous result for the object side and eliminating <var>ψ</var> gives

: <math>\sin \theta = \frac {f} {J} \,.</math>

Again from Figure 7,

: <math>\sin \theta = \frac {d} {J} \,,</math>

so the distance <var>d</var> is the lens focal length <var>f</var>, and the point G is at the intersection the lens's front focal plane with a line parallel to the image plane. The distance <var>J</var> depends only on the lens tilt and the lens focal length; in particular, it is not affected by changes in focus. From Figure&nbsp;7,

: <math>\tan \theta = \frac {v'} {S} \,,</math>

so the distance to the Scheimpflug intersection at S varies as the focus is changed. Thus the PoF rotates about the axis at G as focus is adjusted.