Editing Scheimpflug principle (section)

=== 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>