Editing Perspective distortion (section)

== Optics ==

Consider an idealised [[Gaussian optics|Gaussian optical system]], with the image and the object in the same medium. Thus, for an object in focus, the distance between the lens and image plane {{nowrap|<math>s_\mathrm{i}</math>,}} the distance between lens and the object {{nowrap|<math>s_\mathrm{o}</math>,}}  and the focal length <math>f</math> are related by the [[Lens (optics)#Imaging properties|thin-lens equation]]:

<math display="block">\frac{1}{s_\mathrm{i}} + \frac{1}{s_\mathrm{o}} = \frac{1}{f}\,.</math>
The transverse magnification <math>M</math> is related by

<math display="block">M = \frac{s_\mathrm{i}}{s_\mathrm{o}}.</math>

Assume the distance between the object and the image plane being {{nowrap|<math>D</math>,}}
<math display="block">D={s_\mathrm{i}} + {s_\mathrm{o}}.</math>

For <math>D \geq 4f</math>, let <math>D_\mathrm{f} = D/f</math>,

<math display="block">M = \frac{D -2 f -\sqrt{{D}^{2} -4 D f}}{2f} = \frac{D_f}{2} -1-\sqrt{\frac{{D_f}^{2}}{4} - D_f}\,.</math>

The ''axial magnification'' <math>M_\text{ax}</math> of an object at <math>s_\mathrm{o}</math> is the rate of change of the lens–image distance <math>s_\mathrm{i}</math> as the lens–object distance <math>s_\mathrm{o}</math> changes. For an object of finite depth, one can conceive of the {{em|average}} axial magnification as the ratio of the depth of the image and the depth of the object:

<math display="block">\begin{align}
M_\text{ax} &= \left| \frac{\mathrm{d}s_\mathrm{i}}{\mathrm{d}s_\mathrm{o}} \right| = \left| \frac{\mathrm{d}}{\mathrm{d}s_\mathrm{o}} \left(\frac{1}{f} - \frac{1}{s_\mathrm{o}}\right)^{-1} \right| \\
&= \left| -\left(\frac{1}{f} - \frac{1}{s_\mathrm{o}}\right)^{-2} s_\mathrm{o}^{-2} \right| \\
&= \frac{s_\mathrm{i}^2}{s_\mathrm{o}^2} = M^2\,.
\end{align}</math>

When <math>M > 1</math>, as the magnification increases, the ratio of axial to transverse magnification also increases. For example, if {{nowrap|<math>M = 2</math>,}} then {{nowrap|<math>M_\text{ax} = 4</math>,}} and the image will be twice as tall and four times as deep as the object.

When <math>D \geq 4f, M \leq 1</math> as for most of the camera, if {{nowrap|<math>M = 0.5</math>,}} then {{nowrap|<math>M_\text{ax} = 0.25</math>,}} and the image will be half as tall and quarter as deep as the object.

At <math>s_{o}</math>, the rate of change of the transverse magnification <math>M</math> as <math>s_{o}</math>changes：

<math>{M_{s_{o}}=\left|{d \over d(s_{o})}{s_{i} \over s_{o}}\right|=\left|{d \over d(s_{o})}{f \over (s_{o}-f)}\right|=\left|{-f \over (s_{o}-f)^{2}}\right|={M^{2} \over f}}</math>

It can be seen that, for a fixed <math>M</math>, the longer focal length leads to small change of <math>M</math>, the shorter focal length leads to big change of <math>M</math>. That means to get the same <math>M</math> by moving to the longer focal length, the object changes less, looks more shallow, vice versa.