Editing Dolly zoom (section)

== Optics ==
{{see also|Perspective distortion}}

For most purposes, it can be assumed that the image space and the object space are in the same medium. Thus, for an object in focus, the distance between the lens and [[image plane]] <math>s_\text{i}</math>, the distance between lens and the object <math>s_\text{o}</math>, and the [[focal length]] <math>f</math> are related by

:<math>{1 \over s_i} + {1 \over s_o} = {1 \over f}.</math>

Then the transverse magnification is

:<math>M = {s_\text{i} \over s_\text{o}} = {f \over (s_\text{o} - f)}.</math>

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

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

One can see that if magnification remains constant, a longer focal length results in a smaller axial magnification, and a smaller focal length in a larger axial magnification. That is, when using a longer focal length while moving the camera/lens away from the object to maintain the same magnification ''M'', objects seem shallower, and the axial distances between objects seem shorter. The opposite—increased axial magnification—happens with shorter focal lengths while moving the camera/lens towards the object.

===Calculating distances===

To achieve the effect, the camera needs to be positioned at a certain distance from the object that is supposed to remain still during the dolly zoom. The distance depends on how wide the scene is to be filmed and on the [[field of view]] (FOV) of the camera lens. Before calculating the distances needed at the different fields of view, the constant width of the scene has to be calculated:

:<math> \text{distance} = \frac{\text{width}}{2\tan\left(\frac{1}{2}\text{FOV}\right)}.</math>

For example, a FOV of 90° and a distance of 2 meters yield a constant width of 4 meters, allowing a 4-meter-wide object to remain still inside the frame during the effect.