Editing Optical telescope (section)

===Field of view and magnification relationship===
Finding what can be seen through the optical system begins with the [[eyepiece]] providing the field of view and [[magnification]]; the magnification is given by the division of the telescope and eyepiece focal lengths. Using an example of an amateur telescope such as a [[Newtonian telescope]] with an aperture <math>D</math> of 130&nbsp;mm (5") and focal length <math>f</math> of 650&nbsp;mm (25.5 inches), one uses an eyepiece with a focal length <math>d</math> of 8&nbsp;mm and apparent FOV <math>v_{a}</math> of 52°. The magnification at which the observable world is viewed is given by: <math>M = \frac {f}{d} = \frac {650}{8} = 81.25</math>. The field of view <math>v_{t}</math> requires the magnification, which is formulated by its division over the apparent field of view: <math>v_{t} = \frac {v_{a}}{M} = \frac {52}{81.25} = 0.64</math>. The resulting true field of view is 0.64°, not allowing an object such as the [[Orion nebula]], which appears elliptical with an [[angular diameter]] of 65 × 60 [[arcminutes]], to be viewable through the telescope in its entirety, where the whole of the [[nebula]] is within the observable world. Using methods such as this can greatly increase one's viewing potential ensuring the observable world can contain the entire object, or whether to increase or decrease magnification viewing the object in a different aspect.