Editing Optical telescope (section)

==Observing through a telescope==
There are many properties of optical telescopes and the complexity of observation using one can be a daunting task; experience and experimentation are the major contributors to understanding how to maximize one's observations. In practice, only two main properties of a telescope determine how observation differs: the focal length and aperture. These relate as to how the optical system views an object or range and how much light is gathered through an ocular [[eyepiece]]. Eyepieces further determine how the field of view and [[magnification]] of the observable world change.

===Observable world===
The observable world is what can be seen using a telescope. When viewing an object or range, the observer may use many different techniques. Understanding what can be viewed and how to view it depends on the field of view. Viewing an object at a size that fits entirely in the field of view is measured using the two telescope properties—focal length and aperture, with the inclusion of an ocular [[eyepiece]] with suitable focal length (or diameter). Comparing the observable world and the [[angular diameter]] of an object shows how much of the object we see. However, the relationship with the optical system may not result in high [[surface brightness]]. Celestial objects are often dim because of their vast distance, and detail may be limited by [[diffraction]] or unsuitable optical properties.

===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.

===Brightness factor===
The [[surface brightness]] at such a magnification significantly reduces, resulting in a far dimmer appearance. A dimmer appearance results in less visual detail of the object. Details such as matter, rings, spiral arms, and gases may be completely hidden from the observer, giving a far less ''complete'' view of the object or range. Physics dictates that at the theoretical minimum magnification of the telescope, the surface brightness is at 100%. Practically, however, various factors prevent 100% brightness; these include telescope limitations (focal length, [[eyepiece]] focal length, etc.) and the age of the observer.

Age plays a role in brightness, as a contributing factor is the observer's [[pupil]]. With age the pupil naturally shrinks in diameter; generally accepted a young adult may have a 7&nbsp;mm diameter pupil, an older adult as little as 5&nbsp;mm, and a younger person larger at 9&nbsp;mm. The [[magnification|minimum magnification]] <math>m</math> can be expressed as the division of the aperture <math>D</math> and [[pupil]] <math>p</math> diameter given by: <math>m = \frac {D}{d} = \frac {130}{7} \approx 18.6</math>. A problematic instance may be apparent, achieving a theoretical surface brightness of 100%, as the required effective focal length of the optical system may require an [[eyepiece]] with too large a diameter.

Some telescopes cannot achieve the theoretical surface brightness of 100%, while some telescopes can achieve it using a very small-diameter eyepiece. To find what eyepiece is required to get [[magnification|minimum magnification]] one can rearrange the magnification formula, where it is now the division of the telescope's focal length over the minimum magnification: <math> \frac {F}{m} = \frac {650}{18.6} \approx 35</math>. An eyepiece of 35&nbsp;mm is a non-standard size and would not be purchasable; in this scenario
to achieve 100% one would require a standard manufactured eyepiece size of 40&nbsp;mm. As the eyepiece has a larger focal length than the minimum magnification, an abundance of wasted light is not received through the eyes.

===Exit pupil===
The limit to the increase in [[surface brightness]] as one reduces magnification is the [[exit pupil]]: a cylinder of light that projects out the eyepiece to the observer. An exit pupil must match or be smaller in diameter than one's [[pupil]] to receive the full amount of projected light; a larger exit pupil results in the wasted light. The exit pupil <math>e</math> can be derived with from division of the telescope aperture <math>D</math> and the [[magnification|minimum magnification]] <math>m</math>, derived by: <math>e = \frac {D}{m} = \frac {130}{18.6} \approx 7</math>. The pupil and exit pupil are almost identical in diameter, giving no wasted observable light with the optical system. A 7&nbsp;mm pupil falls slightly short of 100% brightness, where the [[surface brightness]] <math>B</math> can be measured from the product of the constant 2, by the square of the pupil <math>p</math> resulting in: <math>B = 2*p^2 = 2*7^2 = 98</math>. The limitation here is the pupil diameter; it is an unfortunate result and degrades with age. Some observable light loss is expected and decreasing the magnification cannot increase surface brightness once the system has reached its minimum usable magnification, hence why the term is referred to as ''usable''.

[[File:Comparison of exit pupils for astronomy.png|thumb|center|These eyes represent a scaled figure of the [[human eye]] where 15 px = 1&nbsp;mm, they have a [[pupil]] diameter of 7&nbsp;mm. ''Figure A'' has an [[exit pupil]] diameter of 14&nbsp;mm, which for [[astronomy]] purposes results in a 75% loss of light. ''Figure B'' has an exit pupil of 6.4&nbsp;mm, which allows the full 100% of observable light to be perceived by the observer.]]

===Image Scale===
When using a CCD to record observations, the CCD is placed in the focal plane. Image scale (sometimes called ''plate scale'') is how the angular size of the object being observed is related to the physical size of the projected image in the focal plane

:<math>i = \frac{\alpha}{s},</math>

where <math>i</math> is the image scale, <math>\alpha</math> is the angular size of the observed object, and <math>s</math> is the physical size of the projected image. In terms of focal length image scale is

:<math>i = \frac{1}{f},</math>

where <math>i</math> is measured in radians per meter (rad/m), and <math>f</math> is measured in meters. Normally <math>i</math> is given in units of arcseconds per millimeter ("/mm). So if the focal length is measured in millimeters, the image scale is

:<math>i\ (''/\mathrm{mm}) = \frac{1}{f\ (\mathrm{mm})}\left[\frac{180 \times 3600}{\pi}\right].</math>

The derivation of this equation is fairly straightforward and the result is the same for reflecting or refracting telescopes. However, conceptually it is easier to derive by considering a reflecting telescope. If an extended object with angular size <math>\alpha</math> is observed through a telescope, then due to the [[Laws of reflection]] and [[Trigonometry]] the size of the image projected onto the focal plane will be

:<math>s = \tan(\alpha) f.</math>

The image scale (angular size of object divided by size of projected image) will be

:<math>i = \frac{\alpha}{s} = \frac{\alpha}{\tan(\alpha) f},</math>

and by using the small angle relation <math>\tan(a) \approx a</math>, when <math>a \ll 1</math> (N.B. only valid if <math>a</math> is in radians), we obtain

:<math>i = \frac{\alpha}{\alpha f} = \frac{1}{f}.</math>