Editing Magnification (section)

==By instrument==

===Single lens===
The linear magnification of a [[thin lens]] is

<math display="block">M = {f \over f-d_\mathrm{o}} = - \frac{f}{x_o}</math>

where <math display="inline">f</math> is the [[focal length]], <math display="inline">d_\mathrm{o}</math> is the distance from the lens to the object, and <math display="inline">x_0 = d_0 - f</math> as the distance of the object with respect to the front focal point. A [[Lens#Sign convention for other parameters|sign convention]] is used such that <math display="inline">d_0</math> and <math>d_i</math> (the image distance from the lens) are positive for real object and image, respectively, and negative for virtual object and images, respectively. <math display="inline">f</math> of a converging lens is positive while for a diverging lens it is negative.

For [[real image]]s, <math display="inline">M</math> is negative and the image is inverted. For [[virtual image]]s, <math display="inline">M</math> is positive and the image is upright.

With <math display="inline">d_\mathrm{i}</math> being the distance from the lens to the image, <math display="inline">h_\mathrm{i}</math> the height of the image and <math display="inline">h_\mathrm{o}</math> the height of the object, the magnification can also be written as:

<math display="block">M = -{d_\mathrm{i} \over d_\mathrm{o}} = {h_\mathrm{i} \over h_\mathrm{o}}</math>

Note again that a negative magnification implies an inverted image.

The image magnification along the optical axis direction <math>M_L</math>, called longitudinal magnification, can also be defined. [[Lens#Lens equation|The Newtonian lens equation]] is stated as <math>f^2 = x_0 x_i</math>, where <math display="inline">x_0 = d_0 - f</math> and <math display="inline">x_i = d_i - f</math> as on-axis distances of an object and the image with respect to respective focal points, respectively. <math>M_L</math> is defined as 

<math display="block">M_L = \frac{dx_i}{dx_0},</math>

and by using the Newtonian lens equation, 

<math display="block">M_L = - \frac{f^2}{x_o^2} = - M^2.</math>

The longitudinal magnification is always negative, means that, the object and the image move toward the same direction along the optical axis. The longitudinal magnification varies much faster than the transverse magnification, so the 3-dimensional image is distorted.

===Photography===
The image recorded by a [[photographic film]] or [[image sensor]] is always a [[real image]] and is usually inverted. When measuring the height of an inverted image using the [[Cartesian coordinate system|cartesian]] sign convention (where the x-axis is the optical axis) the value for {{math|''h''{{sub|i}}}} will be negative, and as a result {{mvar|M}} will also be negative. However, the traditional sign convention used in photography is "[[real image|real]] is positive, [[virtual image|virtual]] is negative".<ref>{{cite book |first=Sidney F. |last=Ray |title=Applied Photographic Optics: Lenses and Optical Systems for Photography, Film, Video, Electronic and Digital Imaging |publisher=Focal Press |year=2002 |isbn=0-240-51540-4 |page=40 |url=https://books.google.com/books?id=cuzYl4hx-B8C&pg=PA40 }}</ref> Therefore, in photography: Object height and distance are always {{em|real}} and positive. When the focal length is positive the image's height, distance and magnification are {{em|real}} and positive. Only if the focal length is negative, the image's height, distance and magnification are {{em|virtual}} and negative. Therefore, the ''{{dfn|photographic magnification}}'' formulae are traditionally presented as<ref>{{cite book |last= Kingslake |first= Rudolph |date= 1992 |title= Optics in Photography |page= 32 |publisher= SPIE Optical Engineering Press |location= Bellingham, Washington |isbn= 0-8194-0763-1 }} "If a lens is thin, or if we can guess at the position of the principal planes, we can readily construct from [{{math|1=1/''d''{{sub|i}} + 1/''d''{{sub|o}} = 1/''f''}} and {{math|1= M = ''d''{{sub|i}}/''d''{{sub|o}}}}] the following simple rules that it is well to bear in mind. They refer specifically to the case of a positive lens forming a real image of a real object, all distances and the magnification being assumed to be positive quantities. If virtual images are involved, it is better to return to the original formulas, [previously stated]. The equations are [{{math|1= ''d''{{sub|o}} = ''f''(1 + 1/M)}} and {{math|1= ''d''{{sub|i}} = ''f''(1 + M)}}]."</ref>

<math display="block">\begin{align}
M &= {d_\mathrm{i} \over d_\mathrm{o}} = {h_\mathrm{i} \over h_\mathrm{o}} \\
  &= {f \over d_\mathrm{o}-f} = {d_\mathrm{i}-f \over f}
\end{align}</math>

===Magnifying glass===
The maximum angular magnification (compared to the naked eye) of a [[magnifying glass]] depends on how the glass and the object are held, relative to the eye. If the lens is held at a distance from the object such that its front focal point is on the object being viewed, the relaxed eye (focused to infinity) can view the image with angular magnification

<math display="block">M_\mathrm{A}={25\ \mathrm{cm}\over f}</math>

Here, <math display="inline">f</math> is the [[focal length]] of the [[lens (optics)|lens]] in centimeters. The constant 25&nbsp;cm is an estimate of the "near point" distance of the eye&mdash;the closest distance at which the healthy naked eye can focus. In this case the angular magnification is independent from the distance kept between the eye and the magnifying glass.

If instead the lens is held very close to the eye and the object is placed closer to the lens than its focal point so that the observer focuses on the near point, a larger angular magnification can be obtained, approaching

<math display="block">M_\mathrm{A}={25\ \mathrm{cm}\over f}+1</math>

A different interpretation of the working of the latter case is that the magnifying glass changes the diopter of the eye (making it myopic) so that the object can be placed closer to the eye resulting in a larger angular magnification.

===Microscope===
The angular magnification of a [[microscope]] is given by

<math display="block">M_\mathrm{A} = M_\mathrm{o} \times M_\mathrm{e}</math>

where <math display="inline">M_\mathrm{o}</math> is the magnification of the objective and <math display="inline">M_\mathrm{e}</math> the magnification of the eyepiece. The magnification of the objective depends on its [[focal length]] <math display="inline">f_\mathrm{o}</math> and on the distance <math display="inline">d</math> between objective back focal plane and the [[focal plane]] of the [[eyepiece]] (called the tube length):

<math display="block">M_\mathrm{o}={d \over f_\mathrm{o}}</math>

The magnification of the eyepiece depends upon its focal length <math display="inline">f_\mathrm{e}</math> and is calculated by the same equation as that of a magnifying glass:

<math display="block">M_\mathrm{e}={25\ \mathrm{cm} \over f_\mathrm{e}}</math>

Note that both astronomical telescopes as well as simple microscopes produce an inverted image, thus the equation for the magnification of a telescope or microscope is often given with a [[Plus and minus signs|minus sign]].{{Citation needed|date=January 2009}}

===Telescope===
The angular magnification of an [[optical telescope]] is given by

<math display="block">M_\mathrm{A}= {f_\mathrm{o} \over f_\mathrm{e}}</math>

in which <math display="inline">f_\mathrm{o}</math> is the [[focal length]] of the [[objective (optics)|objective]] [[lens (optics)|lens]] in a [[refracting telescope|refractor]] or of the [[primary mirror]] in a [[reflecting telescope|reflector]], and <math display="inline">f_\mathrm{e}</math> is the focal length of the [[eyepiece]].

====Measurement of telescope magnification====
Measuring the actual angular magnification of a telescope is difficult, but it is possible to use the reciprocal relationship between the linear magnification and the angular magnification, since the linear magnification is constant for all objects.

The telescope is focused correctly for viewing objects at the distance for which the angular magnification is to be determined and then the object glass is used as an object the image of which is known as the [[exit pupil]]. The diameter of this may be measured using an instrument known as a Ramsden [[dynameter]] which consists of a Ramsden eyepiece with micrometer hairs in the back focal plane. This is mounted in front of the telescope eyepiece and used to evaluate the diameter of the exit pupil. This will be much smaller than the object glass diameter, which gives the linear magnification (actually a reduction), the angular magnification can be determined from

<math display="block">M_\mathrm{A} = {1 \over M} = {D_{\mathrm{Objective}} \over {D_\mathrm{Ramsden}}}\,.</math>