Magnification

For real images, such as images projected on a screen, size means a linear dimension (measured, for example, in millimeters or inches).

For optical instruments with an eyepiece, the linear dimension of the image seen in the eyepiece (virtual image at infinite distance) cannot be given, thus size means the angle subtended by the object at the focal point (angular size). Strictly speaking, one should take the tangent of that angle (in practice, this makes a difference only if the angle is larger than a few degrees). Thus, angular magnification is given by:

$${\displaystyle M_{A}={\frac {\tan \varepsilon }{\tan \varepsilon _{0}}}\approx {\frac {\varepsilon }{\varepsilon _{0}}}}$$

where ${\textstyle \varepsilon _{0}}$ is the angle subtended by the object at the front focal point of the objective and ${\textstyle \varepsilon }$ is the angle subtended by the image at the rear focal point of the eyepiece.

For example, the mean angular size of the Moon's disk as viewed from Earth's surface is about 0.52°. Thus, through binoculars with 10× magnification, the Moon appears to subtend an angle of about 5.2°.

By convention, for magnifying glasses and optical microscopes, where the size of the object is a linear dimension and the apparent size is an angle, the magnification is the ratio between the apparent (angular) size as seen in the eyepiece and the angular size of the object when placed at the conventional closest distance of distinct vision: 25 cm from the eye.

A Thin lens where black dimensions are real, grey are virtual. The direction of the arrows can be used to describe cartesian +/− signage: from the centre of the lens, left or down = negative, right or up = positive.

The linear magnification of a thin lens is

$${\displaystyle M={f \over f-d_{\mathrm {o} }}}$$

where ${\textstyle f}$ is the focal length and ${\textstyle d_{\mathrm {o} }}$ is the distance from the lens to the object. For real images, ${\textstyle M}$ is negative and the image is inverted. For virtual images, ${\textstyle M}$ is positive and the image is upright.

With ${\textstyle d_{\mathrm {i} }}$ being the distance from the lens to the image, ${\textstyle h_{\mathrm {i} }}$ the height of the image and ${\textstyle h_{\mathrm {o} }}$ the height of the object, the magnification can also be written as:

$${\displaystyle M=-{d_{\mathrm {i} } \over d_{\mathrm {o} }}={h_{\mathrm {i} } \over h_{\mathrm {o} }}}$$

Note again that a negative magnification implies an inverted image.

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 sign convention (where the x-axis is the optical axis) the value for h_i will be negative, and as a result M will also be negative. However, the traditional sign convention used in photography is "real is positive, virtual is negative".[1] Therefore, in photography: Object height and distance are always real and positive. When the focal length is positive the image's height, distance and magnification are real and positive. Only if the focal length is negative, the image's height, distance and magnification are virtual and negative. Therefore, the photographic magnification formulae are traditionally presented as

$${\displaystyle {\begin{aligned}M&={d_{\mathrm {i} } \over d_{\mathrm {o} }}={h_{\mathrm {i} } \over h_{\mathrm {o} }}\\&={f \over d_{\mathrm {o} }-f}={d_{\mathrm {i} }-f \over f}\end{aligned}}}$$

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

$${\displaystyle M_{\mathrm {A} }={25\ \mathrm {cm} \over f}}$$

Here, ${\textstyle f}$ is the focal length of the lens in centimeters. The constant 25 cm is an estimate of the "near point" distance of the eye—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

$${\displaystyle M_{\mathrm {A} }={25\ \mathrm {cm} \over f}+1}$$

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.

The angular magnification of a microscope is given by

$${\displaystyle M_{\mathrm {A} }=M_{\mathrm {o} }\times M_{\mathrm {e} }}$$

where ${\textstyle M_{\mathrm {o} }}$ is the magnification of the objective and ${\textstyle M_{\mathrm {e} }}$ the magnification of the eyepiece. The magnification of the objective depends on its focal length ${\textstyle f_{\mathrm {o} }}$ and on the distance ${\textstyle d}$ between objective back focal plane and the focal plane of the eyepiece (called the tube length):

$${\displaystyle M_{\mathrm {o} }={d \over f_{\mathrm {o} }}}$$

The magnification of the eyepiece depends upon its focal length ${\textstyle f_{\mathrm {e} }}$ and is calculated by the same equation as that of a magnifying glass (above).

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 minus sign.

The angular magnification of an optical telescope is given by

$${\displaystyle M_{\mathrm {A} }={f_{\mathrm {o} } \over f_{\mathrm {e} }}}$$

in which ${\textstyle f_{\mathrm {o} }}$ is the focal length of the objective lens in a refractor or of the primary mirror in a reflector, and ${\textstyle f_{\mathrm {e} }}$ is the focal length of the eyepiece.

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

$${\displaystyle M_{\mathrm {A} }={1 \over M}={D_{\mathrm {Objective} } \over {D_{\mathrm {Ramsden} }}}\,.}$$