Magnification (lateral, angular, longitudinal)

Magnification describes how an optical system scales what passes through it, and three distinct ratios go by the name. Lateral (transverse) magnification is the height of an image divided by the height of the object, $m = h'/h$; a 50 mm lens imaging an object 150 mm away forms an image 75 mm behind it at $m = -0.5$, inverted and half size. Longitudinal magnification is the ratio of a small axial image displacement to the axial object displacement that causes it, equal to $m^2$ in air, 0.25 in the same example. Angular magnification is the ratio of the angles a ray or an apparent object subtends after and before the system, the quantity that describes telescopes and magnifying glasses, whose images are at infinity and have no finite height.

Lateral magnification

For a thin lens in air with object distance $s$ and image distance $s'$, both positive for a real object and a real image,

$$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}, \qquad m = -\frac{s'}{s}.$$

The negative sign records inversion: a real image formed by a single positive lens is upside down. An object at $2f$ images at $2f$ with $m = -1$; as the object recedes, the image approaches the focal plane and $m$ tends to zero; as the object approaches $f$, the image recedes and $|m|$ grows without bound. For a thick lens or a multi-element system the same relations hold with $s$ and $s'$ measured from the principal planes. The location of the image is covered under image plane.

Lateral magnification is measured by imaging a scale of known pitch, such as a stage micrometer, and comparing the image pitch on the camera with it. In a camera system it sets the object-side pixel size: a 6.5 µm pixel behind a system with $|m| = 20$ samples 0.325 µm of the object.

Longitudinal magnification

Differentiating the imaging equation gives $ds'/ds = -s'^2/s^2$, so a small axial movement of the object moves the image by $m^2$ times as much, in the same direction along the light. Between media of indices $n$ and $n'$ the result becomes $m^2 n'/n$. A three-dimensional object is therefore imaged with its depth scaled by the square of its lateral scale: at $m = -0.5$ the depth shrinks by a factor of 4, and at $m = -20$ it is stretched by a factor of 400. This is why a microscope's depth of field in the specimen is tiny while the tolerance on camera position is large, the relation set out in the depth of focus entry.

Angular magnification

An afocal system such as a beam expander or a telescope takes parallel light in and gives parallel light out, so it has no finite image distance. Its angular magnification is the ratio of output to input ray angles, $M = -f_1/f_2$ for two lenses of focal lengths $f_1$ and $f_2$ separated by $f_1 + f_2$. The beam diameter scales by the inverse, $f_2/f_1$: a system that magnifies angles by 10 shrinks the beam by 10. The product of lateral and angular magnification at a pair of conjugate planes is fixed by the Lagrange invariant, $n h u = n' h' u'$, so any gain in size is paid for with a loss in angle.

A simple magnifier is a positive lens held close to the eye with the object at or just inside its focal point. With the image at infinity its angular magnification, relative to the object viewed unaided at the conventional near point of 250 mm, is

$$M = \frac{250\ \text{mm}}{f},$$

10× for a 25 mm lens; placing the virtual image at the near point instead gives $1 + 250/f$, or 11×. The same formula rates microscope eyepieces, and the total visual magnification of a microscope is the product of the objective's lateral magnification and the eyepiece's angular magnification, discussed under microscope objective.

Pitfalls

Magnification does not determine resolution. Beyond the point where the Airy disk of the optics spans a few pixels or a few minutes of arc at the eye, further magnification enlarges blur without revealing detail ("empty magnification"). Distortion makes lateral magnification vary across the field, so precise measurements calibrate it at several field positions.

Common questions

What does a negative magnification mean?

The image is inverted relative to the object. A single positive lens forming a real image always gives negative $m$; a second imaging stage, or a mirror pair, restores an upright image.

How is magnification related to focal length?

For an object at distance $s$, $m = -f/(s - f)$. For a distant object $|m| \approx f/s$, so a longer focal length gives a proportionally larger image; a 100 mm lens gives twice the image size of a 50 mm lens of a subject far away.

Why is longitudinal magnification the square of lateral magnification?

Because image distance depends on object distance through the reciprocal relation of the lens equation, whose derivative is $-(s'/s)^2$. Depth is scaled by $m^2$ while width and height are scaled by $m$.

References: E. Hecht, Optics, 5th ed. (Pearson, 2017); W. J. Smith, Modern Optical Engineering, 4th ed. (McGraw-Hill, 2008); M. Born and E. Wolf, Principles of Optics, 7th ed. (Cambridge University Press, 1999).