Magnification from Heights (m = h_i/h_o)

This is magnification in its most direct form: how tall the image is compared with the thing it is an image of. Everything else said about magnification is derived from this ratio, including the $-d_i/d_o$ form on the companion page, which is really just this quantity re-expressed in terms of distances via similar triangles. Because it is a ratio of two lengths it is dimensionless, and the only unit discipline required is that both heights be measured in the same unit — millimetres over millimetres, micrometres over micrometres, it does not matter which so long as they match.

A microscope objective marked 40× projects a 5 µm red blood cell as a 200 µm image for the eyepiece to work on. Run it backwards for a photographic problem: a full-frame sensor is 24 mm tall, so to fit a person 1.8 m tall into the frame you need $|m| = 24/1800 = 0.0133$, and the companion page then tells you which lens and distance deliver it. Or take the Moon, 3474 km across at a distance of 384 400 km: a 500 mm lens images it $500 \times 3474/384400 = 4.5$ mm across, which is why lunar photography needs a very long lens to fill any part of a frame.

The two magnification relations are designed to be chained. Set them equal and you get $h_i/h_o = -d_i/d_o$, which is the practical heart of thin-lens problem solving: the thin-lens equation hands you $d_i$, the ratio form converts that into a magnification, and this page converts the magnification into a physical size you can compare against a sensor, a screen or a detector. Almost every lens problem you will meet is those three steps in some order.

The sign convention lives in $h_i$, and this is where it bites. Heights measured above the optical axis are positive and heights below it are negative, so an inverted image has a genuinely negative $h_i$ — not a height smaller than nothing, but a height measured downward from the axis. The usual mistake is to enter the physical size of an inverted image as a positive number, which produces a magnification of the right magnitude and the wrong sign, and then to carry that error into whatever comes next. The object height $h_o$ is conventionally taken as positive, with the object upright; making it negative inverts your own reference frame and flips everything. And keep this linear magnification separate from the other things called magnification: a telescope's or microscope's angular magnification is a ratio of apparent angles, not of heights, and the "3×" on a zoom lens is a ratio of focal lengths and is not a magnification at all.