Magnification

Also called Linear magnification, M

The ratio of image height to object height for a mirror or lens. A magnitude above 1 means the image is enlarged, below 1 means reduced, and the ratio can also be found from the image and object distances.

The AP Physics 2 sheet prints it wrapped in absolute value bars:

|M| = \left|\frac{h_i}{h_o}\right| = \left|\frac{s_i}{s_o}\right|

so the printed relation answers size only: enlarged, reduced, or the same. It cannot tell you upright from inverted, because the sign that would carry that has been discarded before you see it.

Most textbooks define a signed version, $M = -s_i/s_o$ , where a negative $M$ means inverted. The convention is standard and consistent with the rest of the sign scheme, but the AP Physics 2 sheet does not print it, so a free-response answer that leans on it is defending a convention the reference sheet did not supply. Safer on this exam: take $|M|$ for the size and get orientation from the sign of $s_i$ , since a real image is inverted and a virtual one upright for a single optic.

Reading it.

$|M| > 1$ enlarged, $|M| < 1$ reduced, $|M| = 1$ same size, as a plane mirror gives.
$|M|$ is a ratio of lengths, so it has no units.
Size and image type are independent. A converging optic can give an enlarged real image or an enlarged virtual one, and a diverging lens gives $|M| < 1$ always.

The second equality is the useful one, because $s_i$ and $s_o$ are what $\dfrac{1}{s_i} + \dfrac{1}{s_o} = \dfrac{1}{f}$ hands you. Topic 13.4 works the cases.

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