Q.A concave mirror forms a real image, three times the size of the object, of an object placed $20\ \text{cm}$ in front of it. Find the focal length of the mirror.

A real, inverted image has negative magnification, so $m=-3$. From $m=-v/u$: $-3=-v/(-20)\Rightarrow v/20=-3\Rightarrow v=-60\ \text{cm}$ (real image, negative as expected). Substituting into the mirror formula, $\dfrac1f=\dfrac1v+\dfrac1u=\dfrac1{-60}+\dfrac1{-20}=-\dfrac1{60}-\dfrac3{60}=-\dfrac4{60}=-\dfrac1{15}$, so $f=-15\ \text{cm}$, i.e. a concave mirror of focal length $15\ \text{cm}$. This is consistent physically: with $f=15\ \text{cm}$, the object at $u=20\ \text{cm}$ lies between $f$ ($15\ \text{cm}$) and $2f$ ($30\ \text{cm}$), exactly the range in which a concave mirror produces a real, inverted, magnified image, as found.

Use $m=-v/u$ with $m=-3$ (real image) to find $v$, then substitute $u$ and $v$ into the mirror formula to find $f$.

• Using $m=+3$ instead of $-3$ (forgetting a real image from a concave mirror is inverted, hence negative $m$).
• Sign errors when adding the two negative reciprocals in the final mirror-formula step.