Q.A concave mirror of focal length $15\ \text{cm}$ forms an image of an object placed $10\ \text{cm}$ from the mirror. Find the position, nature and magnification of the image.

Using the sign convention, $u=-10\ \text{cm}$ (object always negative) and $f=-15\ \text{cm}$ (concave mirror, focus in front of the mirror). Substituting into the mirror formula, $\frac1v=\frac1f-\frac1u=\frac1{-15}-\frac1{-10}=-\frac1{15}+\frac1{10}=\frac{-2+3}{30}=\frac1{30}$, so $v=+30\ \text{cm}$. The positive sign of $v$ means the image forms behind the mirror, i.e. it is virtual. The magnification is $m=-v/u=-\frac{30}{-10}=+3$; the positive sign confirms the image is erect (same orientation as the object), and $|m|=3$ means it is three times the size of the object. This is exactly the physical behaviour expected: since the object ($10\ \text{cm}$) lies between the pole and the focus ($15\ \text{cm}$) of a concave mirror, the mirror must produce a virtual, erect, magnified image — precisely the working principle used, later in this chapter, for a shaving/make-up mirror.

Substitute the correctly-signed $u$ and $f$ into the mirror formula, solve for $v$, then use $m=-v/u$ to find the nature and size of the image.

• Using $f=+15\ \text{cm}$ for a concave mirror instead of the correct negative sign.
• Reading a positive $v$ as a real image (for a mirror, positive $v$ means virtual).
• Forgetting the minus sign in $m=-v/u$ and getting the wrong orientation.