Q.A convex mirror used as a car's rear-view mirror has a focal length of $20\ \text{cm}$. A second car is $10\ \text{m}$ behind it. Find the position and magnification of the image formed.

For the convex mirror, $f=+20\ \text{cm}$ (positive, since a convex mirror's focus is behind the mirror) and $u=-1000\ \text{cm}$ (the car $10\ \text{m}=1000\ \text{cm}$ away). From the mirror formula, $\frac1v=\frac1f-\frac1u=\frac1{20}-\frac1{-1000}=\frac1{20}+\frac1{1000}=\frac{50+1}{1000}=\frac{51}{1000}$, so $v=\frac{1000}{51}\approx19.6\ \text{cm}$. The positive $v$ confirms a virtual image behind the mirror. The magnification is $m=-v/u=-\frac{19.6}{-1000}\approx0.0196\approx\frac1{51}$, positive (erect) and much less than 1 (greatly diminished). This is exactly why a convex mirror is used as a rear-view mirror: whatever the object's actual distance, the image always stays virtual, erect and diminished, and the diminishe …

Substitute the (positive) convex-mirror focal length and the (negative) object distance into the mirro …

• Assigning a negative focal length to a convex mirror.
• Forgetting to convert the object distance to the same unit (cm) as the focal length before substituting. …