Question

An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.

Answer 1

Distance of the object (u) = $-$6 cm
Focal length of the convex mirror (f) = 12 cm
We have to find the position of the image (v) and the nature of the image.
Using the mirror formula, we get:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{1}{12}=\frac{1}{v}+\frac{1}{-6} \\ \Rightarrow \frac{1}{12}=\frac{1}{v}-\frac{1}{6} \\ \Rightarrow \frac{1}{12}+\frac{1}{6}=\frac{1}{v} \\ \Rightarrow \frac{1}{12}+\frac{2}{12}=\frac{1}{v} \\ \Rightarrow \frac{1}{v}=\frac{3}{12} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{4} \\ \Rightarrow \mathrm{v}=4\mathrm{cm} \end{gathered}$$

Thus, the distance of the image (v) is 4 cm behind the mirror.
Now, using the magnification formula, we get:
$$\begin{gathered} \mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}} \\ \mathrm{m}=\frac{-4}{-6}=\frac{2}{3} \\ \mathrm{m}=0.6 \end{gathered}$$

Thus, the image is virtual, erect and small.