Question

An object is placed at a distance of $10cm$ from a convex mirror of a focal length of $15cm$. Find the position and nature of the image.

Answer 1

Step 1: Given data

The focal length of the convex mirror, $f=15cm$

Distance from the mirror of an object, $u=-10cm$(negative sign due to sign convection )

Step 2: Calculation of v using mirror formula

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \frac{1}{v}=\frac{1}{15}-\frac{1}{-10} \\ \frac{1}{v}=\frac{5}{30} \\ \frac{1}{v}=\frac{1}{6} \\ v=6cm \end{gathered}$$

The position of the image is at a distance of 6 cm from the convex mirror on its right side (which is behind the mirror) i.e., a virtual and erect image is formed.

so m = $\frac{-v}{u}=\frac{-6}{-10}=0.6$

Hence, it will be erect and diminished. Thus, the position of the image is at a distance of 6 cm (which is behind the mirror) and the nature of the image is virtual, erect, and diminished.