Question

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

Answer 1

Step 1 - Given data

The distance of the object from the pole of the mirror $u=-10cm$ (Negative sign shows that the object is placed in the front of the mirror)

Focal length $f=15cm$ (For convex mirror focal length is positive)

Step 2 - Finding the formula for the position of the image

The mirror equation is given by

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Where $v$ is the distance of the image from the pole of the mirror.

Rearrange the above equation.

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \frac{1}{v}=\frac{u-f}{uf} \\ v=\frac{uf}{u-f} \end{gathered}$$

Step 3 - Finding the position of the image

Substitute the values into the above formula.

$$\begin{gathered} v=\frac{(-10cm)(15cm)}{(-10cm)-(15cm)} \\ v=\frac{-150}{-25} \\ v=+6cm \end{gathered}$$

Therefore, the image is formed behind the mirror at $6cm$ from the pole of the mirror.

Step 4 - Finding the nature of the object

The magnification $\left(m\right)$ can be calculated as

$$\begin{gathered} m=-\frac{v}{u} \\ m=-\frac{(6cm)}{(-10cm)} \\ m=\frac{6}{10} \\ m=0.6 \end{gathered}$$

Therefore, the image is erect, virtual, and smaller in size by a factor of $0.6$.

Final answer- The image is formed behind the mirror at $6cm$ from the pole of the mirror. The nature of the image is virtual, erect and smaller in size than the object.