A divergent lens has a focal length of $20cm$. At what distance should an object of height $4cm$ from the optical centre of the lens be placed so that its image is formed $10cm$ away from the lens. Find the size of the image also.

Step 1 - Given data

The focal length $\left(f\right)$ of the divergent lens is, $f=-20cm$. (The focal length of the divergent or concave lens is negative)
The image distance $\left(v\right)$ from the lens is, $v=-10cm$. (In the case of concave lens, the image is always formed in front of the lens, so the image distance will be negative)

The height of the object $\left(h_o\right)$ is $4cm$.

Step 2 - Formulae used
The object distance $\left(u\right)$ can be found using the lens formula, given as: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$.
The height of the image $\left(h_i\right)$ can be found using the magnification formula, given as: $m=\frac{h_i}{h_o}=\frac{v}{u}$. Where $m$ is the magnification of the lens.

Step 3 - Finding the object distance.

Substitute all the values into the lens formula to calculate the object distance.
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{-20}=\frac{1}{-10}-\frac{1}{u} \\ \frac{1}{u}=\frac{-1}{10}+\frac{1}{20} \\ \frac{1}{u}=\frac{-1}{10}+\frac{1}{20} \\ \frac{1}{u}=\frac{-1}{20} \\ u=-20cm \end{gathered}$$

Step 4 - Finding the size of the image.

Substitute all the values into the formula for magnification to calculate the size of the image.

$$\begin{gathered} \frac{h_i}{h_o}=\frac{v}{u} \\ {h}_i=\frac{v}{u}{h}_o \\ {h}_i=\frac{-10}{-20}\left(4\right) \\ {h}_i=2cm \end{gathered}$$

1. Hence, the object should be placed $-20cm$ away from the lens. The negative sign shows that the object must be placed in front of the lens.
2. The size of the image is $2cm$.
3. The nature of the image is virtual, erect, and diminished.