An object of length 9 𝑐𝑚 kept at a distance 30 𝑐𝑚 from a concave lens of focal length 60 𝑐𝑚. Determine the nature, position and length of the image formed using lens formula. Also, draw a labelled ray diagram to show the image formation in the above case.

For concave lens 

Focal length 𝑓 =− 60 𝑐𝑚 

Height of the object ℎ =+ 9 𝑐𝑚 

Object distance 𝑢 =− 30 𝑐𝑚 

Image distance, 𝑣 =? 

Length of an image formed ℎ' =? 

According to lens formula

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

$\Rightarrow \frac{1}{-60}=\frac{1}{v}-\frac{1}{-30}$

$\Rightarrow \frac{1}{-60}=\frac{1}{\mathrm{v}}+\frac{1}{30}$

$\Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{60}-\frac{1}{30}$

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{-1-2}{60}$

$\Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{20}$

$\Rightarrow \mathrm{v}=-20 \mathrm{cm}$

According to the linear magnification of the lens,

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{{\mathrm{h}}^{\text{'}}}{\mathrm{h}} \\ \Rightarrow \frac{-20}{-30}=\frac{{\mathrm{h}}^{\text{'}}}{9} \\ \Rightarrow \frac{2}{3}=\frac{{\mathrm{h}}^{\text{'}}}{9} \\ \Rightarrow {\mathrm{h}}^{\text{'}}=\frac{18}{3} \\ \Rightarrow {\mathrm{h}}^{\text{'}}=6 \mathrm{cm} \end{gathered}$$

The labelled ray diagram,