Draw ray diagrams to show the formation of a three times magnified
(i) real image
(ii) virtual image of an object kept in front of a converging lens. Mark the positions of object, F, 2F, O and position of image clearly in the diagram.
An object of size 5 cm is kept at a distance of 25 cm from the optical centre of a converging lens of focal length 10 cm. Calculate the distance of the image from the lens and size of the image.

(i) The ray diagram for three times magnified real image is shown below: 





(ii) The ray diagram for three times magnifield virtual image is shown below: 



In the question given,

Size of the object, h = + 5 cm
Object distance, u = - 25 cm
Focal length, f = + 10 cm
Image distance, v = ?
Size of the image, h' = ? 

Using lens formula, 

               $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}} \\ = \frac{1}{10}+\frac{1}{-25} \\ = \frac{5-2}{50} = \frac{3}{50} \end{gathered}$$

Image distance, $\mathrm{v} = \frac{50}{3} \mathrm{cm} = 16.67 \mathrm{cm}$ 

Also,
Magnification, $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = \frac{\mathrm{v}}{\mathrm{u}}$

$\Rightarrow$   $\frac{\mathrm{h}\text{'}}{+5 \mathrm{cm}} = \frac{50/3}{-25} = -\frac{2}{3}$

i.e., $\mathrm{h}\text{'} = -\frac{2\times 5}{3}\mathrm{cm} = -3.33 \mathrm{cm}$ is the size of the image.