A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the nature, position and size of the image. Also find its magnification.

We are given a convex lens.
Here,
Object size, h = + 5 cm
Focal length, f = +20 cm   [f is +ve for a convex lens]
Object distance, u = -30 cm
Image distance, v = ?
Image size = ? 


For a lens using the lens formula,

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

$\therefore$                 $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{u}}+\frac{1}{\mathrm{f}} \\ = \frac{1}{-30}+\frac{1}{20} \\ = \frac{-2+3}{60} \\ = \frac{1}{60} \end{gathered}$$

i.e., Image distance,  $\mathrm{v}= + 60 \mathrm{cm}$ 

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

$\therefore$$\mathrm{m} = \frac{\mathrm{v}}{\mathrm{u}} = \frac{+60 \mathrm{cm}}{-30 \mathrm{cm}} = -2.$

Also, Image size,  
                           $$\begin{gathered} \mathrm{h}\text{'} = \frac{\mathrm{vh}}{\mathrm{u}} \\ = \frac{(+60) \times (+5)}{(-30)} \\ = -10 \mathrm{cm}. \end{gathered}$$

The positive sign of v shows that the image is formed at a distance of 60 cm to the right of optical centre of the lens. Negaticve sign of h' implies that the image formed is inverted. 

Therefore, the image is real and inverted. 

Thus, a real and inverted image which is 10 cm tall, is formed at a distance of 60 cm on the right side of the lens.
The image is two times enlarged in size than the object.