An object of size 3 cm is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Calculate the distance and size of the image so formed. What will be the nature of the image?

Given,
Size of the object, h = + 3 cm
Object distance, u = - 15 cm
Focal length, f = + 10 cm   [f is +ve for a convex lens]
By lens formula,           
                          $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} = \frac{1}{\mathrm{f}}$ 

$\therefore$                      $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{u}}+\frac{1}{\mathrm{f}} \\ = \frac{1}{-15}+\frac{1}{10} \\ = \frac{-2+3}{30} \\ = \frac{1}{30} \end{gathered}$$ 

i.e., Image distance, $\mathrm{v} = + 30 \mathrm{cm}.$

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

$\therefore$ Image size, $\mathrm{h}\text{'} = \frac{\mathrm{vh}}{\mathrm{u}} = \frac{(+30) (+3)}{-15} = -6 \mathrm{cm}.$

As v is +ve and h' is negative, so a real and inverted image is formed at 30 cm behind the lens.