An object 2 cm high is placed at a distance of 64 cm from the screen. On placing a convex lens 32 cm from the object it is found that a distinct image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.

Given, a convex lens.

Size of the object, h = + 2 cm
Object distance, u = -32 cm
Image distance, v = 64 - 32 = + 32 cm    [v is +ve for a real image]

Now, using the lens formula,  

                       $$\begin{gathered} \frac{1}{\mathrm{f}} = \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{32}-\frac{1}{-32} \\ = \frac{1}{32}+\frac{1}{32} \\ =\frac{2}{32} =\frac{1}{16} \end{gathered}$$ 

$\therefore$ Focal length of the lens, f = + 16 cm.
 
Magnification, m = $\frac{\mathrm{h}\text{'}}{\mathrm{h}} =\frac{\mathrm{v}}{\mathrm{u}}$ 

$\therefore$  Image size, $\mathrm{h}\text{'} = \frac{\mathrm{vh}}{\mathrm{u}} = \frac{(+32) (+2)}{(-32)} = -2 \mathrm{cm}.$ 

The negative sign of the image implies that the image is inverted and since the image distance, v is positive it is real.

The ray diagram below shows us how the image is formed after refraction by the lens: