An object is placed at a distance of 10 cm from a convex lens of focal length 20 cm.

Given,

• Object distance (u) = - 10 cm
• Focal Length (f) = + 20 cm

Let, image distance be 'v'.

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1em] ⇒\frac{1}{\text v} = \frac{1}{\text f} + \frac{1}{\text u} \\[1em] ⇒ \frac{1}{\text v} = \frac{1}{20}+\frac{1}{(-10)}=\frac{1}{20}-\frac{1}{10}= \frac{1-2}{20}\\[1em] \frac{1}{\text v} = -\frac{1}{20} \\[1em] ⇒ \text v = -\frac{20}{1} = -20 \text { cm}\end{aligned}$$

So, the image will form at a distance of 20 cm in front of the convex lens.

(b) As,

$\text {magnification} = \frac {\text v}{\text u}= \frac {-20}{-10} = +\ 2$

Here, sign of magnification is positive implying virtual and erect image formation and since magnification is 2 so size of the image is 2 times the size of the object.

Hence, formed images is virtual, erect and 2 times the size of the object.