According to the question;

Object distance = u;

Image distance (v) = 20cm;

Focal length = 10cm

**By lens formula;**

**\(\frac1v-\frac1u=\frac1f\)**

**⇒ ****\(\frac1{20}-\frac1u=\frac1{10}\)**

⇒ **\(\frac1{u}=\frac1{20}-\frac1{10}\)**

⇒ **\(\frac1{u}=\frac{1-2}{20}-\frac{-1}{20}\)**

⇒ **\(\frac1{u}=\frac{-1}{10}\)**

⇒ u = -20cm.

**Therefore, object is placed at 20 cm in front of lens. **

Now;

Height of object h_{1}= 2cm;

Magnification = \(\frac{h_2}{h_1}=\frac{v}u\)

Putting values of v and u

Magnification = \(\frac{h_2}{2}=\frac{20}{-20}\)

⇒ \(\frac{h_2}{h_1}= -1;\)

⇒ h_{2} = 2 x -1 = -2

Height of image is 2 cm.

Negative sign means image is inverted.

**Thus, the image is real, inverted and same as size of object. **

Diagram below shows the image formation.