According to the question;

Object distance = u;

Image distance (v) = -15cm;

Focal length = -20cm

**By lens formula;**

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

**⇒ ****\(\frac1{15}-\frac1u=\frac1{-20}\)**

⇒ **\(\frac1{u}=\frac1{20}-\frac1{15}\)**

⇒ **\(\frac1{u}=\frac{3-4}{60}=\frac{-1}{60}\)**

⇒ u = -60cm.

**Therefore, object is placed at 60 cm in front of lens. **

Now;

Height of object h_{1}= 5cm;

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

Putting values of v and u

Magnification = \(\frac{h_2}{5}=\frac{-15}{-60}\)

⇒ \(\frac{h_2}{5}=\frac{1}{4};\)

⇒ h_{2} = \(\frac54\) = 1.25

Height of image is 1.25 cm.

**Positive sign means image is virtual.**