**Question:**

An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

**Solution:**

Size of the object, *h*1 = 3 cm

Object distance, *u* = −14 cm

Focal length of the concave lens, *f *= −21 cm

Image distance = *v*

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}=\frac{-2-3}{42}=\frac{-5}{42}$

$\therefore v=-\frac{42}{5}=-8.4 \mathrm{~cm}$

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

$m=\frac{\text { Image height }\left(h_{2}\right)}{\text { Object height }\left(h_{1}\right)}=\frac{v}{u}$

$\therefore h_{2}=\frac{-8.4}{-14} \times 3=0.6 \times 3=1.8 \mathrm{~cm}$

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

According to the lens formula, we have the relation:

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