An object of height 4 cm is placed at a distance of 15 cm in front of a concave lens of power, −10 dioptres. Find the size of the image.

Object height, $h$ = 4 cm

Object distance, $u$ = $-$15 cm   (object distance is always taken negative)

Power of the lens, $P$ = $-$10 D

To find: Size of the image $h'$. 

Solution:

Power of the lens is given by-

$P=\frac {1}{f}$

Substituting the value of power, $P$ we get-

$-10=\frac {1}{f}$

$f=-\frac {1}{10}$

$f=-0.1m=10cm$

Thus, the focal length $f$ of the lens is 10cm.

From the lens formula, we know that-

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

Substituting the given values we get-

$\frac {1}{v}-\frac {1}{(-15)}=\frac {1}{(-10)}$

$\frac {1}{v}+\frac {1}{15}=-\frac {1}{10}$

$\frac {1}{v}=-\frac {1}{10}-\frac {1}{15}$

$\frac {1}{v}=\frac {-3-2}{30}$

$\frac {1}{v}=-\frac {5}{30}$

$\frac {1}{v}=-\frac {1}{6}$

$v=-6cm$

Thus, the distance of image $v$ is 6 cm in front of the lens.

Now, 

From the magnification formula, we know that-

$m=\frac {v}{u}=\frac {h'}{h}$

Sustituting the given value, we get-

$\frac {-6}{-15}=\frac {h'}{4}$

$\frac {2}{5}=\frac {h'}{4}$

$5h'=2\times {4}$

$h'=\frac {8}{5}$

$h'=+1.6cm$

Thus, the size of the image $h'$ is 1.6cm, and the positive sign implies that the image is erect and virtual in nature.