An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) focal length of the lens, and (ii) size of the image.

Object height, $h$ = 4.25 mm = 0.425 cm    $(\because 1cm=10mm)$

Object distance, $u$ = $-$10 cm

Power, $P$ = $+$5 D

To find: (i)Focal length $f$,  (ii) Size of the image $h'$.              

Solution: (i) 

Power of the lens is given by-

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

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

$5=\frac {1}{f}$

$f=\frac {1}{5}$

$f=+0.2m=20cm$

Thus, the focal length $f$ of the lens is 20cm.

Solution: (ii) 

From the lens formula, we know that-

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

Substituting the given values we get-

$\frac {1}{v}-\frac {1}{(-10)}=\frac {1}{20}$

$\frac {1}{v}+\frac {1}{10}=\frac {1}{20}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{10}$

$\frac {1}{v}=\frac {1-2}{20}$

$\frac {1}{v}=-\frac {1}{20}$

$v=-20cm$

Thus, the distance of image $v$ is 20cm 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-

$\frac {-20}{-10}=\frac {h'}{0.425}$

$2=\frac {h'}{0.425}$

$h'=2\times {0.425}$

$h'=+0.85cm=+8.5mm$

Thus, the size of the image $h'$ is 8.5cm, and the positive sign implies that the image is erect and virtual in nature.