A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Calculate the position, nature and size of the image.

 Given:

Focal length, $f$ = 0.10 m = 10 cm

Object distance, $u$ = $-$0.08 m = $-$8 cm

Height of the object, $h$ = 5 mm = 0.5 cm

To find: Position and nature of the image $v$, and size of the image $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the value of $f$ and $v$ in the formula we get-

$\frac {1}{v}-\frac {1}{(-8)}=\frac {1}{10}$

$\frac {1}{v}+\frac {1}{8}=\frac {1}{10}$

$\frac {1}{v}=\frac {1}{10}-\frac {1}{8}$

$\frac {1}{v}=\frac {4-5}{40}$

$\frac {1}{v}=-\frac {1}{40}$

$v=-40cm$

Thus, the image $v$ is formed at a distance of 40 cm from the convex lens, and the positive $-$ sign for image distance implies that the image is placed on the left side of the convex lens.

Now,

From the magnification formula, we know that-

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

Substituting the given values in the formula we get-

$\frac {-40}{-8}=\frac {h'}{0.5}$

$5=\frac {h'}{0.5}$

$h'=0.5\times {5}$

$h'=+2.5cm\ or\ 25mm$

Thus, size of the image $h'$, is 2.5 cm, and the positive sign implies that the image is erect (upward) or above the principal axis.

Hence, the size of the image is 2.5cm, and the nature of the image is virtual and erect as the image is formed on the left side and above the principal axis.