An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.

Object distance, $u$ = $-$10 cm     (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = $+$20 cm

Object height, $h$ = 4 cm

To find: Position and nature of the image $(v)$, Size of the image $(h')$.

Solution:

According to the lens formula, we know that:

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

Substituting the given values in the formula 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 image $v$ is formed at a distance of 20 cm from the convex lens, and the negative $(-)$ sign for image distance implies that the image is formed on the left side of the convex lens.

As the image is formed on the left side of the lens, the nature of the image will be virtual and erect.

Now,

According to magnification formula, we know that:

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

$m=\frac {-20}{-10}$

$m=+2$

Thus, the magnification of the image is 2, which is more than 1, so the image will be larger than the object. And, the positive sign $(+)$ for magnification implies that the image is formed above the principal axis.

Again,

According to the magnification formula, we know that:

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

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

$h'=4\times {2}$

$h'=8cm$

Thus, the height or size of the image, $h'$ is 8cm.

Hence, the position of the image is on the left side of the lens, the nature of the image is virtual and erect, and the size of the image is 8cm.