Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.

Given:

Object Height, $h$ = 5 cm

Focal length, $f$ = $+$6 cm

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

To find: Position and nature of the image $(v)$.

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-

$\frac {1}{v}-\frac {1}{(-10)}=\frac {1}{6}$

$\frac {1}{v}+\frac {1}{10}=\frac {1}{6}$

$\frac {1}{v}=\frac {1}{6}-\frac {1}{10}$

$\frac {1}{v}=\frac {5-3}{30}$

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

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

$v=+15cm$

Thus, the image $v$ is formed at a distance of 15 cm from the convex lens, and the positive $(+)$ sign for image distance implies that the image is placed on the right side of the convex lens.

As the image is formed on the right side of the lens, the nature of the image will be real and inverted.

Hence, the position of the image is on the right side of the lens, and the nature of the image is real and inverted.