Describe the nature of image formed when an object is placed at a distance of 30 cm from a convex lens of focal length 15 cm.

Object distance, $u$ = $-$30 cm     (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = $+$15 cm

To find: Nature of 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-

$\frac {1}{v}-\frac {1}{(-30)}=\frac {1}{15}$

$\frac {1}{v}+\frac {1}{30}=\frac {1}{15}$

$\frac {1}{v}=\frac {1}{15}-\frac {1}{30}$

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

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

$v=+30cm$

Thus, the image $v$ is formed at a distance of 30 cm from the convex lens, and the positive $(+)$ sign for image distance implies that the image is formed 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.