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# An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm.**(i) **What is the nature of image?**(ii)** What is the position of image?

Object distance, $u$ = $-$100 cm (negative sign shows that the object is placed on the left side of the lens)

Focal length, $(f)$ =40 cm

To find: (i) Nature of image.

(ii) Position of image.

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-

$\frac {1}{v}-\frac {1}{(-100)}=\frac {1}{40}$

$\frac {1}{v}+\frac {1}{100}=\frac {1}{40}$

$\frac {1}{v}=\frac {1}{40}-\frac {1}{100}$

$\frac {1}{v}=\frac {5-2}{200}$

$\frac {1}{v}=\frac {3}{200}$

$v=\frac {200}{3}$

$v=+66.6cm$

Thus, the image $v$ is formed at a distance of 66.6 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,

(i) The nature of the image is real and inverted.

(ii) Position of the image is on the right side of the convex lens.