An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.

Object distance, $u$ = $-$25 cm   (since object is always placed on the left side of the lens, it is taken as negative)

Focal length, $f$ = 10 cm

Height of the object $h$ = $+$5 cm

To find: Position, nature, $v$ of the image, and 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-

$\frac {1}{v}-\frac {1}{(-25)}=\frac {1}{10}$

$\frac {1}{v}+\frac {1}{25}=\frac {1}{10}$

$\frac {1}{v}=\frac {1}{10}-\frac {1}{25}$

$\frac {1}{v}=\frac {5-2}{50}$

$\frac {1}{v}=\frac {3}{50}$

$v=\frac {50}{3}$

$v=+16.6\ cm$

Thus, the image distance $v$ is 16.6 cm from the lens, and the positive $(+)$ sign for image distance implies that the image is formed on the right side of the lens (behind the lens). And, we know that on the right side of the lens real image forms.

Now,

From the magnification formula, we know that:

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

Substituting the given values in the formula we get-

$\frac {\frac {50}{3}}{-25}=\frac {h'}{5}$

$\frac {50}{-25\times {3}}=\frac {h'}{5}$

$-\frac {2}{3}=\frac {h'}{5}$

$h'=-\frac {10}{3}$                            [by cross multiplication]

$h'=-3.3cm$

Thus, the size of the image $h'$ is 3.3 cm, and the negative sign $(-)$ implies that the image is inverted (below the principal axis).

Hence, the position of the image is behind the lens (on the right side), the nature of the image is real and inverted, and the size of the image is smaller than the object (3.3 cm).

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Updated on: 10-Oct-2022

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