A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image formed.

Given:

Object height, $h$ = $+$10 cm

Focal length, $f$ = $+$12 cm

 Object distance, $u$ = $-$18 cm

To find: The position and nature of the image, $v$, size of the image, $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

$\frac {1}{v}-\frac {1}{(-18)}=\frac {1}{12}$

$\frac {1}{v}+\frac {1}{18}=\frac {1}{12}$

$\frac {1}{v}=\frac {1}{12}-\frac {1}{18}$

$\frac {1}{v}=\frac {3-2}{36}$

$\frac {1}{v}=\frac {1}{36}$

$v=+36cm$

Thus, the position of the image, $v$ is 36 cm away from the lens, and the plus sign $(+)$ implies that it forms behind the lens (on the right side). So, the image is real.

Now, 

From the magnification formula of the lens, we know that-

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

Putting the values in the formula we get-

$\frac {h'}{10}=\frac {36}{-18}$

$h'=-2\times 10$

$h'=-20cm$

Thus, the size of the image, $h'$ is 20 cm.

As the size of the image is negative then it is inverted (below the principal axis), and its value is more than the size of the object, so the image is magnified.

Thus, we can conclude that the nature of the image is real and inverted. The size of the image is mgnified.

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Updated on: 10-Oct-2022

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