An object 2 cm tall is placed on the axis of a convex lens of focal length 5 cm at a distance of 10 m from the optical centre of the lens. Find the nature, position and size of the image formed. Which case of image formation by convex lenses is illustrated by this example?

Height of the object $h$ = 2 cm

Focal length, $f$ = 5 cm

Object distance, $u$ = $-$10 m = $-$1000 cm(since object is always placed on the left side of the lens, it is taken as negative)

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-

$\frac {1}{v}-\frac {1}{(-1000)}=\frac {1}{5}$

$\frac {1}{v}+\frac {1}{1000}=\frac {1}{5}$

$\frac {1}{v}=\frac {1}{5}-\frac {1}{1000}$

$\frac {1}{v}=\frac {200-1}{1000}$

$\frac {1}{v}=\frac {199}{1000}$

$v=\frac {1000}{199}$

$v=+5.02cm$

Thus, the image distance $v$ is 5.02 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-

$\frac {5.02}{-1000}=\frac {h'}{2}$

$-\frac {502}{100000}=\frac {h'}{2}$

$h'=-\frac {502\times {2}}{100000}$

$h'=-\frac {1004}{100000}$

$h'=-0.01cm$

Thus, the size of the image $h'$ is 0.01 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), nature of the image is real and inverted, and size of the image is highly diminished (0.01 cm).

From the above results, we conclude that:

When an object is placed beyond $2f$, the image is real, inverted, and diminished and is formed between $f$ and $2f$.