Question

An object $5cm$ in length is held $25cm$ away from converging lens of focal length $10cm$. Draw the ray diagram and find the position, size and nature of the image formed.

Answer 1

Given,

Height of object $=5cm$

Position of object, $u=-25cm$

Focal length of the lens, $f=10cm$

Position of image, $v=?$

We know that,

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

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

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

So,

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

That is,

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

So,

$v=\frac{50}{3}=16.66cm$

Thus, distance of image is $16.66cm$ on the opposite side of lens.

Now, magnification $=\frac{v}{u}$

That is,

$m=\frac{16.66}{-25}=-0.66$

Also,

$m=\frac{heightofimage}{heightofobject}$

or

$-0.66=\frac{heightofimage}{5cm}$

Therefore, Height of image $=3.3cm$

The negative sign of the height of image shows that an inverted image is formed.

Thus, position of image is at $16.66cm$ on opposite side of lens.

Size of image $=-3.3cm$ at the opposite side of lens

Nature of image is real and inverted.