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

Given,

Object height (h_o) = 5 cm

Object distance (u) = -25 cm

focal length (f) = 10 cm

Ray diagram is shown below:

Image distance (v) = ?

Image height (h_i) = ?

According to the Lens formula,

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

Substituting the values we get,

$\dfrac{1}{v} - \dfrac{1}{-25} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{25} = \dfrac{1}{10} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} - \dfrac{1}{25} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{5-2}{50} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3}{50} \\[0.5em] \Rightarrow v = 16.66 \text{ cm}$

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

Magnification (m) = $\dfrac{\text{v}}{\text{u}}$ = $\dfrac{\text{height of image}}{\text{height of object}}$

Substituting the values we get,

$\dfrac{16.66}{-25}$ = $\dfrac{\text{height of image}}{5}$

height of image = $\dfrac{16.66 \times 5}{-25}$ = -3.3

Negative sign of the height of image means that an inverted image is formed.

The image is reduced, real and inverted.