Question

An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image distance and height of the image formed.

Answer 1

Object height, O = +10 cm

Focal length, f = +15 cm

Object distance, u = −25 cm

Image distance, v = ?

Image height, I = ?

Using lens formula,

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

${\displaystyle \frac{1}{15}=\frac{1}{\text{v}}-\frac{1}{(-25)}}$

Or, ${\displaystyle \frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{25}}$

${\displaystyle \frac{1}{\text{v}}=\frac{5-3}{75}}$

= ${\displaystyle \frac{1}{\text{v}}=\frac{2}{75}}$

v = ${\displaystyle \frac{75}{2}}$

Also, ${\displaystyle \frac{-\text{v}}{\text{u}}=\frac{\text{I}}{\text{O}}}$

Hence, ${\displaystyle \frac{\frac{75}{2}}{-25}=\frac{\text{I}}{10}}$

${\displaystyle \frac{75}{2}\times \frac{1}{-25} =\frac{\text{I}}{10}}$

${\displaystyle \frac{3}{-2}=\frac{\text{I}}{10}}$

${\displaystyle -1.5 =\frac{\text{I}}{10}}$

I = (-1.5) × 10 =  −15 cm

Height of image is −15 cm, inverted.