Question

5 cm high object is placed at a distance of 10 cm from a converging lens of focal length of 20 cm. Determine the position, size and type of the image.

Answer 1

Given: Focal length (f) = 20 cm,
object distance (u) = – 10 cm,
height of the object (h₁) = 5 cm

To find: Image distance (v), height of the image (h₂)

Formulae: 

1. `1/"f" = 1/"v" - 1/"u"`
2. `"h"_2/"h"_1 = "v"/"u"`

Calculation: From formula (i),

`1/20 = 1/"v" - 1/(- 10)`

∴ `1/ "v" = 1/20 - 1/10`

`= (1 - 2)/20`

∴ `1/"v" = - 1/20`

∴ v = - 20 cm

As the image distance is negative, the image formed is virtual and on the same side of lens as that of the object.

From formula (ii),

`"h"_2/5 = (- 20)/- 10`

∴ `"h"_2 = 20/10 xx 5`

∴ h₂ = 10 cm

The positive sign indicates that the image formed is erect.