Q1. 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.
object distance (u) = – 10 cm,
To find: Image distance (v), height of the image (h2)
Formulae:
`1/"f" = 1/"v" - 1/"u"`
`"h"_2/"h"_1 = "v"/"u"`
`= (1 - 2)/20`
As the image distance is negative, the image formed is virtual and on the same side of lens as that of the object.
The positive sign indicates that the image formed is erect.