5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.

Answer

The real, inverted image of height 3.3 cm is formed at 16. 7 cm.

Explanation

Given: Focal length (f) = 10 cm,
object distance (u) = - 25 cm,
height of the object `(h_(1)) = 5 cm`
To find: Image distance (v), height of the image `(h_(2))`
Formulae: i. `(1)/(f)=(1)/(v)-(1)/(u)` ii. `(h_(2))/(h_(1))=(v)/(u)`
Calculation: From formula (i),
`(1)/(10)=(1)/(v)-(1)/(-25)`
`therefore" "(1)/(v)=(1)/(10)-(1)/(25)=(5-2)/(50)`
`therefore" "(1)/(v)=(3)/(50)`
`therefore" "v=16.7 cm`
As the image distance is positive, the image formed is real. From formula (ii),
`(h_(2))/(5)=(16.7)/(-25)`
`therefore" "h_(2)=(16.7)/(25)xx5=-(16.7)/(5)`
`therefore" "h_(2)=-3.3 cm`
The negative sign indicates that the image formed is inverted.