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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.
Solution
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/"f" = 1/"v" - 1/"u"`
- `"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.
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