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Question
An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger as compared to the object?
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Solution
Given: Height of the object (h1) = 5 cm,
focal length (f) = 10 cm,
distance of the object (u) = –20 cm
To find: Image distance (v), height of the image (h2), magnification (M)
Formulae:
- `1/"v" - 1/"u" = 1/"f"`
- Magnification (M) = `"h"_2/"h"_1 = "v"/"u"`
Calculation: From formula (i),
`1/"v" = 1/"u" + 1/"f"`
∴ `1/"v" = 1/(-20) + 1/10`
∴ `1/"v" = (- 1 + 2)/20`
∴ `1/"v" = 1/20`
∴ v = 20 cm
From formula (ii),
`"h"_2 = "v"/"u" xx "h"_1`
∴ `"h"_2 = 20/(- 20) xx 5`
∴ `"h"_2 = (- 1) xx 5`
∴ `"h"_2` = - 5 cm
∴ M = `"v"/"u" = 20/(-20)` = - 1
The negative sign of the height of the image and the magnification shows that the image is inverted and real. It is below the principal axis and is of the same size as the object.
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