Page 211 - Physics_Form_2

Refraction and dispersion of light

Example 5.5 Example 5.6

An object is placed 10 cm from a An object 2 cm high is placed 24 cm
concave lens of focal length 15 cm. from a converging lens. An erect
Using the lens formula, determine the image, which is 6 cm high, is formed.
nature and the position of the image. Determine the focal length of the lens.
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u  10 cm, f  15 cm Solution
1 = 1 + 1 Using the lens formula:
f u v h = 0 2 cm, h = i 6 cm, u = 24 cm

1  1  1 v v
 15cm 10cm v m  u  24cm

1   1  1 h
v 15cm 10cm Also, m = i
h 0
1   2 cm 3 cm   5 cm 6 cm
v 30 cm 2 30 cm 2 = 2 cm = 3

v  6 cm v
Thus, m   3
A virtual image is formed 6 cm from 24cm
the lens, on the same side as the object. v  72 cm

1 1 1
Magnification by a lens f = u + v

Magnification is a measure of the extent
to which an optical system changes the  1  1
size of an image of an object. The linear 24cm  72cm
or lateral magnification produced by a 3 cm 1 cm 2 cm
lens is the ratio of the height of the image  72 cm 2  72 cm 2
to the height of the object. If the image
size is bigger than the object size, then the f = 36 cm
image is said to be enlarged. On the other Therefore, the focal length of the lens
hand, when the image size is smaller is 36 cm.
than the object size, the image is said to
be diminished. Magnification is usually Example 5.7
denoted by the letter m. Thus,
A vertical object, 10 cm high, is placed
image height ()h
m = i 30 cm in front of a diverging lens. An
object height ()h 0 image of the object is formed 7.5 cm in

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Physics Form 2 Final.indd 205 25/10/2025 10:28