Page 162 - NCERT Science Class 10 English Medium

Solution
Height of the object h = + 2.0 cm;
Focal length f = + 10 cm;
object-distance u = –15 cm;
Image-distance v = ?
Height of the image h′ = ?

1 1 1
Since − =
v u f
1 1 1
or, = +
v u f
1 1 1 1 1
= + = − +
v ( 15) 10 15 10
−
1 − 2 + 3 1
= =
v 30 30
or, v = + 30 cm
The positive sign of v shows that the image is formed at a distance of
30 cm on the other side of the optical centre. The image is real and
inverted.

' h v
Magnification m = =
h u
or, h′ = h (v/u)
Height of the image, h′ = (2.0) (+30/–15) = – 4.0 cm
Magnification m = v/u

+ 30cm
or, m = = − 2
− 15cm
The negative signs of m and h′ show that the image is inverted and
real. It is formed below the principal axis. Thus, a real, inverted image,
4 cm tall, is formed at a distance of 30 cm on the other side of the
lens. The image is two times enlarged.

9.3.8 Power of a Lens

You have already learnt that the ability of a lens to converge or diverge
light rays depends on its focal length. For example, a convex lens of
short focal length bends the light rays through large angles, by focussing
them closer to the optical centre. Similarly, concave lens of very short
focal length causes higher divergence than the one with longer focal
length. The degree of convergence or divergence of light rays achieved
by a lens is expressed in terms of its power. The power of a lens is defined
as the reciprocal of its focal length. It is represented by the letter P. The
power P of a lens of focal length f is given by
1
P = (9.11)
f

Light – Reflection and Refraction 157

2024-25