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Example 9.3 An object is placed at (i) 10 cm, (ii) 5 cm in front of a
concave mirror of radius of curvature 15 cm. Find the position, nature,
and magnification of the image in each case.

Solution
The focal length f = –15/2 cm = –7.5 cm
(i)The object distance u = –10 cm. Then Eq. (9.7) gives
– – .
1 1 1
v 10 7 5
=
or
.
.
10 7 5
2 5
v
×
−
= = – 30 cm
The image is 30 cm from the mirror on the same side as the object.
Also, magnification m =
( 30) – – – 3
( 10)
v
u
−
= =
−
The image is magnified, real and inverted.
(ii) The object distance u = –5 cm. Then from Eq. (9.7),
1 1 1
v 5 7.5
=
− −
or ( )
.
. –
5 7 5 15 cm
7 5 5
v
×
= =
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m =
15 – – 3
( 5)
v
u
= =
−
The image is magnified, virtual and erect.