(a) Draw a diagram to represent a convex mirror. On this diagram mark principal axis, principal focus $F$ and the centre of curvature $C$ if the focal length of convex mirror is 3 cm.(b) An object 1 cm tall is placed 30 cm in front of a convex mirror of focal length 20 cm. Find the size and position of the image formed by the convex mirror.

Distance of the object from the mirror, $u$ = $-$30 cm

Focal length of the mirror, $f$ = 20 cm

To find: Position (or, distance) of the image $v$, and size of the image $h_2$.

Solution:

From the mirror formula, we know that-

$\frac {1}{f}=\frac {1}{v}+\frac {1}{u}$

Substituting the given values we get-

$\frac {1}{20}=\frac {1}{v}+\frac {1}{(-30)}$

$\frac {1}{20}=\frac {1}{v}-\frac {1}{30}$

$\frac {1}{20}+\frac {1}{30}=\frac {1}{v}$

$\frac {1}{v}=\frac {3+2}{60}$

$\frac {1}{v}=\frac {5}{60}$

$\frac {1}{v}=\frac {1}{12}$

$v=+12cm$

Thus, the distance of the image $v$ is 12 cm from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, from the magnification formula, we know that-

$m=-\frac {v}{u}$

Substituting the given values we get-

$m=-\frac {12}{(-30)}$

$m=\frac {12}{30}$

$m=\frac {2}{5}$

$m=+0.4$

Thus, the magnification is $0.4$ which is less than 1, which means the image is small in size, and the positive sign implies that the image is virtual and erect.

Also, using the magnification formula we get-

$m=\frac {h_2}{h_1}$

$0.4=\frac {h_2}{1}$

$h_2=0.4cm$

Therefore, the height of the image h_2 is 0.4 cm.