A perging mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image positions.

The mirror is Convex.

Magnification, $m$ = $\frac {1}{2}$

Radius of curvature, $R$ = 40 cm

Focal length of the mirror, $f$ = 20 cm             $(\because f=\frac{R}{2})$

To find: Distance of the image from the mirror $v$, and distance of the object from the mirror $u$.

Solution:

From the magnification formula, we know that-

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

Substituting the given values we get-

$\frac {1}{2}=-\frac {v}{u}$

$u=-2v$ ---------- (i)

Now, from the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{20}=\frac {1}{v}+\frac {1}{(-2v)}$

$\frac {1}{20}=\frac {1}{v}-\frac {1}{2v}$

$\frac {1}{20}=\frac {2-1}{2v}$

$\frac {1}{20}=\frac {1}{2v}$

$2v=20$

$v=\frac {20}{2}$

$v=+10cm$

Thus, the distance of the image $v$ is 10 cm from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, putting the value of v in the equation (i) we get-

$u=-2/times {10}$

$u=-20cm$

Thus, the distance of the object $u$ is 20 cm from the mirror, and the negative sign implies that the object is placed in front of the mirror (on the left).

Hence, if an object is placed at a distance of 20 cm in front of the mirror, the image will be formed 10 cm behind the mirror.