(a) A concave mirror of focal length 10 cm can produce a magnified real as well as virtual image of an object placed in front of it. Draw ray diagrams to justify this statement.(b) An object is placed perpendicular to the principal axis of a convex mirror of focal length 10 cm. The distance of the object from the pole of the mirror is 10 cm. Find the position of the image formed.

(a) A concave mirror of focal length 10 cm can produce a magnified real image when the object is placed between the centre of curvature $C$ and principal focus $F$.

A concave mirror of focal length 10 cm can produce a magnified virtual image when the object is placed between the pole $P$ and principal focus $F$.

(b) Given:

Focal length, $f$ = 10cm

Object distance, $u$ = $-$10cm   (object distance is always taken negative as it is on the left side)

To find: Position of the image or image distance, $v$.

Solution:

From the mirror formula, we know that-

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

Substituting the given values in the formula, we get-

$\frac {1}{v}+\frac {1}{(-10)}=\frac {1}{10}$

$\frac {1}{v}-\frac {1}{10}=\frac {1}{10}$

$\frac {1}{v}=\frac {1}{10}+\frac {1}{10}$

$\frac {1}{v}=\frac {1+1}{10}$

 $\frac {1}{v}=\frac {2}{10}$

 $\frac {1}{v}=\frac {1}{5}$

 $v=+5cm$

Thus, the position of the image $v$ is 5cm from the lens, and the positive sign $(+)$ implies that it forms behind the lens (on the right side).

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Updated on: 10-Oct-2022

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