# At what distance from a concave mirror of focal length 10 cm should an object be placed so that:(a) its real image is formed 20 cm from the mirror?(b) its virtual image is formed 20 cm from the mirror?

(a) Given:

It is a concave mirror

Distance of the image from the mirror, $v$ = $-$20 cm       (real image)

Focal length of the mirror, $f$ = $-$10 cm

To find: Distance of the object, $u$.

Solution:

From the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-10)}=\frac{1}{(-20)}+\frac{1}{u}$

$-\frac{1}{10}=-\frac{1}{20}+\frac{1}{u}$

$\frac{1}{20}-\frac{1}{10}=\frac{1}{u}$

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

$\frac{1}{u}=\frac{-1}{20}$

$u=-20cm$

Therefore, the object should be placed at a distance of 20 cm from the mirror to form a real image.

(b) Given:

It is a concave mirror

Distance of the image from the mirror, $v$ = $+$20 cm    (virtual image)

Focal length of the mirror, $f$ = $-$10 cm

To find: Distance of the object, $u$.

Solution:

From the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-10)}=\frac{1}{20}+\frac{1}{u}$

$-\frac{1}{10}=\frac{1}{20}+\frac{1}{u}$

$-\frac{1}{20}-\frac{1}{10}=\frac{1}{u}$

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

$\frac{1}{u}=\frac{-3}{20}$

$u=-\frac{20}{3}$

$u=-6.6cm$

Therefore, the object should be placed at a distance of 6.6 cm from the mirror to form a virtual image.

Updated on: 10-Oct-2022

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