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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.
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