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An object is placed (a) 20 cm, (b) 4 cm, in front of a concave mirror of focal length 12 cm. Find the nature and position of the image formed in each case.
CASE (a)
Given:
Distance of the object from the mirror $u$ = $-$20 cm
Focal length of the mirror, $f$ = $-$12 cm
To find: Distance of the image, $(v)$ from the mirror.
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-
$\frac{1}{(-12)}=\frac{1}{v}+\frac{1}{(-20)}$
$-\frac{1}{12}=\frac{1}{v}-\frac{1}{20}$
$\frac{1}{20}-\frac{1}{12}=\frac{1}{v}$
$\frac{1}{v}=\frac{3-5}{60}$
$\frac{1}{v}=\frac{-2}{60}$
$v=-30cm$
Thus, the distance of the image, $v$ is 30 cm from the mirror, and the negative sign implies that the image forms in front of the mirror (on the left).
Now, from the magnification formula, we know that-
$m=-\frac{v}{u}$
Substituting the given values in the magnification formula we get-
$m=-\frac{(-30)}{(-20)}$
$m=-\frac{30}{20}$
$m=-\frac{3}{2}$
$m=-1.5$
Thus, the magnification of the image $m$ is 1.5, enlarged and its negative sign implies that the image is real and inverted.
CASE (b)
Given:
Distance of the object from the mirror $u$ = $-$4 cm
Focal length of the mirror, $f$ = $-$12 cm
To find: Distance of the image, $(v)$ from the mirror.
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-
$\frac{1}{(-12)}=\frac{1}{v}+\frac{1}{(-4)}$
$-\frac{1}{12}=\frac{1}{v}-\frac{1}{4}$
$\frac{1}{4}-\frac{1}{12}=\frac{1}{v}$
$\frac{1}{v}=\frac{3-1}{12}$
$\frac{1}{v}=\frac{2}{12}$
$\frac{1}{v}=\frac{1}{6}$
$v=6cm$
Thus, the distance of the image, $v$ is 6 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 in the magnification formula we get-
$m=-\frac{(6)}{(-4)}$
$m=\frac{6}{4}$
$m=\frac{3}{2}$
$m=1.5$
Thus, the magnification of the image $m$ is 1.5, enlarged and its positive sign implies that the image is virtual and erect.
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