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(a) An object 2 cm tall stands on the principal axis of a converging lens of focal length 8 cm. Find the position, nature and size of the image formed if the object is:(i) 12 cm from the lens(ii) 6 cm from the lens(b) State one practical application each of the use of such a lens with the object in position (i) and (ii).
(a) (i) Given:
Converging lens is a convex lens.
Focal length, $f$ = 8 cm
Height of the object $h$ = $+$2
Object distance, $u$ = $-$12 cm (since object is always placed on the left side of the lens, it is taken as negative)
To find: Position, nature, $v$ of the image, and size of the image $h'$.
Solution:
According to the lens formula, we know that:
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values in the formula we get-
$\frac {1}{v}-\frac {1}{(-12)}=\frac {1}{8}$
$\frac {1}{v}+\frac {1}{12}=\frac {1}{8}$
$\frac {1}{v}=\frac {1}{8}-\frac {1}{12}$
$\frac {1}{v}=\frac {3-2}{24}$
$\frac {1}{v}=\frac {1}{24}$
$v=+24cm$
Thus, the image distance $v$ is 24 cm from the lens, and the positive $(+)$ sign for image distance implies that the image is formed on the right side of the lens (behind the lens). And, we know that on the right side of the lens real image forms.
Now,
From the magnification formula, we know that:
$m=\frac {v}{u}=\frac {h'}{h}$
Substituting the given values in the formula we get-
$\frac {24}{-12}=\frac {h'}{2}$
$-2=\frac {h'}{2}$
$h'=2\times {(-2)}$
$h'=-4cm$
Thus, the size of the image $h'$ is 4 cm, and the negative sign $(-)$ implies that the image is inverted (below the principal axis).
Hence, the position of the image is behind the lens (on the right side), nature of the image is real and inverted, and size of the image is larger than the object (4cm).
(ii) Given:
Focal length, $f$ = 8 cm
Height of the object $h$ = $+$2
Object distance, $u$ = $-$6 cm
To find: Position, nature, $v$of the image, and size of the image $h'$.
Solution:
According to the lens formula, we know that:
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values in the formula we get-
$\frac {1}{v}-\frac {1}{(-6)}=\frac {1}{8}$
$\frac {1}{v}+\frac {1}{6}=\frac {1}{8}$
$\frac {1}{v}=\frac {1}{8}-\frac {1}{6}$
$\frac {1}{v}=\frac {3-4}{24}$
$\frac {1}{v}=\frac {-1}{24}$
$v=-24cm$
Thus, the image distance $v$ is 24 cm from the lens, and the positive $(-)$ sign for image distance implies that the image is formed on the left side of the lens (in front of the lens). And, we know that on the left side of the lens virtual image forms.
Now,
From the magnification formula, we know that:
$m=\frac {v}{u}=\frac {h'}{h}$
Substituting the given values in the formula we get-
$\frac {-24}{-6}=\frac {h'}{2}$
$4=\frac {h'}{2}$
$h'=2\times {4}$
$h'=8cm$
Thus, the size of the image $h'$ is 8 cm, and the negative sign $(+)$ implies that the image is erect (above the principal axis).
Hence, the position of the image is in front of the lens (on the left side), nature of the image is virtual and erect, and size of the image is magnified (8cm).
(b) One practical application or the use of such a lens with the object-
In position (i) $-$ as a film projector.
In position (ii) $-$ as a magnifying lens.
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