A real image 2/3rd of the size of an object is formed by a convex lens when the object is at a distance of 12 cm from it. Find the focal length of the lens.
Given:
It is a convex lens.
Object distance, $u=-12cm$
Magnification, $m=\frac {-2}{3}$
To find: Focal length, $f$ .
Solution:
From magnification formula we know that-
$m=\frac {v}{u}$
Substituting the required values, we get-
$\frac {-2}{3}=\frac {v}{-12}$
$v=\frac {24}{3}$ (by cross multiplication)
$v=+8cm$
Thus, the image distance, $v$ is 8 cm from the lens, and the plus sign implies that it is formed on the right side of the lens.
Now,
From lens formula, we know that-
$\frac {1}{f}=\frac {1}{v}-\frac {1}{u}$
Substituting the given values, we get-
$\frac {1}{f}=\frac {1}{8}-\frac {1}{(-12)}$
$\frac {1}{f}=\frac {1}{8}+\frac {1}{12}$
$\frac {1}{f}=\frac {3+2}{24}$
$\frac {1}{f}=\frac {5}{24}$
$f=\frac {24}{5}$
$f=+4.8cm$
Thus, the focal length, $f$ of the lens is 4.8cm.
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