# A real image 2/3^{rd} 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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