Distance between an object and its image formed by a lens is 30 cm with magnification m = 3. Find the focal length of lens (in cm)

The correct option is 22.5 cm.

Let's find the focal length of the lens step-by-step using the lens formula and the properties of magnification.

1. Identify the given parameters:
The distance between the object and its image is 30 cm.
The magnification is
$m=3$
Since the magnification magnitude is greater than 1 and positive for an erect, virtual image formed by a convex lens, we can write the magnification formula for a lens as:
$m=\frac{v}{u}=3\Rightarrow v=3u$

Here, both the object distance ($u$) and the virtual image distance ($v$) are measured in the direction opposite to the incident light, meaning they are both negative. Let's define:
$u=-x$
Then, the image distance is:
$v=-3x$

2. Express the distance between the object and the image:
The distance between the object and its image is given as 30 cm:
$|v-u|=30\text{cm}$
Substituting the expressions for $u$ and $v$:
$|-3x-(-x\left)\right|=30$
$|-2x|=30\Rightarrow 2x=30\Rightarrow x=15\text{cm}$

Thus, using the sign convention:
$u=-15\text{cm}$
$v=-45\text{cm}$

3. Calculate the focal length using the lens formula:
The lens formula is given by:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substituting the values of $u$ and $v$ into the lens formula:
$\frac{1}{f}=\frac{1}{-45}-\frac{1}{-15}$
$\frac{1}{f}=-\frac{1}{45}+\frac{1}{15}$
Taking the common denominator:
$\frac{1}{f}=\frac{-1+3}{45}=\frac{2}{45}$
Solving for $f$:
$f=\frac{45}{2}=22.5\text{cm}$

Therefore, the focal length of the lens is 22.5 cm.