An object placed in front of a concave mirror at a distance x cm from the pole gives 3 times magnified real image. If it is moved to a distance (x + 5) cm, the magnification becomes 2. The focal length of the mirror is:

The mirror formula may be used to determine a mirror's focal length. Using the magnification, we may substitute image distance in terms of object distance, and then utilise the two sets of data supplied to solve for focal length.

If an image is removed from a concave mirror at a distance that is, then we know that $v$ and that of the screen's item is $u$, then the mirror's focal length comes next $f$ is related to $v$ and $u$ through the mirror equation

${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}$ ... (1)
We also understand the formula for magnification, which indicates the amount by which an image is enlarged. Additionally, the magnification m and object distance are connected.
$m={\displaystyle \frac{-v}{u}}$ … (2)

We see that in order to determine the focal length, we must understand $u$ and $v$. But since magnification is known, we only need to be aware of one of them $v$ and $u$.
Let's thus first build up an equation that provides $f$ in terms of $u$ and $m$
Let's split (1) all of by u.
${\displaystyle \frac{u}{f}}={\displaystyle \frac{u}{v}}+{\displaystyle \frac{u}{u}}$
Now that we can swap ${\displaystyle \frac{u}{v}}$ using (2)
${\displaystyle \frac{u}{f}}={\displaystyle \frac{-1}{m}}+1$
Let's now enter the values supplied into the equation.
Here, while measuring, sign conventions must be employed. $u$, $v$, and $f$. $u$ is the distance to the item is always negative. Since a true picture is created in this case, m is likewise negative.

${\displaystyle \frac{-x}{f}}={\displaystyle \frac{-1}{-3}}+1$ ... (3)

${\displaystyle \frac{-(x+5)}{f}}={\displaystyle \frac{-1}{(-2)}}+1$ ...(4)

By equating the value of X, we may resolve (3) and (4).

$f\left({\displaystyle \frac{1}{3}}+1\right)=f\left({\displaystyle \frac{1}{2}}+1\right)+5$

$f{\displaystyle \frac{4}{3}}=f{\displaystyle \frac{3}{2}}+5$
${\displaystyle \frac{-1}{6}}f=5$
$f=-30cm$
Consequently, the mirror's focal length is 30 cm.

Hence, the correct answer is option D.