An object is placed at a distance of $12cm$ from a concave mirror. The image formed will be real and four times larger than the object. Find out the distance of the image from the mirror. An object is placed $24cm$ from a concave mirror. The image formed will be inverted and double the size of the object. [[1]] the focal length of the mirror and the position where the image will be formed.

The growth formula can be written as,
$m=\frac{h_i}{h_o}=\frac{-v}{u}$
When $h_i$ is the frame of the built-in image, $h_o$ is the height of the object set, $v$ be the distance of the built image and $u$ be the distance of the object placed in front of the mirror.
It is said that the distance to the image in the first case is,
u = −12cm
And the growth produced will be,
m = −4

So, let's place the values ​​in it,
$m=-4=\frac{-v}{u}$

⇒$-4=\frac{-v}{-12}$

From this point, the image range will be available as,

$v=-48cm$
In the next case, the object is set as far away as,

$u=-24cm$
Image enlargement will be

$m=\frac{h_i}{h_o}=\frac{-2}{1}=\frac{-v}{u}$
So let's change the distance of the object to it,
$m=-2=\frac{-v}{-24}$

In these figures we will find the final speed as,
$v=-48cm$
Now let's find the focus length using the screen equation provided as,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
When f be the focal length of the mirror. Let's change the values ​​in it,
$\frac{1}{f}=\frac{1}{-48}+\frac{1}{-24}$

$\frac{1}{f}=\frac{-3}{48}=\frac{-1}{16}$
By taking a reciprocal equation,
$f=-16cm$