Where should the object be placed in front of the concave mirror so that image is as the same size as the object?

The mirror formula, which connects the object's position to the image's position and focal length, must be understood by students. Understanding sign norms is also crucial. understanding of mirror enlargement A spherical mirror has a radius of curvature that is double its focal length.

Formula used:
The mirror Formula:${\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}$
A mirror's magnification is described as,$m=-\frac{v}{u}$

Let's say that the item is located on the mirror's left side. The picture should be the exact same size as the item. As a result, 1 must be the magnitude of the magnification.
And using the formula for magnification, $m=-\frac{v}{u}$, For $m=-1................\left(1\right)$, 

We set the value of magnification as a negative number since the image created by the mirror in reality must be inverted.
Using the mirror formula, we can now:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
From equation (1)
$\frac{1}{u}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{2}{u}=\frac{1}{f}$
$\Rightarrow u=2f$

We now know that the radius of curvature is twice as far from the mirror as the focal length.
As a result, the item must be positioned at the middle of the curve. The configuration for the same is shown in the following ray diagram.

Therefore, option C is the best choice; that is, the item should be positioned in the concave mirror's centre of curvature such that the image is the same size as the thing.

Hence, the correct answer is option C.