A $2.0\text{cm}$ high object is placed at a distance of $20\text{cm}$ from a concave mirror. A real image is formed at $40\text{cm}$ from the mirror. Calculate the focal length of the mirror and the size of the image.

The correct option is A $-13.3\text{cm},4\text{cm}$
The situation is shown in Figure. As the image is real, it is formed on the same side as reflected rays.

Here, $u=-20\text{cm}$ and $v=-40\text{cm}$.

Thus, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{1}{-20\text{cm}}+\frac{1}{-40\text{cm}}$
$f=13.3\text{cm}$
The magnification is$m=-\frac{v}{u}=\frac{-40\text{cm}}{-20\text{cm}}=-2$
Thus, $\frac{h_e}{h_o}=-2h_e=-2h_o=2\times 2.0\text{cm}=-4.0\text{cm}$
The height of the image is $4.0\text{cm}$, and the negative sign of $h_e$ shows that the image is inverted.