A concave mirror of focal length $8cm$ forms an inverted image of an object placed at a certain distance. If the image is twice as large as the object, what is the distance of the object and the image from the pole of the mirror?
I started off with the relation: $\frac{h_o}{h_i} = \frac{d_o}{d_i}$ since the image is inverted, $h_i = -2h_o$, and $\frac{h_o}{h_i} = \frac{1}{-2}$.
$\implies \frac{d_o}{d_i} = \frac{1}{-2} \implies -2d_o = d_i$.
Using the mirror formula $\frac1f = \frac{1}{d_i} + \frac{1}{d_o}$ and substituting $d_i$ for $-2d_o$, $\frac1f = \frac{1}{-2d_o} + \frac{1}{d_o}$
$\implies \frac1f = \frac{1 + -2}{-2d_o} = \frac{1}{2d_o} \implies d_o = \frac{f}{2}$
Since $f = -8cm$ (using Cartesian sign convention),
$d_o = -4cm$, but how can a real image be formed when the object is placed at less than the focal length?