A concave makeup mirror is designed so that a person 25 cm in front of it sees an upright image magnified by a factor of two. What is the radius of curvature of the mirror?

For the concave mirror, we deduce that magnification $M>0$. Therefore applying equation:

$$\begin{align*} M=-\frac{q}{p}=2 \end{align*}$$

We solving for the object image $q$, we get:

$$\begin{align*} q=-2\left[p\right]=-2\left[25\textrm{cm}\right]=-50\textrm{cm} \end{align*}$$

For the focal length $f$, following relationship occurs:

$$\begin{align*} \frac{1}{f}=\frac{2}{R}=\frac{1}{p}+\frac{1}{q}=\left[\frac{1}{25}-\frac{1}{50}\right]\left(\textrm{cm}\right)=\frac{1}{50}\left(\textrm{cm}\right) \end{align*}$$

Thus focal length radius $R$ yields:

$$\begin{align*} R=2\left[0.5\right]\left(\textrm{m}\right) \end{align*}$$

Solution is:

$$\begin{align*} \boxed{R=1\textrm{m}} \end{align*}$$

In this problem, we are given a concave mirror. A person $25~\mathrm{cm}$ in front of the mirror sees an upright image with twice the size of the person. We calculate the radius of curvature of the mirror.

The image is upright and the magnification is 2, therefore:

$q = -2 \ p = -(2) \ (25) = -50 \ cm$

We have:

$\dfrac{2}{R} = \dfrac{1}{25} + \dfrac{1}{-50}$

$\dfrac{2}{R} = 0.02$

$$==> R = 100 \ cm$$