Question

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?

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

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

We solving for the object image qq, we get:

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

For the focal length ff, following relationship occurs:

1f=2R=1p+1q=[125150](cm)=150(cm)\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 RR yields:

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

Solution is:

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

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