91Ó°ÊÓ

The radius of curvature is$r=35.0\text{cm}$

Magnification is $m=2.50$

Use the concept of magnification and focal length of the mirror. Find the expression for image distance from the formula for magnification and substitute it in the mirror formula. Substitute the given values to find the distance of the mirror from the face.

Formulae are as follows:

$m=-\frac{i}{p}$

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$

where, $m$ is the magnification, $p$ is the pole, $f$is the focal length.

The magnification is given by,

$\begin{array}{rcl}m& =& -\frac{i}{p}\\ i& =& -mp\end{array}$

Plugging this in the mirror equation,

$\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{p}+\frac{1}{i}\\ \frac{1}{f}& =& \frac{1}{p}-\frac{1}{mp}\end{array}$

Use the equation of focal length.

For spherical mirrors,$f=\frac{r}{2}$

Therefore,

$\frac{2}{r}=\frac{1}{p}-\frac{1}{mp}$

Solve it for p:

$\begin{array}{rcl}\frac{2}{r}& =& \frac{1}{p}\left(1-\frac{1}{m}\right)\\ p& =& \frac{r}{2}\left(1-\frac{1}{m}\right)\end{array}$

Plugging the values,

$\begin{array}{rcl}p& =& \frac{35.0}{2}\left(1-\frac{1}{2.50}\right)\\ p& =& 10.5\text{cm}\end{array}$

Therefore, the distance of the mirror from the face is $p=10.5\text{cm}$.