**Problem 6.82: A concave spherical mirror has radius of curvature of $20.0 \:cm$. A linear object of height $3.5 \:cm$ is placed $15.0 \:cm$ from the center of the mirror along the optic axis, as shown in the figure. Calculate the location of the image.**

**Solution:**

Focal length of a spherical mirror of radius of curvature $R$ is $f\:=\:\frac{R}{2}$.

Given that radius of curvature of = $20.0 \:cm$.

Therefore focal length $f\:=\:\frac{20}{2}\:=\:10\:cm$.

**Cartesian sign convention for spherical mirror.**

Mirror equation $\frac{1}{f}\:=\:\frac{1}{v}\:+\:\frac{1}{u}$.

In the given problem, object distance, $u\:=\:-15\:cm$.

Foal length, $f\:=\:-10\:cm$.

Therefore,

$\frac{1}{-10}\:=\:\frac{1}{v}\:+\:\frac{1}{-15}$.

Or, $\frac{1}{v}\:=\:\frac{1}{-10}\:-\:\frac{1}{-15}$.

Or $\frac{1}{v}\:=\:\frac{1}{-10}\:+\:\frac{1}{15}$.

Or $\frac{1}{v}\:=\:\frac{10-15}{150}\:=\:\frac{-5}{150}$.

Thus image distance $v\:=\:-30\:cm$.

So the image is formed at a distance of $30\:cm$ from the mirror.