Question

Find the image distance of an object placed 5.00 cm in front of a convex mirror whose focal length is 8.00 cm.

Answer 1

The mirror equation is given by:

$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$

Where d0 is the distance to the object, di is the distance of the image and f is the focal length. In this case we have that:

• The distance of the object is 5 cm

,

• The focal length is 8 cm.

Plugging these and solving for di we have that:

$\begin{gathered} \frac{1}{5}+\frac{1}{d_i}=\frac{1}{8} \\ \frac{1}{d_i}=\frac{1}{8}-\frac{1}{5} \\ \frac{1}{d_i}=\frac{5-8}{40} \\ \frac{1}{d_i}=-\frac{3}{40} \\ d_i=-\frac{40}{3} \\ d_i=-13.33 \end{gathered}$

Therefore, the image distance is -13.33 cm. Note: the minus sign indicates that the image is behind the mirror, that is, the image is virtual.