Answer
$d_o=2f$
Work Step by Step
The new object distance is 30.0 cm. The focal length is 15.0 cm.
Use the mirror equation, 23–2.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
Solve for the image distance.
$$d_i=\frac{d_of}{d_o-f}=\frac{(30.0cm)(15.0cm)}{30.0cm-15.0cm}$$
$$=30.0cm$$
Calculate the magnification.
$$m=-\frac{d_i}{d_o}=-\frac{30.0cm}{30.0cm}=-1.00$$
The magnification is negative one, so the image is inverted and the same size as the object.
In this situation, we see that the object distance (30.0 cm) is twice the focal length of 15.0 cm, i.e., $d_o=2f$.
