The equation needed to solve the problem is
[tex] \frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d_{i}} [/tex]
where
f = focal length, 8.0 cm
do = object distance, 25.0 cm
di = image distance
Therefore
[tex] \frac{1}{8} = \frac{1}{25} + \frac{1}{d_{i}} \\ 0.125 = 0.04 + \frac{1}{d_{i}} \\ \frac{1}{d_{i}} =0.125-0.04=0.085\\d_{i}= \frac{1}{0.085} =11.765 \, cm[/tex]
The closest answer is 12 cm.
Answer: 12 cm
