Answer
Object distance of 6.2 cm, magnification of 4.0.
Work Step by Step
Comparing equations 25-2a and 25-2b, we see that to have maximum magnification, the image should be located at the near point (the image distance is negative because the image is virtual).
Use the lens equation to find the object distance.
$$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$
$$d_o=\frac{fd_i}{d_i-f}=\frac{(8.20cm) (-25cm) }{-25cm-8.20cm}=6.17cm\approx 6.2cm$$
Find the magnification from equation 25–2b.
$$M=\frac{N}{f}+1=\frac{25cm}{8.20cm}+1= 4.0\times$$
