Answer
Choice A.
Work Step by Step
Apply the thin-lens formula in both cases.
$$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$
Case 1
$$\frac{1}{10cm}+\frac{1}{0.8cm}=\frac{1}{f}$$
Case 2, object farther away
$$\frac{1}{15cm}+\frac{1}{s’}=\frac{1}{f}$$
The focal length stays the same, so equate the left hand sides and solve for s’=0.779cm.
The lens must move closer to the retina by 0.8 cm – 0.779 cm = 0.021 cm.
