Answer
A focal length of 49.5cm, or power of 2.02D is required.
Work Step by Step
The corrective lens forms a virtual image at the near point of the eye. The distances from the corrective lens are s=23.0 cm and s’=-43.0 cm.
Use the lens equation.
$$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$
$$\frac{1}{23.0cm}+\frac{1}{-43.0cm}=\frac{1}{f}$$
$$f=49.45cm$$
$$P=\frac{1}{f}=\frac{1}{0.4945m}=2.02D$$
