Answer
A power of -1.37D is required.
Work Step by Step
The corrective lens forms a virtual image at the far point of the eye. The distances from the corrective lens are s=infinity and s’=-73.0 cm.
Use the lens equation.
$$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$
$$\frac{1}{\infty}+\frac{1}{-73.0cm}=\frac{1}{f}$$
$$f=-73.0cm$$
$$P=\frac{1}{f}=\frac{1}{-0.730m}=-1.37D$$
