Answer
-0.75D for the upper portion of the bifocals, and 2.0D for the lower.
Work Step by Step
Start with the upper portion of the bifocals. The glasses are designed to take an object at infinity and make a virtual image 135 cm away from the eye that the person can focus upon.
The glasses are 2.0 cm away from the eye, so the distance of the image from the lens is only 135 cm – 2.0 cm = 133 cm. The image is virtual, so the image distance is negative.
$$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{\infty}+\frac{1}{-1.33m}=-0.75D$$
Now consider the lower portion. The person’s current near point is 45 cm. With the reading glasses, an object placed 25 cm from the face (23 cm from the lens) should produce a virtual image that is 45 cm from the eye (43 cm from the lens).
In other words, $d_o=0.23 m$ and $ d_i=-0.43 m $. Find the power of the lens from the lens equation.
$$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$
$$P=\frac{1}{0.23m}+\frac{1}{-0.43m}=2.0D$$
