Answer
a. Converging; 3.08 D
b. Diverging; -0.148 m
Work Step by Step
a. The power of the lens is defined by equation 23–7.
$$P=\frac{1}{f}=\frac{1}{0.325m}=3.08D$$
The lens power is positive; this is a converging lens.
b. Use equation 23–7.
$$P=\frac{1}{f}$$
$$f=\frac{1}{D}=\frac{1}{-6.75D}=-0.148m$$
The focal length is negative; this is a diverging lens.