Answer
Converging, real, 0.368 m
Work Step by Step
The image is on the opposite side as the lens so $ d_i\gt0$ and it is a real image.
Use the lens equation, 23–8.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
Solve for the focal length.
$$f=\frac{d_od_i}{d_o+d_i}$$
$$=\frac{(1.55m)(0.483m)}{1.55m+0.483m}=0.368m$$
This is a converging lens because the focal length is positive.
