Answer
a. The image is 37 cm behind the lens.
b. +2.3
Work Step by Step
Use the lens equation, 23–8. The focal length and the object distance are positive.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
a. Solve for the image distance.
$$d_i=\frac{d_of}{d_o-f}=\frac{(0.16m)(0.28m)}{0.16m-0.28m}$$
$$=-0.37m$$
The negative sign means that the image is 37 cm behind the lens, i.e., it is a virtual image.
b. Find the magnification from Eq. 23–9.
$$m=-\frac{d_i}{d_o}=-\frac{-0.3733m}{0.16m}=+2.3$$
The magnification is positive. The image is upright.