Answer
a. 0.106 m
b. 0.109 m
c. 0.117m
d. 0.513m
Work Step by Step
Use the lens equation, 23–8.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
a. Solve for the image distance.
$$d_i=\frac{d_of}{d_o-f}=\frac{(10.0m)(0.105m)}{10.0m-0.105m}$$
$$=0.106m$$
b. Solve for the image distance.
$$d_i=\frac{d_of}{d_o-f}=\frac{(3.0m)(0.105m)}{3.0m-0.105m}$$
$$=0.109m$$
c. Solve for the image distance.
$$d_i=\frac{d_of}{d_o-f}=\frac{(1.0m)(0.105m)}{1.0m-0.105m}$$
$$=0.117m$$
d. Solve for the object distance.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
$$d_o=\frac{d_if}{d_i-f}$$
The smallest object distance corresponds to the largest image distance.
$$d_o=\frac{(0.132m)(0.105m)}{0.132m-0.105m}$$
$$=0.513m$$