Chapter 4.6, Problem 36E

(a)

To determine

To find: The rational equation to model the situation.

Answer to Problem 36E

The rational equation to model the situationis $\frac{1}{d_1}=\frac{1}{8}-\frac{1}{32}$ .

Explanation of Solution

Given:

The general equation is $\frac{1}{f}=\frac{1}{d_1}+\frac{1}{d_0}$ and the object is $32\text{cm}$ from the lens and the focal length is $8\text{cm}$ .

Calculation:

Substitute $32\text{cm}$ for $d_0$ and $8\text{cm}$ $f$ in the equation $\frac{1}{f}=\frac{1}{d_1}+\frac{1}{d_0}$ to find the model.

  $\begin{array}{c}\frac{1}{8}=\frac{1}{d_1}+\frac{1}{32}\\ \frac{1}{d_1}=\frac{1}{8}-\frac{1}{32}\end{array}$

Therefore, therational equation to model the situationis $\frac{1}{d_1}=\frac{1}{8}-\frac{1}{32}$ .

(b)

To determine

To find: The distance from the lens to the image.

Answer to Problem 36E

The distance from the lens to the imageis $\frac{3}{32}$ .

Explanation of Solution

Given:

The general equation is $\frac{1}{f}=\frac{1}{d_1}+\frac{1}{d_0}$ and the object is $32\text{cm}$ from the lens and the focal length is $8\text{cm}$ .

Calculation:

Solve the equation $\frac{1}{d_1}=\frac{1}{8}-\frac{1}{32}$ .

  $\begin{array}{c}\frac{1}{d_1}=\frac{1}{8}-\frac{1}{32}\\ =\frac{4-1}{32}\\ =\frac{3}{32}\end{array}$

Therefore, thedistance from the lens to the imageis $\frac{3}{32}$ .