Chapter 5.3, Problem 101AYU

105. Optics If a single pane of glass obliterates $3\%$ of the light passing through it, the percent p of light that passes through n successive panes is given approximately by the function

$p(n)=100{\left(0.97\right)}^n$

(a) What percent of fight will pass through 10 panes?

(b) What percent of light will pass through 25 panes?

(c) Explain the meaning of the base 0.97 in this problem.

To determine

To find:

a. Percent of light will pass through 10 panes.

Answer to Problem 101AYU

a. $73.74$

Explanation of Solution

Given:

$p\left(n\right)=100{\left(0.97\right)}^n$

a. Percent of light will pass through 10 panes $p\left(10\right)=100{\left(0.97\right)}^{10}=73.74$.

To determine

To find:

b. Percent of light will pass through 25 panes.

Answer to Problem 101AYU

b. $46.70$

Explanation of Solution

Given:

$p\left(n\right)=100{\left(0.97\right)}^n$

b. Percent of light will pass through 25 panes $p\left(25\right)=100{\left(0.97\right)}^{25}=46.70$.

To determine

To find:

c. The meaning of the base $\text{0}\text{.97}$ in this problem.

Answer to Problem 101AYU

c. The base $\text{0}\text{.97}$ stands for the probability of light pass through the single pane.

Explanation of Solution

Given:

$p\left(n\right)=100{\left(0.97\right)}^n$

c. The base $\text{0}\text{.97}$ stands for the probability of light pass through the single pane.