Chapter 16, Problem 38A

To determine

To Find:The minimum setting that should be used.

Explanation of Solution

Given:

The minimum illuminance is $E=175\text{lx}$

Distance of the bulb above the paper is $r=80\text{cm}=0.8\text{m}$

Formula:

The expression for illumination is $$

  $\begin{array}{l}E=\frac{\text{luminous}\text{flux}}{4\pi r^2}\\ E=\frac{P}{4\pi r^2}\end{array}$

Here, $P$ is luminous flux and $r$ is the distance.

Calculation:

Luminous flux P can be written from equation (1) as

  $\text{P}=\text{E}\times 4\pi r^2$  ...... (2)

Substitute $175\text{lx}$ for $E$ and $0.8\text{m}$ for $r$ in equation (2)

  $\begin{array}{c}\text{}P= 175\times 4\times 3.14\times {0.8}^2\\ \text{}P= 1407 \mathrm{lm}\end{array}$

Conclusion:

Hence, the required minimum luminous flux required is $1407\text{lm}$ . The setting required for the three-way bulb is $100\text{W}$ of electric power to deliver $1620\text{lm}$ .