Chapter 16, Problem 36A

To determine

To Find: Illumination of a lamp.

Answer to Problem 36A

  $2.02\text{lx}$ .

Explanation of Solution

Given:

Distance to the screen, $r=4\text{m}$ .

Luminous flux of the bulb, $P=405\text{lm}$

Formula:

The expression for illumination is $$

  $\begin{array}{l}E=\frac{\text{luminous}\text{flux}}{4\pi r^2}\\ E=\frac{P}{4\pi r^2}\end{array}$  ...... (1)

Calculation:

Substitute $405\text{lm}$ for $P$ and $4\text{m}$ for $r$ in equation (1):

  $\begin{array}{c}\text{}\text{E}= \frac{405}{4\times 3.14\times 4^2}\\ \text{}\text{E}=\frac{405}{200.96}\text{}\\ \text{E}=2.02\text{ lx}\end{array}$

Conclusion:

Hence, the required illumination is $2.02\text{lx}$ .