Chapter 16.1, Problem 5PP

To determine

Illumination on a surface that is $3.0\text{m}$ below a 150- W incandescent lamp that emits a luminous flux of $2275\text{lm}$ .

Answer to Problem 5PP

  $\boxed{E=20.13\text{lx}}$

Explanation of Solution

Given:

The distance is $d=3.0\text{m}$ .

Luminous flux is $P=2275\text{lm}$ .

Formula used:

The expression for illumination is given by

  $E=\frac{P}{4\pi d^2}$  ...... (1)

Here, $P$ and $d$ are luminous flux and radius of the sphere which is equal to the distance of the object from the light source respectively.

Calculation:

Substitute 3.0 m for $r$ and $2275\text{lm}$ for $P$ in equation (1)

  $\begin{array}{l}E=\frac{2275\text{lm}}{4\pi {\left(3.0\text{m}\right)}^2}\\ \boxed{E=20.13\text{lx}}\end{array}$

Conclusion:

Hence, the required value of illumination is $\boxed{E=20.13\text{lx}}$ .