Chapter 16.1, Problem 3PP

To determine

The illumination on the painting in lx if the 64-cd source of light is 3.0 m away from a painting.

Answer to Problem 3PP

  $\boxed{E=7.11\text{lx}}$

Explanation of Solution

Given:

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

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:

For a point source, distance is $d=1\text{m}$ .

Therefore, from equation (1), after rearranging the terms

  $E=\frac{P}{4\pi }$

  $P=E\times 4\pi$  ...... (2)

Substitute $64\text{cd}$ for $E$ in equation (2)

  $\begin{array}{l}P=4\pi \left(64\text{cd}\right)\\ P=\left(256\pi \right)\text{lm}\end{array}$

Substitute $\left(256\pi \right)\text{lm}$ for $P$ and $3\text{m}$ for $d$ in equation (1)

  $\begin{array}{l}E=\frac{256\pi \text{lm}}{4\pi {\left(3.0\text{m}\right)}^2}\\ \boxed{E=7.11\text{lx}}\end{array}$

Conclusion:

Hence, the required value of illumination is $\boxed{E=7.11\text{lx}}$ .