Chapter 16.1, Problem 2PP

To determine

To Draw: A graph of the illuminance produced by a lamp with a luminous flux of 2275 lm at a distance from 0.50 m and 5.0 m.

Explanation of Solution

Given:

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

The range of distance is $0.50\text{m}\text{to}\text{5}\text{.0}\text{m}$ .

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 the distance of the object from the light source respectively.

Calculation:

Substitute $2275\text{lm}$ for $P$ in equation (1)

  $E=\frac{2275}{4\pi d^2}$  ...... (2)

Now for each value of $d$ , we get different value of $E$ and substitute $0.50\text{m}$ for $d$ in equation (2)

  $\begin{array}{l}E=\frac{2275}{4\pi {\left(0.50\right)}^2}\\ E=724\text{lx}\end{array}$

Substitute $1.50\text{m}$ for $d$ in equation (2)

  $\begin{array}{l}E=\frac{2275}{4\pi {\left(1.50\right)}^2}\\ E=80.5\text{lx}\end{array}$

Substitute $2.50\text{m}$ for $d$ in equation (2)

  $\begin{array}{l}E=\frac{2275}{4\pi {\left(2.50\right)}^2}\\ E=28.9\text{lx}\end{array}$

Substitute $3.50\text{m}$ for $d$ in equation (2)

  $\begin{array}{l}E=\frac{2275}{4\pi {\left(3.50\right)}^2}\\ E=14.7\text{lx}\end{array}$

Substitute $3.50\text{m}$ for $d$ in equation (2)

  $\begin{array}{l}E=\frac{2275}{4\pi {\left(4.50\right)}^2}\\ E=8.94\text{lx}\end{array}$

Substitute $5.00\text{m}$ for $d$ in equation (2)

  $\begin{array}{l}E=\frac{2275}{4\pi {\left(5.00\right)}^2}\\ E=7.24\text{lx}\end{array}$

These values can be shown in a tabular form as

d (m)	E(lx)
0.50	724
1.50	80.5
2.50	28.9
3.50	14.7
4.50	8.94
5.00	7.24

Label the vertical axis as $E$ and the horizontal axis as $d$ , then plot between $E$ and $d$ can be shown below

  

Conclusion:

Hence, the required graph is drawn.