Chapter 16, Problem 74A

To determine

Illumination on the screen for each case.

To sketch: The graph of a obtained data.

To find: The shape of the graph. Also, the relationship between the illuminance and distance.

Answer to Problem 74A

Illumination on the screen for each case is shown in Table 1.

The graph of the obtained data is shown in Figure 1.

The shape of the graph is hyperbola.

The relationship between illuminance and distance in the graph is inverse square.

Explanation of Solution

Given:

The distance of $110\text{ cd}$ light source from screen is $1.0\text{ m}$ .

Formula used:

The expression for the point source illuminance is,

$E=\frac{P}{4\pi r^2}$

Here, $P$ is the luminous flux of the light source and $r$ is the distance of the object is from the light source.

Calculation:

The luminous flux of the light source is,

$\begin{array}{c}P=4\pi \left(\text{intensity of flux}\right)\\ P=4\pi \times 110\text{ cd}\times \frac{1\text{ lm}}{1\text{ cd}}\\ P=1.4\times {10}^3\text{ lm}\end{array}$

The point source illuminance at $1.0\text{ m}$ is,

$\begin{array}{l}E=\frac{P}{4\pi r^2}\\ E=\frac{1.4\times {10}^3\text{ lm}}{4\pi \times {\left(1.0\text{m}\right)}^2}\\ E=111\text{ lm}\end{array}$

Similarly, calculate the point source illuminance for every meter of increasing distance up to $7.0\text{m}$ as shown in table below.

Table 1

Distance of light source from screen	Point of illuminance
1	111
2	28
3	12
4	7
5	4
6	3
7	2

Sketch the graph as shown in Figure below.

Figure 1

The shape of the graph is hyperbola.

The luminous flux $\left(P\right)$ of a given point source of light (say a lamp) is always a constant.

The illuminance of the point source (lamp) is inversely proportional to the square of the distance $\left(r\right)$ from the source.

$E\propto \frac{1}{r^2}$

The illuminance decreases with the increase in distance of the light source from screen.Hence, the relationship between illuminance and distance in the graph is Inverse Square.