Question

The intensity Iof light from an isotropic point source is determined as a function of distance r from the source. The Figure gives intensity I versus the inverse square of that square r^-2. The vertical axis scale is set by I_s=200 w/m², and the horizontal axis scale is set by r_s^-2 = 8.0 m^-2. What is the power of the source?

Figure:

Short answer

Power of the source P_s is $2.5\times {10}^2 \text{W}$.

Step-by-step solution

Listing the given quantities

Maximum intensity on the vertical axis I_s=200 w/m²

Maximum inverse square of the distance on horizontal axis r_s^-2 = 8.0 m^-2

Understanding the concepts of power and intensity relation

We use the intensity relation to find the power of the source. Fromthegraph, we can calculate the slope, and substituting this slope value inthecalculated slope, we can find the power of the source.

Formula:

Calculations of the power of the source

The intensity I_s of an isotropic point source at a point whose radial distance from the source is, r_s.

The intensity is nothing but power per unit area.

Hence the intensity of source is,

$I_s=\frac{P_s}{A_s}$

As the source is spherical, its surface area is

$A_s=4\pi r_s^2$

$$\begin{gathered} I_s=\frac{P_s}{4\pi r_s^2} \\ {I}_s{r}_s^2=\frac{P_s}{4\pi } \\ \frac{I_s}{r_s^{-2}}=\frac{P_s}{4\pi } \end{gathered}$$

…(1)

The plot in the problem is I_s versus r_s^-2.

Hence the relation 1) shows the slope of the above plot.

Now we can find the slope from the above plot.

The maximum value of intensity on axis is I_s=200 w/m² and the minimum is zero.

The maximum value of inverse square of the distance on the axis is r_s^-2 = 8.0 m^-2, and minimum value is zero.

$\begin{array}{c}\frac{\Delta y}{\Delta x}=\frac{200\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}-0}{10{\text{\hspace{0.17em}m}}^{\text{-2}}-0}\\ =20\text{\hspace{0.17em}W}\\ \frac{I_s}{r_s^{-2}}=20\text{\hspace{0.17em}W}\end{array}$

Substituting this slope value in equation (1), we get

$$\begin{gathered} 20 \text{W}=\frac{P_s}{4\pi } \\ {P}_s=20 \text{W}\times 4\pi \\ =2.5\times {10}^2 \text{W} \end{gathered}$$

Hence the power of the source is, P_s is $2.5\times {10}^2 \text{W}$.