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18P

Expert-verifiedFound in: Page 1002

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The intensity I**** of 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/m^{2}, and the horizontal axis scale is set by r_{s}^{-2} = 8.0 m^{-2}. What is the power of the source?**

**Figure:**

Power of the source P_{s} is $2.5\times {10}^{2}\u200a\text{W}$.

Maximum intensity on the vertical axis I_{s}=200 w/m^{2}

Maximum inverse square of the distance on horizontal axis r_{s}^{-2} = 8.0 m^{-2}

**We use the intensity relation to find the power of the source. From ****the ****graph, we can calculate the slope, and substituting this slope value in ****the ****calculated slope, we can find the power of the source.**

**Formula:**

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

${A}_{s}=4\pi {r}_{s}^{2}$

${I}_{s}=\frac{{P}_{s}}{4\pi {r}_{s}^{2}}\phantom{\rule{0ex}{0ex}}{I}_{s}{r}_{s}^{2}=\frac{{P}_{s}}{4\pi}\phantom{\rule{0ex}{0ex}}\frac{{I}_{s}}{{r}_{s}^{-2}}=\frac{{P}_{s}}{4\pi}$

…(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/m^{2}** 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

$20\u200a\text{W}=\frac{{P}_{s}}{4\pi}\phantom{\rule{0ex}{0ex}}{P}_{s}=20\u200a\text{W}\times 4\pi \phantom{\rule{0ex}{0ex}}=2.5\times {10}^{2}\u200a\text{W}$

Hence the power of the source is, P_{s} is $2.5\times {10}^{2}\u200a\text{W}$.

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