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Fundamentals Of Physics
Found in: Page 1002

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Short Answer

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 Is=200 w/m2, and the horizontal axis scale is set by rs-2 = 8.0 m-2. What is the power of the source?


Power of the source Ps is 2.5×102W.

See the step by step solution

Step by Step Solution

Step 1: Listing the given quantities

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

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

Step 2: Understanding the concepts of power and intensity relation

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.


Step 3: Calculations of the power of the source

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

The intensity is nothing but power per unit area.

Hence the intensity of source is,


As the source is spherical, its surface area is




The plot in the problem is Is versus rs-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 Is=200 w/m2 and the minimum is zero.

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

ΔyΔx=200 W/m2010 m-20=20 WIsrs2=20 W

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


Hence the power of the source is, Ps is 2.5×102W.

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