Question

electromagnetic waves from a light bulb
a 60 - w light bulb radiates electromagnetic waves uniformly in all directions. at a distance of 1.0 m from the bulb, the light intensity is ( i_0 ), the average energy density of the waves is ( u_0 ), and the rms electric and magnetic field values are ( e_0 ) and ( b_0 ), respectively.
part b
at 2.0 m from the bulb, what is the average energy density of the waves?
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( u_0 )
( sqrt{\frac{1}{2}}u_0 )
( \frac{1}{2}u_0 )
( \frac{1}{4}u_0 )
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the average energy density does decrease, but not by this factor

Explanation:

Step1: <Intensity formula>

The intensity \(I\) of electromagnetic waves is given by \(I = \frac{P}{4\pi r^{2}}\), where \(P\) is the power of the source and \(r\) is the distance from the source. Also, the intensity \(I\) is related to the average energy density \(u\) by \(I = cu\), where \(c\) is the speed of light.

Step2: <Find ratio of energy densities>

Let \(u_0\) be the energy density at \(r_0 = 1.0\space m\) and \(u\) be the energy density at \(r = 2.0\space m\).
Since \(I_0=cu_0=\frac{P}{4\pi r_0^{2}}\) and \(I = cu=\frac{P}{4\pi r^{2}}\)
Dividing the two equations: \(\frac{u}{u_0}=\frac{r_0^{2}}{r^{2}}\)
Substitute \(r_0 = 1.0\space m\) and \(r = 2.0\space m\)
\(\frac{u}{u_0}=\frac{1^{2}}{2^{2}}=\frac{1}{4}\)

Answer:

\(\frac{1}{4}u_0\)