Question 125 of 165

A red $LED$ emits light at $0.1watt$ uniformly around it. The amplitude of the electric field of the light at a distance of $1m$ from the diode is:

A$7.75V{m}^{-1}$B$1.73V{m}^{-1}$C$2.45V{m}^{-1}$D$5.48V{m}^{-1}$

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Answer: (C) $2.45V{m}^{-1}$

For a point source of power = P, then intensity at a point at a separation x from the source is, $I=\dfrac{Power}{Area}=\dfrac{P}{4\pi {x}^{2}}$$∵$ The average intensity of E.M. wave is given by, $I=\dfrac{1}{2}c{\epsilon }_{0}{E}_{o}^{2}$$\Rightarrow {E}_{0}=\sqrt{\dfrac{2P}{4\pi {\epsilon }_{0}c{x}^{2}}}$

$∵\dfrac{1}{4\pi {\epsilon }_{0}}=9\times {10}^{9},P=0.1W,x=1m$, $c=3\times {10}^{8}m{s}^{-1}$

$\Rightarrow {E}_{0}=\sqrt{\dfrac{2\times 0.1\times 9\times {10}^{9}}{3\times {10}^{8}\times {1}^{2}}}=\sqrt{6}=2.45V{m}^{-1}$

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