A yellow LED of power 0.2 W emits light uniformly around it. The amplitude of the electric field of light at a distance of 2 m from the LED is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.73 V/m
The LED radiates uniformly in all directions, so at distance \( r \) its power is spread over a sphere of area \( 4\pi r^{2} \). The intensity of an electromagnetic wave is related to the amplitude \( E_0 \) of its electric field by \( I = \frac{1}{2}c\,\varepsilon_0 E_0^{2} \).
\( \displaystyle I = \frac{P}{4\pi r^{2}} = \frac{0.2}{4\pi \times 2^{2}} = \frac{0.2}{50.27} = 3.98\times10^{-3}\,\mathrm{W\,m^{-2}} \)
\( \displaystyle E_0 = \sqrt{\frac{2I}{c\,\varepsilon_0}} \)
\( c\,\varepsilon_0 = 3\times10^{8} \times 8.85\times10^{-12} = 2.655\times10^{-3} \)
\( \displaystyle E_0 = \sqrt{\frac{2 \times 3.98\times10^{-3}}{2.655\times10^{-3}}} = \sqrt{3.0} = 1.73\,\mathrm{V\,m^{-1}} \)