The intensity of a plane monochromatic wave is \( 1380\,\mathrm{W\,m^{-2}} \). What is the amplitude of electric field in this wave? (Take speed of light \( 3\times10^{8}\,\mathrm{m\,s^{-1}} \); \( \varepsilon_0 = 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1020\,\mathrm{N\,C^{-1}} \)
The intensity of an electromagnetic wave is the energy density times the speed of light. The average energy density is \( \frac{1}{2}\varepsilon_0 E_0^{2} \), where \( E_0 \) is the amplitude of the electric field.
\( \displaystyle I = \frac{1}{2}\,c\,\varepsilon_0 E_0^{2} \;\Rightarrow\; E_0 = \sqrt{\frac{2I}{c\,\varepsilon_0}} \)
\( c\,\varepsilon_0 = 3\times10^{8} \times 8.85\times10^{-12} = 2.655\times10^{-3} \)
\( \displaystyle E_0 = \sqrt{\frac{2 \times 1380}{2.655\times10^{-3}}} = \sqrt{1.04\times10^{6}} \approx 1020\,\mathrm{N\,C^{-1}} \)