Electromagnetic Waves

JEE Main level, single correct, medium

109 of 165

Question 109 of 165

A 27 mW27 \mathrm{~mW} laser beam has a cross-sectional area of 10 mm210 \mathrm{~mm}^{2}. The magnitude of the maximum electric field in this electromagnetic wave is given by:

[Given permittivity of space ϵ0=9×10−12SI\epsilon_{0}=9 \times 10^{-12} \mathrm{SI} units, Speed of light c=3×108 m/s]\left.\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]

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Answer: (D) 1.4kV/m1.4 \mathrm{kV} / \mathrm{m}

EM wave intensity

⇒I= Power  Area =12ε0E02c\Rightarrow \mathrm{I}=\frac{\text { Power }}{\text { Area }}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \mathrm{c}

[where E0=\mathrm{E}_{0}= maximum electric field] ]]

⇒27×10−310×10−6=12×9×10−12×E02×3×108\Rightarrow \frac{27 \times 10^{-3}}{10 \times 10^{-6}}=\frac{1}{2} \times 9 \times 10^{-12} \times \mathrm{E}_{0}^{2} \times 3 \times 10^{8}

⇒E0=2×103kV/m=1.4kV/m\Rightarrow \mathrm{E}_{0}=\sqrt{2} \times 10^{3} \mathrm{kV} / \mathrm{m}=1.4 \mathrm{kV} / \mathrm{m}

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