Q79 / 90 · JEE Main 2019 (11 Jan Shift 2) · Physics

A $27 \mathrm{~mW}$ laser beam has a cross-sectional area of $10 \mathrm{~mm}^{2}$. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space $\epsilon_{0}=9 \times 10^{-12} \mathrm{SI}$ units, Speed of light $\left.\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]$

  • A
    $2 \mathrm{kV} / \mathrm{m}$
  • B
    $0.7 \mathrm{kV} / \mathrm{m}$
  • C
    $1 \mathrm{kV} / \mathrm{m}$
  • Correct answer
    D
    $1.4 \mathrm{kV} / \mathrm{m}$

Solution

EM wave intensity $\Rightarrow \mathrm{I}=\frac{\text { Power }}{\text { Area }}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \mathrm{c}$ [where $\mathrm{E}_{0}=$ maximum electric field] $]$ $\Rightarrow \frac{27 \times 10^{-3}}{10 \times 10^{-6}}=\frac{1}{2} \times 9 \times 10^{-12} \times \mathrm{E}_{0}^{2} \times 3 \times 10^{8}$ $\Rightarrow \mathrm{E}_{0}=\sqrt{2} \times 10^{3} \mathrm{kV} / \mathrm{m}=1.4 \mathrm{kV} / \mathrm{m}$

Hence, the correct answer is option: 'D'.

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