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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2019 · 11 January · Evening Shift · Physics · Electromagnetic Waves

A 27 ~mW laser beam has a cross-sectional area of 10 ~mm^ 2 . The...

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]

Options

  1. A2kV/m2 \mathrm{kV} / \mathrm{m}
  2. B0.7kV/m0.7 \mathrm{kV} / \mathrm{m}
  3. C1kV/m1 \mathrm{kV} / \mathrm{m}
  4. D1.4kV/m1.4 \mathrm{kV} / \mathrm{m}

Correct answer

D. 1.4kV/m1.4 \mathrm{kV} / \mathrm{m}

Step-by-step solution

EM wave intensity ⇒\Rightarrow\mathrm{I}= Power  Area =12ε0E02c=\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⇒E0=2×103kV/m=1.4kV/m] \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}
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