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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

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

A laser beam of power 27mW27mW has a cross-sectional area of 10mm210{mm}^{2}. The magnitude of the maximum electric field in this electromagnetic wave is given by

[[Given permittivity of space ε0≈9×10−12C2N−1m−2{ε}_{0} \approx 9 \times {10}^{- 12}C^{2}N^{- 1}m^{- 2}, speed of light \leftc = 3 \times {10}^{8}ms^{- 1}]</p></p>

Options

  1. A1kVm−11kVm^{- 1}
  2. B1.4kVm−11 . 4kVm^{- 1}
  3. C0.7kVm−10.7kVm^{- 1}
  4. D2kVm−12kVm^{- 1}

Correct answer

B. 1.4kVm−11 . 4kVm^{- 1}

Step-by-step solution

I=PA=12ε0E02cI = \dfrac{P}{A} = \dfrac{1}{2}{ε}_{0}{E}_{0}^{2}c

∴E0=2Pε0CA=2×27×10−3×36π×1093×108×10×10−6∴ {E}_{0} = \sqrt{\dfrac{2P}{{ε}_{0}CA}} = \sqrt{\dfrac{2 \times 27 \times {10}^{- 3} \times 36\pi \times {10}^{9}}{3 \times {10}^{8} \times 10 \times {10}^{- 6}}}

E0=1.4kVm−1{E}_{0} = 1 . 4kVm^{- 1}

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