Physics - EM Waves Question with Solution | TestHub

PhysicsEM WavesElectromagnetic wavesEasy2 minPYQ_2020

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A laser beam of power $27 mW$ has a cross-sectional area of $10 {mm}^{2}$ . The magnitude of the maximum electric field in this electromagnetic wave is given by [Take $ε_{0} \approx 9 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ and $c = 3 \times 10^{8} m s^{- 1}$ ]

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Answer:

Solution:

$I = \frac{P}{A} = \frac{1}{2} ε_{0} E_{0}^{2} c$
$∴ E_{0} = \sqrt{\frac{2 P}{ε_{0} c A}} = \sqrt{\frac{2 \times (27 \times 10^{- 3})}{(9 \times 10^{- 12}) \times (3 \times 10^{8}) \times (10 \times 10^{- 6})}}$
$\Rightarrow E_{0} = 1.4 kV / m$

Subject:PhysicsTopic:EM WavesSubtopic:Electromagnetic waves

⏱ 2mℹ️ Source: PYQ_2020

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