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

A laser beam of power $27mW$ 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 $c = 3 \times {10}^{8}ms^{- 1}$ ]

Options

A $1kV / m$
B $1. 4kV / m$
C $0. 7kV / m$
D $2kV / m$

Correct answer

B. $1. 4kV / m$

Step-by-step solution

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

∴ {E}_{0} = \sqrt{\dfrac{2P}{{ε}_{0}cA}} = \sqrt{\dfrac{2 \times \left(27 \times {10}^{- 3}\right)}{\left(9 \times {10}^{- 12}\right) \times \left(3 \times {10}^{8}\right) \times \left(10 \times {10}^{- 6}\right)}}

⇒ {E}_{0} = 1. 4kV / m

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