Amplitude of electric field in a laser beam of power 4.4 mW with cross-sectional area of $2 m m^{2}$ is

\sqrt{\frac{2}{3}} k V / m

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\sqrt{\frac{5}{3}} k V / m

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2 k V / m

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\frac{23}{2} k V / m

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Solution

The correct option is B $\sqrt{\frac{5}{3}} k V / m$
Intensity $I = \frac{P o w e r}{A r e a} = \frac{4.4 \times 10^{- 3}}{2 \times 10^{- 6}} = \frac{4.4}{2} \times 10^{3} W / m^{2}$
$I = \frac{1}{2} ε_{0} C E_{0}^{2} \Rightarrow \frac{4.4}{2} \times 10^{3} = \frac{1}{2} \times 8.8 \times 10^{- 12} \times 3 \times 10^{8} E_{0}^{2}$

$E_{0}^{2} = \frac{4.4 \times 10^{7}}{8.8 \times 3} = \frac{10}{6} \times 10^{6} \Rightarrow E_{0} = \sqrt{\frac{5}{3}} \times 10^{3} V / m$

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