# A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by :[ Given permittivity of space $\epsilon _{0}$ = 9 x 10-12 SI units, Speed of light c = 3 x 108 m/s ]Option 1)2 kV/mOption 2)1.4 kV/mOption 3)1 kV/mOption 4)0.7 kV/m

Intensity of EM wave -

$I = \frac{1}{2} \epsilon _{o} E_{o}^{2}c$

- wherein

$\epsilon _{o}$ = Permittivity of free space

$E _{o}$ = Electric field amplitude

c = Speed of light in vacuum

$I=Intensity\: of\: EM\: wave$

$I=\frac{Power}{Area}=\frac{27\times 10^{-3}}{10\times 10^{-6}}\cdots (1)$

$I=\frac{1}{2}\epsilon _{0}{E_{0}}^{2}C\cdots \left ( 2 \right )$

(1) = (2)

$\frac{27\times 10^{-3}}{10\times 10^{-6}}=\frac{1}{2}\times 9\times 10^{-12}\times E^{2}\times 3\times 10^{8}$

$\Rightarrow E=\sqrt{2}\times 10^{3}$

$E=1.4Kv/m$

Option 1)

2 kV/m

Option 2)

1.4 kV/m

Option 3)

1 kV/m

Option 4)

0.7 kV/m

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