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/m

  • Option 2)

    1.4 kV/m

  • Option 3)

    1 kV/m

  • Option 4)

    0.7 kV/m

Answers (1)
A admin

 

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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