Question 13: EM waves

Question

A 30 mW laser beam has a cross-sectional area of 15 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by
[Permuittivity of space, $$\varepsilon_0=9	imes10^{-12}$$
Speed of light, $$c=3	imes10^8$$ m/s]

Accepted Answer

Given, power of laser beam,
$$P=30~mW=30	imes10^{-3}~W$$
Area of cross section, A = 15 mm$$^2$$ = 15 $$	imes$$ 10$$^{-6}$$ m$$^2$$
Permittivity of free space, $$\varepsilon_0=9	imes10^{-12}$$ SI unit
Speed of light, $$c=3	imes10^8$$ m/s
Intensity of electromagnetic wave is given by
$$I=\frac{1}{2}nc\varepsilon_0E^2$$
where, $$n$$ is refractive index for air, $$n=1$$
$$I=\frac{1}{2}c~.~\varepsilon_0E^2$$ ..... (i)
$$I=\frac{P}{A}$$ ....... (ii)
From Eqs. (i) and (ii), we get
$${1 \over 2}c{\varepsilon _0}{E^2} = {P \over A}$$
$${E^2} = {{2P} \over {Ac{\varepsilon _0}}}$$
$$E = \sqrt {{{2 	imes 30 	imes {{10}^{ - 3}}} \over {15 	imes {{10}^{ - 6}} 	imes 3 	imes {{10}^8} 	imes 9 	imes {{10}^{ - 12}}}}} $$
$$ = 1.22 	imes {10^3}$$ V/m = 1.22 kV/m

Answer

1.22 kV/m

Answer

12 kV/m

Answer

10 kV/m

Answer

201 kV/m

EM waves

A 30 mW laser beam has a cross-sectional area of 15 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Permuittivity of space, $\varepsilon_0=9\times10^{-12} Speed of light, c=3\times10^8$ m/s]

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