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(A) The intensity $I$ of a parallel beam of light is given by the formula:$I = \frac{1}{2} \varepsilon_0 E_0^2 c$ ... $(i)$where $E_0$ is the amplitud

Vedclass · Accepted Answer

$60$

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(A) The intensity $I$ of a parallel beam of light is given by the formula:$I = \frac{1}{2} \varepsilon_0 E_0^2 c$ ... $(i)$where $E_0$ is the amplitude of the electric field,$\varepsilon_0$ is the permittivity of free space,and $c$ is the speed of light.Given:$I = \frac{15}{\pi} 	ext{ W/m}^2$$\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 	ext{ Nm}^2/	ext{C}^2 \implies \varepsilon_0 = \frac{1}{36 \pi 	imes 10^9} 	ext{ F/m}$$c = 3 	imes 10^8 	ext{ m/s}$Rearranging equation $(i)$ for $E_0^2$:$E_0^2 = \frac{2I}{\varepsilon_0 c} = \frac{2I 	imes 4 \pi}{(4 \pi \varepsilon_0) c}$Substituting the values:$E_0^2 = \frac{2 	imes (15/\pi) 	imes 4 \pi}{(1 / (9 	imes 10^9)) 	imes 3 	imes 10^8}$$E_0^2 = \frac{120}{3 	imes 10^8 / 9 	imes 10^9} = \frac{120}{1/30} = 120 	imes 30 = 3600$$E_0 = \sqrt{3600} = 60 	ext{ N/C}$

What is the amplitude of the electric field in a parallel beam of light of intensity $\left(\frac{15}{\pi}\right) \text{ W/m}^2$ (in $\text{ N/C}$)? $\left[\text{Assume} \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\right]$

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