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In the given electromagnetic wave $\mathrm{E}_{\mathrm{y}}=600 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{Vm}^{-1}$, intensity of the associated light beam is (in $\mathrm{W} / \mathrm{m}^2$ : (Given $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )
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486
$\begin{aligned} \text { Intensity } & =\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c} \\ & =\frac{1}{2} \times 9 \times 10^{-12} \times(600)^2 \times 3 \times 10^8 \\ & =\frac{9}{2} \times 36 \times 3=486 \mathrm{w} / \mathrm{m}^2\end{aligned}$
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