In the given electromagnetic wave $E _{ y }=600 \sin (\omega t - kx ) Vm ^{-1}$, intensity of the associated light beam is (in $W / m ^2:\left(\right.$ Given $\left.\epsilon_0=9 \times 10^{-12} C ^2 N ^{-1} m ^{-2}\right)$

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YESPYQ · Accepted Answer

Correct answer: c) 486. $\mathrm{I}=\frac{1}{2}{\epsilon }_{0}{\mathrm{E}}_{0}^{2}\mathrm{c}$ $\mathrm{I}=\frac{1}{2}\times 9\times {10}^{-12}\times (600{)}^{2}\times 3\times {10}^{8}$ $I=\frac{9\times 3}{2}\times 36\times {10}^{4}\times {10}^{-12}\times {10}^{8}$ $I=27\times 18$ $I=486\frac{\text{ watt }}{{\mathrm{m}}^{2}}$

In the given electromagnetic wave \(E _{ y }=600 \sin (\omega t - kx ) Vm ^{-1}\), intensity of the associated light bea…

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