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Electromagnetic Waves

JEE Main level, single correct, medium

31 of 165

Question 31 of 165

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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Answer: (D) 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}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Electromagnetic Waves (Electromagnetic Waves and Maxwell's equations), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.