In the given electromagnetic wave E y =600 sin (ω t-k x) Vm -1 , intensity of the associated light beam is (in W / m 2 ); (Given ε 0 = 9 × 10 -12 C 2 N -1 m -2 )

Physics · Electromagnetic Waves

JEE Main 2024 — 6 April, Shift 2 — Question 34

In the given electromagnetic wave $E_{y}=600 \sin (\omega t-k x) \mathrm{Vm}^{-1}$ , intensity of the associated light beam is

(in $\mathrm{W} / \mathrm{m}^{2}$ ); (Given $\epsilon_{0}=$ $9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )

Option A:
486
Correct
Option B:
243
Option C:
729
Option D:
972

Answer: A

Step-by-step solution

Intensity $=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \mathrm{c}$

$\begin{array}{*{35}{r}}{}&~=\frac{1}{2}\times9\times{{10}^{12}}\times{{(600)}^{2}}\times3\times{{10}^{8}} \\{} & ~=\frac{9}{2}\times 36\times 3=486\text{w}/{{\text{m}}^{2}} \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Power , Energy and Intensity of EM Waves