In the given electromagnetic wave Ey = 600 sin (ω t - kx) V m⁻¹, intensity of the associated light beam is (in W/m²): (Given ε₀ = 9 × 10⁻¹² C² N⁻¹ m⁻²) (a) 486 (b) 729 (c) 243 (d) 972

Answer: (a). I = 1/2ε₀ c E₀² I = 1/2 × 9 × 10⁻¹² × 3 × 10⁸ × 600² = 1.35 × 10⁻³ × 3.6 × 10⁵ I = 486 W/m ² So the answer is (a) 486.

In the given electromagnetic wave Ey = 600 sin (ω t - kx) V m⁻¹… | JEE Main 2024 (6 Apr, Shift 2)

Q6JEE Main 2024 · 6 APR · SHIFT 2Easy

Energy Density & Intensity

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

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Answer: (a)

$I = \tfrac{1}{2}\varepsilon_0 c E_0^2$
$I = \tfrac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^{8} \times 600^2 = 1.35 \times 10^{-3} \times 3.6 \times 10^{5}$
$I = 486$ W/m $^2$
So the answer is (a) 486.

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