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In the given electromagnetic wave E_y = 600 sin (ωt − kx) V m⁻¹, intensity of the associated light beam is in (W/m²); (Given ε₀ = 9 × 10⁻¹² C² N⁻¹ m⁻²)

1. 486
2. 243
3. 729
4. 972

Correct answer: 486

Solution

The intensity of an electromagnetic wave is proportional to the square of the electric field amplitude. Given the electric field E_y = 600 V/m, the intensity can be calculated using the formula I = (1/2) ε₀ c E², where c is the speed of light. Substituting the values yields an intensity of 486 W/m².

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