JEE Main 2024 Physics question
In the given electromagnetic wave = 600 sin(ωt − kx) Vm⁻¹, intensity of the associated light beam is (in W/m²): (Given ε₀ = 9 × 10⁻¹² C²N⁻¹m⁻²)
Answer: (A)
Solution
I = ½ε₀E₀²c = ½ × 9 × 10⁻¹² × 600² × 3 × 10⁸ = 486 W/m².