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

Solution

Physics sir · 4 steps

Read the peak field off the equation, then use I = 1/2ε₀ c E₀². Do not forget the half.

1. Peak field. The number in front of the sine is the amplitude, so E₀ = 600 V m⁻¹.
2. Intensity formula. Intensity is the average energy density times the speed of light. The average of sin² over a cycle is 1/2, which brings in the half.

   I = 1/2ε₀ c E₀²

3. Put in numbers. ε₀ c = 9 × 10⁻¹² × 3 × 10⁸ = 2.7 × 10⁻³ and E₀² = 3.6 × 10⁵. Their product is 972, and half of that is 486.
4. So. The intensity is 486 W m⁻², option (a).

The idea to keep. Intensity of an EM wave is 1/2ε₀ c E₀² with the peak field, or ε₀ c Eᵣₘₛ² with the rms field. Use one or the other, never both.

Answer from the NTA final answer key. Solution by PrepWiser.