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A floodlight filtered to pass only red light produces a sinusoidal plane electromagnetic wave whose electric field component is given by Eₓ = 36 sin(1.20 * 10⁷ * z - 3.6 * 10¹⁵ * t) V/m. The average intensity of this beam is closest to sqrt(n) W/m², where n is an integer. Find n.

1. n = 1716
2. n = 1764
3. n = 1849
4. n = 1600

Correct answer: n = 1716

Solution

Using the formula for average intensity I = (c * epsilon₀ * E₀²) / 2, substituting E₀ = 36 V/m gives I = (3*10⁸ * 8.85*10⁻¹² * 36²) / 2 = (3*10⁸ * 8.85*10⁻¹² * 1296) / 2 ≈ 1.724 / 2 *... Let me recompute: c*epsilon₀ = 3*10⁸ * 8.85*10⁻¹² = 2.655*10⁻³. Then I = 0.5 * 2.655*10⁻³ * 1296 = 0.5 * 3.441 = 1.720 W/m²... Hmm. Actually I = (E₀²)/(2*mu₀*c) = (36²)/(2 * 4*pi*10⁻⁷ * 3*10⁸) = 1296/(2 * 376.7) = 1296/753.4 ≈ 1.720 W/m². So sqrt(n) ≈ 1.720 => n ≈ 2.96... That is not an integer match for a clean answer. Re-examining: perhaps the question means I = sqrt(n) where n ~ 1296*something. Using I_avg = c*epsilon₀*E₀²/2 = 3*10⁸*8.85*10⁻¹²*36²/2 = 1.721 W/m² ≈ sqrt(2.96). So n = 3 is closest integer. But answer options above assume a different interpretation — that the intensity value itself equals sqrt(n) and n is large. With I ≈ 1.72 W/m², sqrt(n) = 1.72 gives n ≈ 2.96, so n = 3.

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