Suppose that intensity of a laser is 315/π W/m².
The rms electric field, in units of V/m associated with this source is close to the nearest integer is __________.
∈₀ = 8.86 × 10 –12 C² Nm –2 ; c = 3 × 10⁸ ms –1 )

Solution

Physics sir · 3 steps

Intensity uses the rms field squared, with no half in front; then just take the square root.

1. Intensity and field. In an EM wave the energy density averages to ε₀ Eᵣₘₛ², and that energy moves at speed c. So intensity is energy density times c.

   I = ε₀ c Eᵣₘₛ² ⇒ Eᵣₘₛ = √(I/(ε₀ c))

2. Put in numbers. I = 315/π = 100.27 W m⁻² and ε₀ c = 8.86 × 10⁻¹² × 3 × 10⁸ = 2.658 × 10⁻³. So Eᵣₘₛ² = 100.27/(2.658 × 10⁻³) ≈ 3.772 × 10⁴ V² m⁻².
3. So. Eᵣₘₛ = √(3.772 × 10⁴) ≈ 194.2 V m⁻¹, which rounds to 194.

The common slip: 275. Using I = 1/2ε₀ c E² and calling that E the rms field gives the peak value, about 275, which is √2 times too big.

The idea to keep. Remember the pair: I = 1/2ε₀ c E₀² = ε₀ c Eᵣₘₛ². The half goes with the peak field, never with the rms field.

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