The electric field in a plane electromagnetic wave is given by
E⃗ = 200 cos [ ( (0.5 × 10³)/m )x - ( 1.5 × 10¹¹rad/s × t ) ]V/mĵ. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm². If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is x/10⁹N/m². Find the value of x .

Solution

Physics sir · 4 steps

For a perfect mirror the momentum flips, so pressure is 2I/c, and that collapses to just ε₀E₀².

1. Read the amplitude. The peak electric field is E₀ = 200 V m⁻¹. The wave number and frequency are not needed for pressure.
2. Intensity of wave. The average intensity is I = 1/2ε₀ E₀² c. A perfectly reflecting surface reverses the wave's momentum, so it feels twice the pressure an absorber would.

   P = 2I/c = 2/c×1/2ε₀ E₀² c = ε₀ E₀²

3. Put in numbers. P = 8.85 × 10⁻¹² × (200)² = 8.85 × 10⁻¹² × 4 × 10⁴ = 3.54 × 10⁻⁷ N m⁻². The area of 100 cm² and the 10 minutes do not change pressure, which is force per unit area and does not depend on time.
4. So. P = 354/10⁹ N m⁻², so x = 354.

The common slip: 177. Using the absorber formula P = I/c instead of 2I/c gives 1/2ε₀ E₀², that is 177, which is half the right value.

The idea to keep. Radiation pressure is I/c on an absorber and 2I/c on a perfect reflector. With I = 1/2ε₀ E₀² c, a reflector feels exactly ε₀ E₀². Area and time only matter for force or impulse.

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