If 2.5 × 10⁻⁶ N average force is exerted by a light wave on a non-reflecting surface of 30 cm² area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm². (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)

Solution

Physics sir · 4 steps

Radiation pressure on an absorbing surface is just intensity over c, so the 40 minutes play no part.

1. Force from light. On a surface that absorbs everything at normal incidence, the light pushes with pressure I/c. So the average force is the pressure times the area.

   F = IA/c ⇒ I = Fc/A

2. Time is extra. Neither the force nor the intensity depends on how long the light shines. The 40 minutes would matter only if you were asked for the total energy or momentum delivered.
3. Put in numbers. Fc = 2.5 × 10⁻⁶ × 3 × 10⁸ = 750 W. Keep the area in cm² so the answer comes out directly in W/cm²: I = 750/30 = 25 W cm⁻².
4. So. The energy flux just before the light hits the surface is 25 W/cm², which is 2.5 × 10⁵ W m⁻² in SI units.

The common slip: 250000. Working in SI and stopping at W/m² gives 250000, not 25. Divide by 10⁴ to convert to W/cm².

The idea to keep. Absorbing surface: F = IA/c. Perfect reflector: F = 2IA/c. Read the surface type first, and ignore the time unless the question asks for energy or momentum.

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