Physics · Electromagnetic Waves · NEET
Yes. In this topic 'average flux' means the same as intensity: the average energy that crosses one square metre of surface each second. Both are measured in W/m². So when NEET says 'average flux of 20 W/cm²', treat it exactly like intensity in U = I × A × t.
Only for a fully reflecting surface, and only when the question asks about momentum or radiation pressure — not energy. For ENERGY received: a non-reflecting (fully absorbing) surface takes ALL the energy, so U = I × A × t with no extra factor. A perfect reflector receives no energy (it sends it all back). The factor of 2 belongs to the radiation pressure topic, not this one.
Because flux is given in W per m² (SI) or W per cm², and the answer is in joules (SI). You must keep units consistent. 1 cm² = 10⁻⁴ m², and 1 minute = 60 s. If you forget these conversions, your answer will be wrong by large powers of ten — a very common NEET slip.
Power P = I × A tells you energy per second (in watts). Energy U = P × t = I × A × t is the total joules collected over the whole time. NEET often gives flux and area (so you can find power) and then a time, so you must do the final × t to get energy. Do not stop at power.
Yes, as long as flux and area use the SAME area unit. 20 W/cm² × 20 cm² × 60 s = 24000 J directly, because the cm² cancel. Converting to m² gives the same answer. Just never mix W/cm² with an area in m².
Light with an average flux of 20 W/cm² falls on a non-reflecting surface at normal incidence having surface area 20 cm². The energy received by the surface during a time span of 1 minute is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
U = I × A × t, where I is the average energy flux (intensity) in W/m², A is the area in m², and t is the time in seconds. This assumes the surface is a normal-incidence, fully absorbing (non-reflecting) surface.
For an answer in joules, use flux in W/m², area in m², and time in seconds. You can also keep flux in W/cm² with area in cm² (they cancel), but never mix cm² flux with m² area.
No. A fully reflecting surface receives no net energy — it returns the wave. A fully absorbing (non-reflecting) surface receives the maximum energy, U = I × A × t. The factor of 2 you may remember applies to radiation pressure and momentum, not to energy.
Average intensity I = (1/2) ε₀ c E₀² = ε₀ c E_rms², measured in W/m². Once you know I, the energy on a surface is just I × A × t. The electric and magnetic parts contribute equally to this intensity.
Normal incidence means the EM wave hits the surface straight on, at 90° to the surface (along the surface normal). Then the full flux crosses the area A, so U = I × A × t with no cosine factor.
