Radiation Pressure

Electromagnetic waves carry momentum as well as energy, so striking a surface they exert a small pressure. For a perfectly absorbing surface at normal incidence the radiation pressure equals the intensity divided by the speed of light, P = I/c, because all the incoming momentum is delivered and none reflected back. Substituting I = 1.4 \times 10^{3}\ \text{W/m}^2 and c = 3.0 \times 10^{8}\ \text{m/s} gives P = (1.4 \times 10^{3})/(3.0 \times 10^{8}) = 4.7 \times 10^{-6}\ \text{Pa}. The 9.3 \times 10^{-6}\ \text{Pa} option uses P = 2I/c, which applies to a perfect reflector rather than an absorber. The 2.3 \times 10^{-6}\ \text{Pa} value halves the correct result without justification. The 4.7 \times 10^{-3}\ \text{Pa} answer is off by three powers of ten from a c misplacement. The factor distinguishing absorption from reflection is central to the JEE treatment of radiation momentum, and recognising whether a surface absorbs or reflects is usually the first decision a student must make. The deep reason a pressure exists at all is that electromagnetic waves carry momentum density equal to their energy density divided by c, so a steady stream of energy means a steady delivery of momentum to whatever stops it. As a magnitude check, sunlight pressure is famously tiny, of order micropascals, which the answer correctly reflects despite the large intensity involved.