Radiation Force On A Reflector

When light reflects, its momentum reverses direction, so the change in momentum is twice that of simple absorption. The momentum delivered per second equals the force, and for a beam of power P fully reflected straight back the force is F = 2P/c. The factor of two arises because the photons arrive with forward momentum and leave with backward momentum, doubling the impulse on the mirror. Substituting P = 30\ \text{W} and c = 3.0 \times 10^{8}\ \text{m/s} gives F = (2 \times 30)/(3.0 \times 10^{8}) = 2.0 \times 10^{-7}\ \text{N}. The 1.0 \times 10^{-7}\ \text{N} option uses F = P/c, valid only for an absorber. The 4.0 \times 10^{-7}\ \text{N} value mistakenly applies a factor of four. The 6.0 \times 10^{-8}\ \text{N} answer arises from arithmetic error in the exponent. This reflection-doubling is a recurring JEE Advanced theme connecting radiation to mechanics, and it parallels the way an elastic ball bouncing off a wall delivers twice the impulse of one that simply sticks. The same reasoning explains why solar sails are designed with highly reflective surfaces, since maximising momentum transfer maximises the thrust available from sunlight. As a check, a reflector always feels twice the force of an equivalent absorber, and 2.0 \times 10^{-7}\ \text{N} is exactly double the absorber value of 1.0 \times 10^{-7}\ \text{N}.