IIT JEE , AIEEE Forum - Physics , Modern Physics

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Let the energy emitted per second = E

Momentum of photons emitted = E/c

70% of photons are absorbed, initial momentum = 0.7E/c and final momentum of photons = 0
Change in momentum of surface = 0.7E/c - 0 = 0.7E/c

30% of photons are reflected, initial momentum = 0.3E/c and after reflecting momentum will be -0.3E/c
so change in momentum of surface = 0.3E/c-(-0.3E/c) = 0.6E/c

Net change in momentum = 0.7E/c + 0.6E/c = 1.3E/c

Force = Rate of change of momentum = d(1.3E/c)/dt

= (1.3/c)(dE/dt)

= 1.3P/c (since dE/dt is the rate of change of energy which is power)

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