A radiation of energy E falls normally on a p | vedclass

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(B) The energy of a photon is given by $E = pc$,where $p$ is the momentum of the photon and $C$ is the speed of light.Therefore,the initial momentum o

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$2E/C$

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(B) The energy of a photon is given by $E = pc$,where $p$ is the momentum of the photon and $C$ is the speed of light.Therefore,the initial momentum of the radiation incident on the surface is $p_i = E/C$.Since the surface is perfectly reflecting,the radiation reflects back with the same magnitude of momentum but in the opposite direction.Thus,the final momentum of the radiation is $p_f = -E/C$.The momentum transferred to the surface is the change in momentum of the radiation,given by $\Delta p = p_i - p_f$.Substituting the values,we get $\Delta p = E/C - (-E/C) = E/C + E/C = 2E/C$.

$A$ radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ($C =$ velocity of light).

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