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(A) The force exerted by a beam of light on a surface is given by the rate of change of momentum.For a beam of power $P$ incident normally:Force due t

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$4.33 	imes 10^{-8} \, N$

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(A) The force exerted by a beam of light on a surface is given by the rate of change of momentum.For a beam of power $P$ incident normally:Force due to absorption $(F_a)$ $= \frac{P_a}{c} = \frac{0.70 	imes P}{c}$Force due to reflection $(F_r)$ $= \frac{2 	imes P_r}{c} = \frac{2 	imes 0.30 	imes P}{c}$Total force $F = F_a + F_r = \frac{P}{c} (0.70 + 0.60) = \frac{1.30 	imes P}{c}$Given $P = 10 \, W$ and $c = 3 	imes 10^8 \, m/s$:$F = \frac{1.30 	imes 10}{3 	imes 10^8} = \frac{13}{3} 	imes 10^{-8} \approx 4.33 	imes 10^{-8} \, N$.

$A$ beam of white light is incident normally on a plane surface absorbing $70\%$ of the light and reflecting the rest. If the incident beam carries $10\,W$ of power,find the force exerted by it on the surface :-

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