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(B) 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:$1$. Force

Vedclass · Accepted Answer

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

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(B) 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:$1$. Force due to absorbed part $(70 \%)$: $F_{abs} = \frac{P_{abs}}{c} = \frac{0.7 P}{c}$$2$. Force due to reflected part $(30 \%)$: $F_{ref} = \frac{2 P_{ref}}{c} = \frac{2 	imes 0.3 P}{c} = \frac{0.6 P}{c}$Total force $F = F_{abs} + F_{ref} = \frac{0.7 P + 0.6 P}{c} = \frac{1.3 P}{c}$Given $P = 10 \ W$ and $c = 3 	imes 10^8 \ m/s$:$F = \frac{1.3 	imes 10}{3 	imes 10^8} = \frac{13}{3} 	imes 10^{-8} \ N \approx 4.33 	imes 10^{-8} \ N$.

$A$ 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,the force exerted by it on the surface is

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