A parallel beam of light is incident normally | vedclass

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Question

(A) The power of the incident beam is $P = 90 	ext{ W}$.The force exerted by the absorbed part of the light is $F_a = \frac{P_a}{C}$,where $P_a = 0.5

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$4.5 	imes 10^{-7} 	ext{ N}$

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(A) The power of the incident beam is $P = 90 	ext{ W}$.The force exerted by the absorbed part of the light is $F_a = \frac{P_a}{C}$,where $P_a = 0.5P = 45 	ext{ W}$.$F_a = \frac{45}{3 	imes 10^8} = 1.5 	imes 10^{-7} 	ext{ N}$.The force exerted by the reflected part of the light is $F_r = \frac{2P_r}{C}$,where $P_r = 0.5P = 45 	ext{ W}$.$F_r = \frac{2 	imes 45}{3 	imes 10^8} = 3.0 	imes 10^{-7} 	ext{ N}$.The total force exerted on the surface is $F = F_a + F_r = 1.5 	imes 10^{-7} + 3.0 	imes 10^{-7} = 4.5 	imes 10^{-7} 	ext{ N}$.

$A$ parallel beam of light is incident normally on a plane surface absorbing $50\%$ of the light and reflecting the rest. If the incident beam carries $90 \text{ W}$ of power,the force exerted by it on the surface is ($C = 3 \times 10^8 \text{ m/s}$ is the velocity of light in air).

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