A laser beam with wavelength lambda = 633 nm | vedclass

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(B) The power of the laser beam is $P = 3 \ mW = 3 	imes 10^{-3} \ W$.The cross-sectional area is $A = 3 \ mm^2 = 3 	imes 10^{-6} \ m^2$.For a perfe

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$6.6 	imes 10^{-6} \ N/m^2$

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(B) The power of the laser beam is $P = 3 \ mW = 3 	imes 10^{-3} \ W$.The cross-sectional area is $A = 3 \ mm^2 = 3 	imes 10^{-6} \ m^2$.For a perfectly reflecting surface with normal incidence,the force exerted by the radiation is given by $F = \frac{2P}{c}$,where $c = 3 	imes 10^8 \ m/s$ is the speed of light.$F = \frac{2 	imes 3 	imes 10^{-3}}{3 	imes 10^8} = 2 	imes 10^{-11} \ N$.The pressure $p$ is defined as force per unit area: $p = \frac{F}{A}$.$p = \frac{2 	imes 10^{-11} \ N}{3 	imes 10^{-6} \ m^2} = \frac{2}{3} 	imes 10^{-5} \ N/m^2 = 0.666... 	imes 10^{-5} \ N/m^2 = 6.66 	imes 10^{-6} \ N/m^2$.Rounding to two significant figures,the pressure is $6.6 	imes 10^{-6} \ N/m^2$.

$A$ laser beam with wavelength $\lambda = 633 \ nm$ has a power of $3 \ mW$. If the cross-sectional area of the beam is $3 \ mm^2$,what is the pressure exerted by the beam on a surface? (Assume the surface is a perfect reflector and the light is incident normally.)

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