24.(a) We note that the cross section area of the beam isπd2/4, wheredis the diameter of the spot(d=2.00λ). The beam intensity isI=P2/4=5.00×10−3Wπ[(2.00)(633×10−9m)]2/4=3.97×109W/m2.(b) The radiation pressure ispr=Ic=3.97×109W/m22.998×108m/s=13.2Pa.(c) In computing the corresponding force, we can use the power and intensity to eliminate the area(mentioned in part (a)). We obtainF
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